maths problem solver indices

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Revision Notes

This section covers Indices revision.

An index number is a number which is raised to a power. The power, also known as the index, tells you how many times you have to multiply the number by itself. For example, 2 5 means that you have to multiply 2 by itself five times = 2×2×2×2×2 = 32.

There are a number of important rules of index numbers:

• y a × y b = y a+b

2 4 × 2 8 = 2 12

5 4 × 5 -2 = 5 2

y a ÷ y b = y a-b

3 9 ÷ 3 4 = 3 5

7 2 ÷ 7 5 = 7 -3

y -b = 1/y b

2 -3 = 1/2 3 = 1/8

y m/n = ( n √y) m

16 1/2 = √16 = 4

8 2/3 = ( 3 √8) 2 = 4

(y n ) m = y nm

= 2 5 + (2 3 ) 4

= 2 5 + 2 12

• Maths Questions

Index Questions

Index or Indices refers to the powers of a number or variable. Index questions are provided here to help students understand how to simplify expressions using simple formulas and techniques. In this article, you can practise various problems on indices of numbers, variables and some special expressions.

What is an Index?

In mathematics, an index is a number that tells us how many times a particular number is multiplied by itself. For instance, 35, where 5 is the index, means 3 is multiplied by itself five times.

Also, read: Index

Index Questions and Answers

1. Evaluate: 20 7 ÷ 20 5

20 7 ÷ 20 5

Using the formula a m /a n = a m-n , we have;

20 7 ÷ 20 5 = 20 7 / 20 5

Therefore, 20 7 ÷ 20 5 = 400.

2. If 3 (x-y) = 27 and 3 (x+y) = 243, find the value of x and y.

3 (x-y) = 27

3 (x-y) = 3 3

Bases are equal hence, equate the powers (indices).

That means, x – y = 3….(i)

3 (x+y) = 243

3 (x+y) = 3 5

That means, x + y = 5….(ii)

Adding equations (i) and (ii), we get;

Substituting x = 4 in equation (ii), we get;

y = 5 – 4 = 1

Therefore, x = 4 and y = 1.

3. Simplify the expression: (p 3 ) 0 × (p 1/2 ) 4.

(p 3 ) 0 × (p 1/2 ) 4

We know that a 0 = 1 and (a m ) n = a mn .

So, (p 3 ) 0 × (p 1/2 ) 4 = p (3×0) × p (½ × 4)

= p 0 × p 2

Therefore, (p 3 ) 0 × (p 1/2 ) 4 = p 2 .

4. Write 8 2x+3 in the form of 2 y and express the relation between x and y.

As we know, 8 = 2 3 .

So, 8 2x+3 = (2 3 ) 2x+3

Using the formula (a m ) n = a mn , we get;

8 2x+3 = 2 3(2x+3)

Thus, y = 6x + 9.

5. Evaluate the expression: 125 2/3 × 625 3/4.

125 2/3 × 625 3/4

As we know, 125 = 5 3 and 625 = 5 4

So, 125 2/3 × 625 3/4 = (5 3 ) ⅔ × (5 4 ) ¾

= 5 2 × 5 3

Therefore, 125 2/3 × 625 3/4 = 150.

6. Simplify: [5 a 5 b 2 × 3(a b 3 ) 2 ]/ (15 a 2 b).

[5 a 5 b 2 × 3(a b 3 ) 2 ]/ (15 a 2 b)

Using the formula (a m ) n = a mn , we have;

= [(5 × 3) × (a 5 b 2 ) × a 2 b 6 ]/ (15 a 2 b)

= (15 a 5+2 b 2+6 )/ (15 a 2 b)

= (a 7 b 8 )/(a 2 b)

= a 7-2 b 8-1

Therefore, [5 a 5 b 2 × 3(a b 3 ) 2 ]/ (15 a 2 b) = a 5 b 7 .

7. If a and b are whole numbers such that a b = 121, find the value of (a – 1) b + 1 .

We know that 11 2 = 121

Thus, ab = 11 2

⇒ a = 11 and b = 2

Now, (a – 1) b + 1

= (11 – 1) 2 + 1

= 10 × 10 × 10

Hence, (a – 1) b + 1 = 1000.

8. If 5 m = 3125, find the value of 4 (m – 2) .

We know that 5 5 = 3125

⇒ 5 m = 5 5

Now, 4 (m – 2) = 4 (5 – 2) = 4 3 = 4 × 4 × 4 = 64

Therefore, 4 (m – 2) = 64.

9. Evaluate: (1/216) -2/3 ÷ (1/27) -4/3.

(1/216) -2/3 ÷ (1/27) -4/3

Using the formula a -m = 1/a m or 1/a -m = a m , we get;

= 216 2/3 ÷ 27 4/3

= (6 3 ) 2/3 ÷ (3 3 ) 4/3

= 6 2 ÷ 3 4

Thus, (1/216) -2/3 ÷ (1/27) -4/3 = 4/9.

