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IPMAT Indore 2019

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Remainder

Medium

The remainder when (2929)29(29^{29){29}} is divided by 99 is

✅ Correct Option: 2
Don't get confused with the two notations:
I. (ab)c=ab×c(a^{b})^c = a^{b \times c}
II. abc=aba^{b^c} = a^b to the power cc.
The question asks for 2929×29=2984129^{29 \times 29} = 29^{841}
This can be solved through cyclicity:
\newline\newline\newline\newline\newline\newline\newline\newline\newline\newline\newline\newline\newline\newline\newline\newline\newline\newline\newline\newline\newline\newline\newline\newline\newline\newline\newline\newline\newline\newline\newline\newline\newline\newline\newline
Power Expression Remainder when ÷ 9
1 29129^1 2
2 29229^2 4
3 29329^3 6
4 29429^4 8
5 29529^5 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:
\newline\newline\newline\newline\newline\newline\newline\newline\newline\newline\newline\newline\newline\newline\newline\newline\newline\newline\newline\newline\newline\newline\newline\newline\newline\newline\newline\newline\newline
Power Form Remainder when ÷ 9
295k+129^{5k+1} 2
295k+229^{5k+2} 4
295k+329^{5k+3} 6
295k+429^{5k+4} 8
295k29^{5k} 1
We need to express 841 in the form 5k+r5k + r: \newline 841=840+1841 = 840 + 1 (because we know that 840840 is a multiple of 55)
\newline\newline\newline\newline\newline\newline\newline\newline\newline\newline\newline\newline\newline
Power Form Remainder when ÷ 9
295k+129^{5k+1} 2\boxed{2}
\newline
Had the question been 292929{29^{29}}^{29}, the answer co-incidentally comes out to be the same:
Finding Euler

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Finding Euler

Euler's Theorem

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Euler's Theorem

Power to a Power

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Power to a Power

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