Number System

Remainder

Medium

The remainder when $(29^{29})^{29}$ is divided by $9$ is

✅ Correct Option: 2

Don't get confused with the two notations:

I. $(a^{b})^c = a^{b \times c}$

II. $a^{b^c} = a^b$ to the power $c$ .

The question asks for $29^{29 \times 29} = 29^{841}$

This can be solved through cyclicity:

Power	Expression	Remainder when ÷ 9
1	$29^1$	2
2	$29^2$	4
3	$29^3$	8
4	$29^4$	7
5	$29^5$	5
6	$29^6$	1

It's in the cyclicity of 5. The same remainders will repeat for powers of 6 to 10, then 11 to 15 and so forth!

Think of it like this:

Power Form	Remainder when ÷ 9
$29^{6k+1}$	2
$29^{6k+2}$	4
$29^{6k+3}$	8
$29^{6k+4}$	7
$29^{6k+5}$	5
$29^{6k}$	1

We need to express 841 in the form $6k + r$ :

$841 = 840 + 1$ (because we know that $840$ is a multiple of $6$ )

Power Form	Remainder when ÷ 9
$29^{6k+1}$	$\boxed{2}$

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The remainder when $(29^{29})^{29}$ is divided by $9$ is