10. Simplify: 1 – {1 + (x 2 – 1) -1 } -1.

1 – {1 + (x 2 – 1) -1 } -1

= 1 – {1 + [1/(x 2 – 1)] } -1

= 1 – [(x 2 – 1 + 1)/ (x 2 – 1)] -1

= 1 – [x 2 /(x 2 – 1)] -1

= 1 – [(x 2 – 1)/x 2 ]

= [x 2 – (x 2 – 1)]/ x 2

= (x 2 – x 2 + 1)/x 2

Therefore, 1 – {1 + (x 2 – 1) -1 } -1 = 1/x 2 .

Practice Questions on Index (Indices)

• If (25) 7.5 x (5) 2.5 ÷ (125) 1.5 = 5 m , find the value of m.
• Simplify the expression: a 3 b −2 × (a 2 b 2 ) 4
• If (a/b) (x-1) = (b/a) (x-3) , then what is the value of x?
• Simplify: (2x) -4 /x 4
• Evaluate: (16) 0.16 × (16) 0.04 × (2) 0.2

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Exam Questions On Indices

These lessons help GCSE/IGCSE Maths students learn about indices by working through some examination questions.

Related Pages Exam Questions On Indices Rules Of Indices More GCSE Maths Lessons Math Worksheets

All You Need To Know About Indices (Revision and Summary)

x n means x multiplied repeatedly by itself n times. x m x n = x m + n (x m ) n = x mn (x m y n ) p = x mp y np x m /x n = x m-n x 0 = 1 x -n = 1/x n

Indices GCSE Maths Revision Higher Level Worked Exam Questions (include fractional and negative powers)

• Work out 56 1 - 56 0
• Explain why 27 1/3 = 3
• Write 27 -1/3 as a fraction.
• Work out the value of 64 2/3
• Work out all solutions of the equation: 8 m = 2 m 2
• Show clearly that 4 3/2 = 8 . Hence, or otherwise, work out the value of y if 4 y = 8 6
• Write down the value of 2 0 .
• Simplify 7 18 ÷ (7 3 ) 2 . Give your answer as a power of 7.
• A patient has a disease. She has 4 3 body cells on day 1. The number of affected cells doubles every day. The disease becomes serious when 2 10 body cells are affected. On which day does the disease become serious? You must show your working.
• Work out 10 -2 × 64 1/2 .
• Write down the square of the cube root of x using index notation.
• Given x = 8, what is the value of the square of the cube root of x?
• Work out the value of 9 -3/2 .
• Evaluate 27 -2/3 .
• Explain why 25 4 = 5 8 .
• Write 4 -1 as a fraction.
• Work out 8 4/3 .

Indices (Exponents) : C1 Edexcel June 2013 Q3 (a) Find the value 8 5/3 (b) Simplify fully (2x 1/2 ) 3 /(4x 2 )

Indices Tricky - GCSE Further Maths revision Exam paper practice & help

Example: x 3/2 = 8 where x > 0 and y -2 = 25/4 where y > 0. Work out the value of x/y.

Indices beastie - Dare you try this difficult question

Indices and Surds - GCSE Maths Higher revision Exam paper practice & help

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Laws Of Indices

Laws of Indices

Here we will learn everything you need to know about the laws of indices for GCSE & iGCSE maths (Edexcel, AQA and OCR). You’ll learn what the laws of indices are and how we can use them. You’ll learn how to multiply indices, divide indices, use brackets and indices, how to raise values to the power of 0 and to the power of 1, as well as fractional and negative indices.

Look out for the index laws worksheet and exam questions at the end.

What are the laws of indices?

Laws of indices provide us with rules for simplifying calculations or expressions involving powers of the same base . This means that the larger number or letter must be the same.

For example,

We cannot use laws of indices to evaluate calculations when the bases are different

We can however evaluate these calculations. Check out our other pages to find out how.

What are the 6 laws of indices?

The 6 laws of indices are:

• Multiplying indices
• Dividing indicies
• Brackets with indices
• Negative indices
• Fractional indices

For examples and practice questions on each of the rules of indices, as well as how to evaluate calculations with indices with different bases, follow the links below.

See also : Index notation

Laws of indices methods

Law 1: multiplying indices.

When multiplying indices with the same base, add the powers.

Step-by-step guide: Multiplying indices

Law 2: Dividing indices

When dividing indices with the same base, subtract the powers.

Step-by-step guide: Dividing indices

Law 3: Brackets with indices

When there is a power outside the bracket multiply the powers.

Step-by-step guide: Brackets with indices

Law 4: Power of 0

Any non-zero value raised to the power of 0 is equal to 1.

Step-by-step guide: Power of 0

Law 5: Negative indices

When the index is negative, put it over 1 and flip (write its reciprocal) to make it positive.

Step-by-step guide: Negative indices

Law 6: Fractional indices

When the index is a fraction, the denominator is the root of the number or letter, then raise the answer to the power of the numerator.

Step-by-step guide: Fractional indices

What are indices?

An index is a small number that tells us how many times a term has been multiplied by itself.

The plural of index is indices.

Below is an example of a term written in index form:

4 is the base and 3 is the index.

We can read this as ‘4 to the power 3’

Another way of expressing 4 3 is

Indices can be positive or negative numbers.

Moving up the rows we get 2 times larger per row.

