GCSE Integer indices Practice Questions

Count the repeated factors

The number $3$ is multiplied by itself, so count how many $3$s there are. There are four of them.

Use index notation

A power is a short way of writing repeated multiplication. The little number (the index) tells you how many $3$s are multiplied together.

State the answer

So four $3$s multiplied together is written as $3$ to the power $4$.

List powers of $2$

Build up the powers of $2$ one at a time, doubling each time, until you reach $32$.

Double once more

Doubling $16$ gives exactly $32$, so five $2$s are needed.

State the value of n

Since $2$ to the power $5$ equals $32$, the missing index n is $5$.

Write $8$ as a power of $2$

To combine everything into one power of $2$, first convert the ordinary number $8$.

Substitute into the expression

Replace $8$ with $2$ cubed so both factors are powers of $2$.

Recall the product law

Multiplying powers of the same base means adding the indices.

Add the indices

Three $2$s times five $2$s gives eight $2$s in total.

Write the single power

$3$ plus $5$ is $8$, so the expression is $2$ to the power $8$.

Sense check

Checking with values: $8$ times $32$ is $256$, and $2^8$ is also $256$. ✓

State the student's claim

Read the claim carefully. The indices have been added correctly, but look at what has happened to the base.

Recall the product law

When multiplying powers of the same base, the indices are added but the base stays exactly the same.

Identify the error

The student multiplied the bases as well as adding the indices. The base must stay as $2$.

Apply the law correctly

Keep the base $2$ and add the indices $3$ and $4$.

Simplify the index

$3$ plus $4$ is $7$, so the correct answer is $2$ to the power $7$.

Evaluate the correct answer

Doubling seven times: $2$, $4$, $8$, $16$, $32$, $64$, $128$.

Evaluate the student's answer

The student's answer is enormously bigger, which shows how serious the error is.

Check the correct answer directly

$2$ cubed is $8$ and $2$ to the power $4$ is $16$; multiplying them gives $128$, matching $2$ to the power $7$.

Summarise the error

In one sentence: the student wrongly changed the base from $2$ to $4$ while adding the indices.

State the correct result

The corrected calculation gives $2$ to the power $7$, which is $128$.

Understand the goal

We need the first power of $3$ that passes $1000$, so build up the powers of $3$ until one goes over.

Start the list

Begin with $3$ to the power $1$.

Next power

Multiply by $3$: nine.

Next power

Multiply by $3$ again: twenty-seven.

Next power

$27$ times $3$ is $81$.

Next power

$81$ times $3$ is $243$. Still well below $1000$.

Next power

$243$ times $3$ is $729$ — close to $1000$, but not past it yet.

Compare with $1000$

$3$ to the power $6$ falls short, so $n = 6$ does not satisfy the inequality.

Next power

$729$ times $3$ is $2187$.

Compare with $1000$

$3$ to the power $7$ is past $1000$ — the inequality is satisfied for the first time.

Bracket the answer

$1000$ is trapped between the sixth and seventh powers of $3$, so the smallest winning index is $7$.

Check the multiplication $729 \times 3$

Checking the key step: $700 \times 3 = 2100$ and $29 \times 3 = 87$, giving $2187$. ✓

Why no smaller n works

Powers of $3$ grow every time n goes up, so every index below $7$ gives an even smaller value than $729$, all below $1000$.

Watch for the common error

It is tempting to stop at $729$ because it is close, but the question needs strictly greater than $1000$.

State the answer

The smallest integer n with $3$ to the power n greater than $1000$ is $7$.