CMS Math Olympiad · 6th Grade

Exponents and Powers

From repeated multiplication to elegant shortcuts. Powers grow very quickly — let’s learn how to tame them.

1
What is a power?

When you multiply the same number by itself many times, writing it out becomes long:

\(2 \times 2 \times 2 \times 2 \times 2 = \;?\)

Instead we write \(2^5\) and read it as “two to the fifth power.”

Definition
\(a^n = \underbrace{a \times a \times a \times \dots \times a}_{n \text{ times}}\)
\(a\) — base. \(n\) — exponent.
How to read it

\(3^4\) = “three to the fourth power” = \(3 \times 3 \times 3 \times 3 = 81\)

\(5^2\) = “five squared” = \(5 \times 5 = 25\)

\(7^1\) = simply \(7\). Any number to the first power is the number itself.

Special cases
\(a^0 = 1\) for any \(a \neq 0\). Yes, even \(100^0 = 1\).
\(a^1 = a\) always. The first power changes nothing.
Try it
What is \(2^5\)?
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2
Powers of small numbers

For olympiad problems these values should be memorised.

n\(2^n\)\(3^n\)\(5^n\)\(10^n\)
123510
24925100
38271251 000
4168162510 000
5322433 125100 000
664729
7128
8256
9512
101 024
Powers of two are everywhere
Memorise \(2^1\) to \(2^{10}\). Many CMS questions ask: “What is the greatest \(n\) such that \(2^n < \text{some value}\)?” If you know the table, you can answer instantly.
A test problem you missed
«The greatest integer \(n\) such that \(2^n < 100\)»

\(2^6 = 64 < 100\) ✓ but \(2^7 = 128 > 100\) ✗
Answer: \(n = 6\).

Trick: do not compute from scratch — scan the table until you exceed the number.
Try it
The greatest integer \(n\) such that \(3^n < 100\)?
Try it
The greatest integer \(n\) such that \(2^n < 500\)?

Squares of numbers are useful powers too:

\(1, 4, 9, 16, 25, 36, 49, 64, 81, 100, 121, 144, 169, 196, 225\)

These are \(1^2, 2^2, 3^2, \dots, 15^2\).

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3
Rules for working with powers

These rules let us simplify expressions without calculating enormous numbers.

Rule 1 — Multiplication
\(a^m \times a^n = a^{m+n}\)
Add the exponents. \(2^3 \times 2^4 = 2^7 = 128\)
Why this works

\(2^3 \times 2^4 = (2 \times 2 \times 2) \times (2 \times 2 \times 2 \times 2) = 2^7\)

3 twos + 4 twos = 7 twos.

Rule 2 — Division
\(\dfrac{a^m}{a^n} = a^{m-n}\)
Subtract the exponents. \(\dfrac{3^5}{3^2} = 3^3 = 27\)
Rule 3 — Power of a power
\((a^m)^n = a^{m \times n}\)
Multiply the exponents. \((2^3)^4 = 2^{12} = 4096\)
Rule 4 — Power of a product
\((a \times b)^n = a^n \times b^n\)
\((2 \times 5)^3 = 2^3 \times 5^3 = 8 \times 125 = 1000\)
Typical mistake
\((a + b)^2 \neq a^2 + b^2\). A power does NOT distribute over addition!
\((3+4)^2 = 49\), but \(3^2 + 4^2 = 25\). These are different things!
Try it
Simplify: \(2^3 \times 2^5\)
Try it
What is \((5^2)^3\)?
Try it
Simplify: \(\dfrac{4^5}{4^3}\)
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4
Comparing powers

CMS problems often ask you to compare powers. Here are three main strategies:

Strategy 1 — Same base
A larger exponent gives a larger number (when the base is greater than 1).
\(3^5 > 3^4\), because \(5 > 4\).
Strategy 2 — Same exponent
A larger base gives a larger number.
\(5^3 = 125 > 64 = 4^3\).
Strategy 3 — Rewrite with one base
Compare \(4^{10}\) and \(8^6\):
\(4^{10} = (2^2)^{10} = 2^{20}\), while \(8^6 = (2^3)^6 = 2^{18}\).
\(20 > 18\), therefore \(4^{10} > 8^6\). ✓
Try it
Which is larger: \(2^{12}\) or \(4^5\)?
\(4^5 = (2^2)^5 = 2^{10} = 1024\)
\(2^{12} = 4096\)
\(2^{12} > 2^{10}\), therefore \(2^{12}\) is larger.
Try it
The greatest integer \(x\) such that \(3^{20} > 9^x\):
\(9^x = (3^2)^x = 3^{2x}\)
We need \(20 > 2x\), so \(x < 10\).
The greatest integer — 9.
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5
CMS-style practice

These tasks match the style of real olympiad problems on powers.

Problem 1 · Easy
The value of \(2^3 + 2^3\) is:
\(2^3 + 2^3 = 8 + 8 = 16 = 2^4\).
Watch out: \(2^3 + 2^3 \neq 2^6\)!
But \(2^3 + 2^3 = 2 \times 2^3 = 2^4\) — this is a common trick.
Problem 2 · Easy
If \(2^n = 64\), then \(n =\)
Problem 3 · Medium
The value of \(\dfrac{4^{2024} \times 3^{2025}}{6^{2025} \times 2^{2024}}\) is:
\(4^{2024} = 2^{4048}\), \(6^{2025} = 2^{2025} \times 3^{2025}\)
The threes cancel: \(\dfrac{2^{4048}}{2^{2025+2024}} = \dfrac{2^{4048}}{2^{4049}} = \dfrac{1}{2}\)
Problem 4 · Medium
If \(8^x = 2^{15}\), then \(x =\)
\(8^x = (2^3)^x = 2^{3x} = 2^{15}\)
\(3x = 15 \Rightarrow x = 5\)
Problem 5 · Medium
The number of digits in \(2^{10} \times 5^{10}\):
\((2 \times 5)^{10} = 10^{10} = 10\,000\,000\,000\) — 11 digits.
Problem 6 · Medium+
Among \(2^{30}\), \(3^{20}\), and \(5^{13}\), the greatest is:
\(2^{30} = (2^3)^{10} = 8^{10}\)
\(3^{20} = (3^2)^{10} = 9^{10}\)
\(9 > 8\), so \(9^{10} > 8^{10}\), therefore \(3^{20}\) is the greatest.
Cheat sheet — Powers
\(a^m \times a^n = a^{m+n}\) — same base → add the exponents
\(\dfrac{a^m}{a^n} = a^{m-n}\) — same base → subtract the exponents
\((a^m)^n = a^{mn}\) — power of a power → multiply the exponents
\((ab)^n = a^n b^n\) — distribute over multiplication
\(a^0 = 1\) — anything to the zero power equals 1
\(2^{10} = 1024 \approx 10^3\) — useful for estimates
For comparisons: rewrite to a common base

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