Moving down the rows we get 2 times smaller per row.

Calculations with indices

In order to calculate with indices we need to be able to use the laws of indices in a variety of different ways. Let’s look at the different ways we can calculate with indices.

Index laws worksheet

Get your free laws of indices worksheet of 20+ questions and answers. Includes reasoning and applied questions.

How to use the laws of indices

We can use the laws of indices to simplify expressions.

Explain how to use the laws of indices

Law 1: multiplying indices

1 Add the powers .

2 Multiply any coefficients .

…………………………………………………………………………………………………………………………………………………….

This can be written as

1 Add together the indices 3 and the 2.

2 Multiply 4 and 7 together.

Law 2: dividing indices

• Subtract the indices .
• Divide any coefficients of the base number or letter .

…………………………………………………………………………………………………………………….………………………………

Subtract the indices 9 and 2.

Divide the coefficients 21 and 7.

Law 3: brackets with indices

• Multiply the indices .
• Raise all coefficients inside the bracket to the power outside the bracket .

Why does this work?

Let’s look at

This algebraic expression has been raised to the power of 4, which means:

We know when we multiply indices with the same base, we must add the powers.

Is there a quicker way to work this out?

Simplifying a 3 × a 3 × a 3 × a 3 to a 3×4 we can work out the simplified answer is a 12 .

When we have brackets and indices we can multiply the powers.

Brackets with indices examples

Example 1: no coefficient in front of base.

Simplify and leave your answer in index form.

Multiply the powers 3 and the 4.

Example 2: coefficient inside bracket

Multiply the indices 3 and the 2.

Raise 10 to the power of 2.

Example 3: negative indices

Multiply the indices -3 and the 2.

Raise -4 to the power of 2.

Law 4: power of 0

Any non-zero value raised to the power of 0 is equal to 1 .

We can see that 12 × 1 = 12 .

Multiplying anything by 1 leaves it unchanged, this is called the multiplicative identity. 

Law 5: negative indices

When the index is negative, put it over 1 and flip to make it positive.

• Put the term over 1 .
• Flip the fraction to make the index positive .
• Simplify if necessary .

Simplify 

Notice how the index affects the entire bracket

Put the term over 1.

Flip and change the power -3 to +3.

Simplify the denominator.

Law 6: fractional indices (H)

• Use the denominator to find the root of the term
• Raise the answer to the power of the numerator

Use the denominator to find the cube root of the number or letter.

Raise the answer to the power of the numerator.

Common misconceptions

• The multiplication and division laws can only be used for terms with the same base

You cannot simplify

as the bases are different

• When a power is outside the bracket you must raise all terms inside the bracket by that power

When we simplify

We must remember to square both the 4 and the a. It is common to forget to square the 4.

The brackets mean the entire term is squared.

• Confusing integer and fractional powers

Raising a term to the power of 2 means we square it

Raising a term to the power of ½ means we find the square root of it

Raising a term to the power of 3 means we cube it

Raising a term to the power of ⅓ means we find the cube root of it

• Confusing positive and negative powers

We can have decimal, fractional, negative or positive integers.

• Indices, powers or exponents

Indices can also be called powers or exponents.

Practice laws of indices questions

1. Simplify

We need to multiply the coefficients

Since the base number is the same in each term, we can add the indices

2. Simplify

3. Simplify

Since the base number is the same in each term, we can subtract the indices

4. Simplify

We need to divide the coefficients

5. Simplify

Anything raised to the power zero is 1 , so

This means we have

6. Simplify

The negative index number indicates we need the reciprocal of

7. Simplify

The index number does not affect the 6 , hence

8. Evaluate

The index number is telling us to square the cube root, so

9. Evaluate

The index number is telling us to find the reciprocal of the square root, so

Laws of indices GCSE questions

h^{7} \text { or } m^{11} seen (evidence of adding powers)

24 h^{7} m^{11}

3.  Simplify

\frac{12 x^{5} y^{7}}{3 x^{2} y}

x^{3} \text { or } y^{6} seen (evidence of subtracting powers)

4 x^{3} y^{6}

Learning checklist

You have now learned how to:

• Simplify expressions involving the laws of indices
• Calculate with roots, and with integer and fractional indices

The next lessons are

• Mathematical proof
• Functions in algebra

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Resources tagged with: Indices

There are 37 NRICH Mathematical resources connected to Indices , you may find related items under Algebraic expressions, equations and formulae .

Negative Power

What does this number mean ? Which order of 1, 2, 3 and 4 makes the highest value ? Which makes the lowest ?

When is $7^n + 3^n$ a multiple of 10? Can you prove the result by two different methods?

Perfectly Square

The sums of the squares of three related numbers is also a perfect square - can you explain why?

Sums of Squares

Can you prove that twice the sum of two squares always gives the sum of two squares?

Power Quady

Find all real solutions of the equation (x^2-7x+11)^(x^2-11x+30) = 1.

Which is the bigger, 9^10 or 10^9 ? Which is the bigger, 99^100 or 100^99 ?

How Many Solutions?

Find all the solutions to the this equation.

