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What is the remainder when 19^561 is divided by 29?

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Bunuel
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What is the remainder when 19^561 is divided by 29? [#permalink] New post   27 Jan 2022, 07:00
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What is the remainder when \(19^{561}\) is divided by 29?

(A) 19
(B) 18
(C) 17
(D) 9
(E) 8

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RajatGMAT777
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Re: What is the remainder when 19^561 is divided by 29? [#permalink] New post   27 Feb 2022, 06:49
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An easy way to work this out would be to se Fermat Little's theorem which says that -
for any prime number p and any integer a such that p does not divide a (the pair are relatively prime), p divides exactly into a^p − a.
\(19^{28}\) ≡ 1 (mod 29)
⟹ \((19^{28})^{28}\) ≡ \(1^{28}\) (mod 29)

⟹ \(19^{560}\) ≡ 1 (mod 29)

⟹ \(19^{560}\) * 19 ≡ 19 (mod 29)

Since 19 is less than 29, the remainder is the same number and the answer is 19(A)
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What is the remainder when 19^561 is divided by 29? [#permalink] New post   20 Feb 2022, 07:51
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Remainder of \(19^{561}\) / 29
=> \(19^{561}\) = \((19^2)^{280}*19 \)= \((348+13)^{280}*19\) =
= \((13^2)^{140}*19 \)= \((174 - 5)^{140}*19 \)= \((- 5)^{140}*19 \) =
= \((5^3)^{46}*5^2*19 \)= \((116+9)^{46}*5^2*19\) =
= \(9^{45}*9*5^2*19\) = \((9^3)^{15}*9*5^2*19 \) = \((725 + 4)^{15}*9*5^2*19\) =
= \(4^{15}*9*5^2*19 \)= \((58+6)^5*9*5^2*19 \)= \(6^5*9*5^2*19 \)=
= \((6*5)^2*6^2*9*6*19\) = \(7*9*6*19 = (58+5)*6*19\) =
= \((5*6)*19 \)= 19

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Ravi86
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Re: What is the remainder when 19^561 is divided by 29? [#permalink] New post   03 Mar 2022, 12:30
Hello Brunel can you please post the solution for this

Posted from my mobile device
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gmatophobia
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What is the remainder when 19^561 is divided by 29? [#permalink] New post   05 Mar 2022, 09:30
chetan2u - Can you please share an approach to solve this question.

I am aware that we can use Fermat Little's theorem. Considering that's something that GMAT doesn't expect one to know (please correct me if I am wrong here) , is there an alternate approach ?
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chetan2u
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Re: What is the remainder when 19^561 is divided by 29? [#permalink] New post   11 Mar 2022, 20:28
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AbhiroopGhosh wrote:
chetan2u - Can you please share an approach to solve this question.

I am aware that we can use Fermat Little's theorem. Considering that's something that GMAT doesn't expect one to know (please correct me if I am wrong here) , is there an alternate approach ?



Immediate response to this question: I can’t think of any other quick method except breaking 19 and further terms in multiples of 19 in accordance with Binomial theorem.

But not likely that you will get such questions that require Fermat littles theorem or excessive calculations and steps through Binomial theorem.
Ignore the question.
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Re: What is the remainder when 19^561 is divided by 29? [#permalink]
11 Mar 2022, 20:28
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