Climbing Powers

$2\wedge 3\wedge 4$ could be $(2^3)^4$ or $2^{(3^4)}$. Does it make any difference? For both definitions, which is bigger: $r\wedge r\wedge r\wedge r\dots$ where the powers of $r$ go on for ever, or $(r^r)^r$, where $r$ is $\sqrt{2}$?

Telescoping Series

Find $S_r = 1^r + 2^r + 3^r + ... + n^r$ where r is any fixed positive integer in terms of $S_1, S_2, ... S_{r-1}$.

Mega Quadratic Equations

What do you get when you raise a quadratic to the power of a quadratic?

Add powers of 3 and powers of 7 and get multiples of 11.

Card Shuffle

This article for students and teachers tries to think about how long would it take someone to create every possible shuffle of a pack of cards, with surprising results.

Public Key Cryptography

An introduction to coding and decoding messages and the maths behind how to secretly share information.

The Public Key

Find 180 to the power 59 (mod 391) to crack the code. To find the secret number with a calculator we work with small numbers like 59 and 391 but very big numbers are used in the real world for this.

Multiplication Magic

Given any 3 digit number you can use the given digits and name another number which is divisible by 37 (e.g. given 628 you say 628371 is divisible by 37 because you know that 6+3 = 2+7 = 8+1 = 9). The question asks you to explain the trick.

Lastly - Well

What are the last two digits of 2^(2^2003)?

Powerful Factors

Use the fact that: x²-y² = (x-y)(x+y) and x³+y³ = (x+y) (x²-xy+y²) to find the highest power of 2 and the highest power of 3 which divide 5^{36}-1.

Pythagoras Mod 5

Prove that for every right angled triangle which has sides with integer lengths: (1) the area of the triangle is even and (2) the length of one of the sides is divisible by 5.

Evaluate without a calculator: (5 sqrt2 + 7)^{1/3} - (5 sqrt2 - 7)^1/3}.

Learn about Number Bases

We are used to writing numbers in base ten, using 0, 1, 2, 3, 4, 5, 6, 7, 8, and 9. Eg. 75 means 7 tens and five units. This article explains how numbers can be written in any number base.

Modulus Arithmetic and a Solution to Dirisibly Yours

Peter Zimmerman from Mill Hill County High School in Barnet, London gives a neat proof that: 5^(2n+1) + 11^(2n+1) + 17^(2n+1) is divisible by 33 for every non negative integer n.

More Sums of Squares

Tom writes about expressing numbers as the sums of three squares.

Sums of Squares and Sums of Cubes

An account of methods for finding whether or not a number can be written as the sum of two or more squares or as the sum of two or more cubes.

Really Mr. Bond

115^2 = (110 x 120) + 25, that is 13225 895^2 = (890 x 900) + 25, that is 801025 Can you explain what is happening and generalise?

Enriching Experience

Find the five distinct digits N, R, I, C and H in the following nomogram

Power Crazy

What can you say about the values of n that make $7^n + 3^n$ a multiple of 10? Are there other pairs of integers between 1 and 10 which have similar properties?

Find some triples of whole numbers a, b and c such that a^2 + b^2 + c^2 is a multiple of 4. Is it necessarily the case that a, b and c must all be even? If so, can you explain why?

Thirty Six Exactly

The number 12 = 2^2 × 3 has 6 factors. What is the smallest natural number with exactly 36 factors?

Rachel's Problem

Is it true that $99^n$ has 2n digits and $999^n$ has 3n digits? Investigate!

Solving the equation x^3 = 3 is easy but what about solving equations with a 'staircase' of powers?

Remainder Hunt

What are the possible remainders when the 100-th power of an integer is divided by 125?

Show without recourse to any calculating aid that 7^{1/2} + 7^{1/3} + 7^{1/4} < 7 and 4^{1/2} + 4^{1/3} + 4^{1/4} > 4 . Sketch the graph of f(x) = x^{1/2} + x^{1/3} + x^{1/4} -x

Big, Bigger, Biggest

Which is the biggest and which the smallest of $2000^{2002}, 2001^{2001} \text{and } 2002^{2000}$?

Find the smallest positive integer N such that N/2 is a perfect cube, N/3 is a perfect fifth power and N/5 is a perfect seventh power.

Novemberish

a) A four digit number (in base 10) aabb is a perfect square. Discuss ways of systematically finding this number. (b) Prove that 11^{10}-1 is divisible by 100.

Which is larger: (a) 1.000001^{1000000} or 2? (b) 100^{300} or 300! (i.e.factorial 300)

What is the units digit for the number 123^(456) ?

Perfect Maths

Useful mathematics tutorials and articles for primary up to spm and igcse, indices and logarithms.

Fractional Indices:

Law of Indices:

How to simplify algebraic expressions

Example 4: 

Answer: 1.2788 [Use Calculator to find the answer]

Answer: m = 0.0027 [Press (shift)(log)(-2.5686)]

Laws of Logarithms: 

Example 8: 

Example 10:

Change the Base of Logarithm 

Example 11: 

The following examples need to be solved using the Laws of Logarithms and change of base. So please remember the laws of logarithms and the change of the base of logarithms.

Example 12:

Example 13: 

Solving Equation involving indices and logarithms

a) Method 1: Expressing the equation to same base and compare the indices.

b) Method 2: Expressing the equation to same indices and compare the base.

d) Method 4: Expressing the equation as a single logarithms form to the same base

About the Authors:

Mr Low – A full time tutor from Malaysia, passionate about teaching Mathematics for PMR Maths, SPM Additional Maths, SPM Modern Maths and International School Maths included SAT,GCSE, Checkpoint, IGCSE syllabus.

Ms Tifeny – A full time tutor from Malaysia teaching Mathematics for SPM Additional Maths, STPM Maths and International School Maths included SAT,GCSE, Checkpoint, IGCSE syllabus, A-level, Pre-U and IB. 

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Solved Problems on Indices

To solve problems related to indices (or exponentials or powers), we need to have the list of laws of indices. So here are the laws/rules of indices.

• $a^{-1}=1/a$

• $a^{-n}=1/a^n$

• $a^{m+n}=a^m . a^n$

• $(a^m)^n=a^{m \times n}$

• $\frac{a^m}{a^n}=a^{m-n}$

Solved Problems

Problem 1:  Simplify $(81)^{3/4}$

Note that $81=$ $3 \times 3 \times 3 \times 3$ $=3^4$

∴ $(81)^{3/4}$ $=(3^4)^{3/4}$

$=3^{4 \times 3/4}$ $[\because (a^m)^n=a^{m \times n}]$

Problem 2: Find the value of $(8)^{-1/3}$ (cube root of 1/8)

We have $8=$ $2 \times 2 \times 2$ $=2^3$

∴ $(8)^{-1/3}$ $=(2^3)^{-1/3}$

$=2^{-3 \times 1/3}$ $[\because (a^m)^n=a^{m \times n}]$

$=\frac{1}{2}$ $[\because a^{-1}=\frac{1}{a}]$

Problem 3: Simplify $(125)^{-1/3}$ (cube root of 1/125)

Note that $125=$ $5 \times 5 \times 5$ $=5^3$

∴ $(125)^{-1/3}$ $=(5^3)^{-1/3}$

$=5^{-3 \times 1/3}$ $[\because (a^m)^n=a^{m \times n}]$

$=\frac{1}{5}$ $[\because a^{-1}=\frac{1}{a}]$

Problem 4: Find $\, 5^0 \times (16)^{-3/4}$

Using $16=2^4,$ we get

$5^0 \times (16)^{-3/4}$

$=1 \times (2^4)^{-3/4}$ $[\because a^0=1]$

$=(2^4)^{-3/4}$

$=2^{4 \times \frac{-3}{4}}$ $[\because (a^m)^n=a^{m \times n}]$

$=\frac{1}{2^3}$ $[\because a^{-n}=\frac{1}{a^n}]$

$=\frac{1}{8}$

Problem 5: Simplify $\, x^{a-b}.x^{b-c}.x^{c-a}$

$x^{a-b}.x^{b-c}.x^{c-a}$

$=x^{a-b+b-c+c-a}$

Problem 5: Solve for $x$

(i) $4^x=8^2$

(ii) $3^x=2^{-x}$

$\Rightarrow (2^2)^x=(2^3)^2$

$\Rightarrow 2^{2x}=2^{3 \times 2}$ $[\because (a^m)^n=a^{m \times n}]$

$\Rightarrow 2^{2x}=2^6$

Comparing the powers on both sides, we get

$\Rightarrow x=\frac{6}{2}=3$

$\Rightarrow 3^x=\frac{1}{2^x}$ $[\because a^{-n}=\frac{1}{a^n}]$

$\Rightarrow 3^x \times 2^x=1$

$\Rightarrow (3 \times 2)^x=1$ $[\because a^n \times b^n=(ab)^n]$

$\Rightarrow 6^x=6^0$

Comparing the powers, we obtain

This article is written by Dr. T. Mandal, Ph.D in Mathematics. On Mathstoon.com you will find Maths from very basic level to advanced level. Thanks for visiting.

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Watch CBS News

Teens come up with trigonometry proof for Pythagorean Theorem, a problem that stumped math world for centuries

By Bill Whitaker

May 5, 2024 / 7:00 PM EDT / CBS News

As the school year ends, many students will be only too happy to see math classes in their rearview mirrors. It may seem to some of us non-mathematicians that geometry and trigonometry were created by the Greeks as a form of torture, so imagine our amazement when we heard two high school seniors had proved a mathematical puzzle that was thought to be impossible for 2,000 years. 

We met Calcea Johnson and Ne'Kiya Jackson at their all-girls Catholic high school in New Orleans. We expected to find two mathematical prodigies.

Instead, we found at St. Mary's Academy , all students are told their possibilities are boundless.

Come Mardi Gras season, New Orleans is alive with colorful parades, replete with floats, and beads, and high school marching bands.

In a city where uniqueness is celebrated, St. Mary's stands out – with young African American women playing trombones and tubas, twirling batons and dancing - doing it all, which defines St. Mary's, students told us.

Junior Christina Blazio says the school instills in them they have the ability to accomplish anything. 

Christina Blazio: That is kinda a standard here. So we aim very high - like, our aim is excellence for all students. 

The private Catholic elementary and high school sits behind the Sisters of the Holy Family Convent in New Orleans East. The academy was started by an African American nun for young Black women just after the Civil War. The church still supports the school with the help of alumni.

In December 2022, seniors Ne'Kiya Jackson and Calcea Johnson were working on a school-wide math contest that came with a cash prize.

Ne'Kiya Jackson: I was motivated because there was a monetary incentive.

Calcea Johnson: 'Cause I was like, "$500 is a lot of money. So I-- I would like to at least try."

Both were staring down the thorny bonus question.

Bill Whitaker: So tell me, what was this bonus question?

Calcea Johnson: It was to create a new proof of the Pythagorean Theorem. And it kind of gave you a few guidelines on how would you start a proof.

The seniors were familiar with the Pythagorean Theorem, a fundamental principle of geometry. You may remember it from high school: a² + b² = c². In plain English, when you know the length of two sides of a right triangle, you can figure out the length of the third.

Both had studied geometry and some trigonometry, and both told us math was not easy. What no one told  them  was there had been more than 300 documented proofs of the Pythagorean Theorem using algebra and geometry, but for 2,000 years a proof using trigonometry was thought to be impossible, … and that was the bonus question facing them.

Bill Whitaker: When you looked at the question did you think, "Boy, this is hard"?

Ne'Kiya Jackson: Yeah. 

Bill Whitaker: What motivated you to say, "Well, I'm going to try this"?

Calcea Johnson: I think I was like, "I started something. I need to finish it." 

Bill Whitaker: So you just kept on going.

Calcea Johnson: Yeah.

For two months that winter, they spent almost all their free time working on the proof.

CeCe Johnson: She was like, "Mom, this is a little bit too much."

CeCe and Cal Johnson are Calcea's parents.

CeCe Johnson:   So then I started looking at what she really was doing. And it was pages and pages and pages of, like, over 20 or 30 pages for this one problem.

Cal Johnson: Yeah, the garbage can was full of papers, which she would, you know, work out the problems and-- if that didn't work she would ball it up, throw it in the trash. 

Bill Whitaker: Did you look at the problem? 

Neliska Jackson is Ne'Kiya's mother.

Neliska Jackson: Personally I did not. 'Cause most of the time I don't understand what she's doing (laughter).

Michelle Blouin Williams: What if we did this, what if I write this? Does this help? ax² plus ….

Their math teacher, Michelle Blouin Williams, initiated the math contest.

Bill Whitaker: And did you think anyone would solve it?

Michelle Blouin Williams: Well, I wasn't necessarily looking for a solve. So, no, I didn't—

Bill Whitaker: What were you looking for?

Michelle Blouin Williams: I was just looking for some ingenuity, you know—

Calcea and Ne'Kiya delivered on that! They tried to explain their groundbreaking work to 60 Minutes. Calcea's proof is appropriately titled the Waffle Cone.

Calcea Johnson: So to start the proof, we start with just a regular right triangle where the angle in the corner is 90°. And the two angles are alpha and beta.

Bill Whitaker: Uh-huh

Calcea Johnson: So then what we do next is we draw a second congruent, which means they're equal in size. But then we start creating similar but smaller right triangles going in a pattern like this. And then it continues for infinity. And eventually it creates this larger waffle cone shape.

Calcea Johnson: Am I going a little too—

Bill Whitaker: You've been beyond me since the beginning. (laughter) 

Bill Whitaker: So how did you figure out the proof?

Ne'Kiya Jackson: Okay. So you have a right triangle, 90° angle, alpha and beta.

Bill Whitaker: Then what did you do?

Ne'Kiya Jackson: Okay, I have a right triangle inside of the circle. And I have a perpendicular bisector at OP to divide the triangle to make that small right triangle. And that's basically what I used for the proof. That's the proof.

Bill Whitaker: That's what I call amazing.

Ne'Kiya Jackson: Well, thank you.

There had been one other documented proof of the theorem using trigonometry by mathematician Jason Zimba in 2009 – one in 2,000 years. Now it seems Ne'Kiya and Calcea have joined perhaps the most exclusive club in mathematics. 

Bill Whitaker: So you both independently came up with proof that only used trigonometry.

Ne'Kiya Jackson: Yes.

Bill Whitaker: So are you math geniuses?

Calcea Johnson: I think that's a stretch. 

Bill Whitaker: If not genius, you're really smart at math.

Ne'Kiya Jackson: Not at all. (laugh) 

To document Calcea and Ne'Kiya's work, math teachers at St. Mary's submitted their proofs to an American Mathematical Society conference in Atlanta in March 2023.

Ne'Kiya Jackson: Well, our teacher approached us and was like, "Hey, you might be able to actually present this," I was like, "Are you joking?" But she wasn't. So we went. I got up there. We presented and it went well, and it blew up.

Bill Whitaker: It blew up.

Calcea Johnson: Yeah. 

Ne'Kiya Jackson: It blew up.

Bill Whitaker: Yeah. What was the blowup like?

Calcea Johnson: Insane, unexpected, crazy, honestly.

It took millenia to prove, but just a minute for word of their accomplishment to go around the world. They got a write-up in South Korea and a shout-out from former first lady Michelle Obama, a commendation from the governor and keys to the city of New Orleans. 

Bill Whitaker: Why do you think so many people found what you did to be so impressive?

Ne'Kiya Jackson: Probably because we're African American, one. And we're also women. So I think-- oh, and our age. Of course our ages probably played a big part.

Bill Whitaker: So you think people were surprised that young African American women, could do such a thing?

Calcea Johnson: Yeah, definitely.

Ne'Kiya Jackson: I'd like to actually be celebrated for what it is. Like, it's a great mathematical achievement.

Achievement, that's a word you hear often around St. Mary's academy. Calcea and Ne'Kiya follow a long line of barrier-breaking graduates. 

The late queen of Creole cooking, Leah Chase , was an alum. so was the first African-American female New Orleans police chief, Michelle Woodfork …

And judge for the Fifth Circuit Court of Appeals, Dana Douglas. Math teacher Michelle Blouin Williams told us Calcea and Ne'Kiya are typical St. Mary's students.  

Bill Whitaker: They're not unicorns.

Michelle Blouin Williams: Oh, no no. If they are unicorns, then every single lady that has matriculated through this school is a beautiful, Black unicorn.

Pamela Rogers: You're good?

Pamela Rogers, St. Mary's president and interim principal, told us the students hear that message from the moment they walk in the door.

Pamela Rogers: We believe all students can succeed, all students can learn. It does not matter the environment that you live in. 

Bill Whitaker: So when word went out that two of your students had solved this almost impossible math problem, were they universally applauded?

Pamela Rogers: In this community, they were greatly applauded. Across the country, there were many naysayers.

Bill Whitaker: What were they saying?

Pamela Rogers: They were saying, "Oh, they could not have done it. African Americans don't have the brains to do it." Of course, we sheltered our girls from that. But we absolutely did not expect it to come in the volume that it came.  

Bill Whitaker: And after such a wonderful achievement.

Pamela Rogers: People-- have a vision of who can be successful. And-- to some people, it is not always an African American female. And to us, it's always an African American female.

Gloria Ladson-Billings: What we know is when teachers lay out some expectations that say, "You can do this," kids will work as hard as they can to do it.

Gloria Ladson-Billings, professor emeritus at the University of Wisconsin, has studied how best to teach African American students. She told us an encouraging teacher can change a life.

Bill Whitaker: And what's the difference, say, between having a teacher like that and a whole school dedicated to the excellence of these students?

Gloria Ladson-Billings: So a whole school is almost like being in Heaven. 

Bill Whitaker: What do you mean by that?

Gloria Ladson-Billings: Many of our young people have their ceilings lowered, that somewhere around fourth or fifth grade, their thoughts are, "I'm not going to be anything special." What I think is probably happening at St. Mary's is young women come in as, perhaps, ninth graders and are told, "Here's what we expect to happen. And here's how we're going to help you get there."

At St. Mary's, half the students get scholarships, subsidized by fundraising to defray the $8,000 a year tuition. Here, there's no test to get in, but expectations are high and rules are strict: no cellphones, modest skirts, hair must be its natural color.

Students Rayah Siddiq, Summer Forde, Carissa Washington, Tatum Williams and Christina Blazio told us they appreciate the rules and rigor.

Rayah Siddiq: Especially the standards that they set for us. They're very high. And I don't think that's ever going to change.

Bill Whitaker: So is there a heart, a philosophy, an essence to St. Mary's?

Summer Forde: The sisterhood—

Carissa Washington: Sisterhood.

Tatum Williams: Sisterhood.

Bill Whitaker: The sisterhood?

Voices: Yes.

Bill Whitaker: And you don't mean the nuns. You mean-- (laughter)

Christina Blazio: I mean, yeah. The community—

Bill Whitaker: So when you're here, there's just no question that you're going to go on to college.

Rayah Siddiq: College is all they talk about. (laughter) 

Pamela Rogers: … and Arizona State University (Cheering)

Principal Rogers announces to her 615 students the colleges where every senior has been accepted.

Bill Whitaker: So for 17 years, you've had a 100% graduation rate—

Pamela Rogers: Yes.

Bill Whitaker: --and a 100% college acceptance rate?

Pamela Rogers: That's correct.

Last year when Ne'Kiya and Calcea graduated, all their classmates went to college and got scholarships. Ne'Kiya got a full ride to the pharmacy school at Xavier University in New Orleans. Calcea, the class valedictorian, is studying environmental engineering at Louisiana State University.

Bill Whitaker: So wait a minute. Neither one of you is going to pursue a career in math?

Both: No. (laugh)

Calcea Johnson: I may take up a minor in math. But I don't want that to be my job job.

Ne'Kiya Jackson: Yeah. People might expect too much out of me if (laugh) I become a mathematician. (laugh)

But math is not completely in their rear-view mirrors. This spring they submitted their high school proofs for final peer review and publication … and are still working on further proofs of the Pythagorean Theorem. Since their first two …

Calcea Johnson: We found five. And then we found a general format that could potentially produce at least five additional proofs.

Bill Whitaker: And you're not math geniuses?

Bill Whitaker: I'm not buying it. (laughs)

Produced by Sara Kuzmarov. Associate producer, Mariah B. Campbell. Edited by Daniel J. Glucksman.

Bill Whitaker is an award-winning journalist and 60 Minutes correspondent who has covered major news stories, domestically and across the globe, for more than four decades with CBS News.

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History Daily

New Orleans teens solve 2,000-year-old math problem

Posted: May 11, 2024 | Last updated: May 11, 2024

Some ingenuity

Teacher Michelle Blouin Williams initiated a math competition with a bonus question tasking students to develop a new proof for the Pythagorean theorem using trigonometry, without anticipating that anyone would successfully tackle the challenge.  

She said, “I was just looking for some ingenuity.”   

Williams' expectations

Calcea Johnson and Ne’Kiya Jackson exceeded Williams' expectations by successfully solving the challenge in 2023.   

St. Mary's Academy

These two teenagers, who were seniors at St. Mary's Academy in New Orleans, a renowned Catholic school for girls with impressive college acceptance and graduation rates, were featured on CBS News' "60 Minutes" to discuss their accomplishment.   

$500 reward

Initially enticed by the math competition's $500 reward, their determination to complete the task intensified as they delved into the complex bonus question.   

Over a period of two months, these high school seniors dedicated themselves to completing their proof.  

Pages and pages

During an interview with "60 Minutes," CeCe Johnson, the mother of Calcea, said “It was pages and pages and pages of, like, over 20 or 30 pages for this one problem.”  

The garbage can

Her father, Cal Johnson, added, “Yeah, the garbage can was full of papers, which she would, you know, work out the problems and — if that didn’t work she would ball it up, throw it in the trash.”  

Upon completion

Upon completion, the teachers at St. Mary's Academy acknowledged the significance of Calcea and Ne'Kiya's achievement and submitted their proof to the American Mathematical Society for recognition at a conference in March 2023, where the students presented their findings.  

The Pythagorean theorem

The Pythagorean theorem, at its core, states that knowing the lengths of two sides of a right triangle allows you to determine the length of the third side using the formula a² + b² = c².   

While commonly attributed to the Greek mathematician Pythagoras, historical evidence suggests its existence in Babylon and Iron Age India. This theorem finds practical applications in construction, architecture, navigation, and surveying.  

A mathematical proof

A mathematical proof is a logical argument that demonstrates the truth of a mathematical theorem. American mathematician Daniel Kane likens proofs to essays, but rooted in mathematical concepts.  

Using trigonometry

As per the "60 Minutes" segment, “there had been more than 300 documented proofs of the Pythagorean Theorem using algebra and geometry, but for 2,000 years a proof using trigonometry was thought to be impossible.”  

Mathematician Elisha Loomis

Back in 1927, mathematician Elisha Loomis made a statement in his book "The Pythagorean Proposition." Loomis contended that a trigonometric proof of the theorem was impossible as it would create a circular argument.  

Stuart Anderson

Stuart Anderson, a retired mathematics professor from Texas A&M University-Commerce, mentioned to Scientific American, “A lot of the basic trig ‘identities’ are nothing more than Pythagoras’ theorem.”  

Trigonometric functions

Loomis contended that attempting to use trigonometric functions to prove the Pythagorean theorem would result in a circular reasoning loop, as the functions themselves are built upon the theorem. This, according to Loomis, would constitute a fundamental mathematical mistake.  

The law of sines

As reported by Scientific American, the teenagers challenged this notion during their presentation in 2023 and stated that “a trigonometric identity called the law of sines didn’t depend on the Pythagorean theorem and that they could use it to prove the theorem.”  

Final peer review

Calcea and Ne'Kiya are now part of a select few who have achieved a similar milestone, such as mathematician Jason Zimba, who developed a new proof in 2009. They submitted their proof for the final peer review earlier this year and are actively working on crafting additional proofs.  

In recognition

In recognition of their accomplishment, the teenagers were honored with the keys to the city of New Orleans and received commendations from the governor of Louisiana, along with other public acknowledgments.  

Widespread recognition

Despite the widespread recognition of their accomplishment which “blew up,” as Ne'Kiya expressed it, the two students maintain their humility and even chuckled at being labeled as geniuses.  

Surprised and skeptical

Upon the revelation of their achievement, there were individuals who appeared surprised and skeptical, dismissing the news as untrue, as mentioned by St. Mary's president Pamela Rogers during the interview.  

African Americans

“They were saying, ‘Oh, they could not have done it. African Americans don’t have the brains to do it.’ ... People — have a vision of who can be successful. And — to some people, it is not always an African American female. And to us, it’s always an African American female.”  

The significant reaction

When questioned by interviewer Bill Whitaker about their thoughts on the significant reaction, Ne’Kiya said, “Probably because we’re African American, one. And we’re also women. So I think — oh, and our age. Of course our ages probably played a big part.”  

Great mathematical achievement

“I’d like to actually be celebrated for what it is. Like, it’s a great mathematical achievement,” she continued.  

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