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Remainder from Sum containing Factorial

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pannalal Junior | Next Rank: 30 Posts Default Avatar
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Remainder from Sum containing Factorial

Post Sun Sep 24, 2017 5:50 am
Let S = (33!/1 + 33!/2 + 33!/3 + ..... + 33!/32 + 33!/33).
What will be remainder when S is divided by 29?
(A) 1
(B) 5
(C) 24
(D) 28
(E) None of the above

OA will be given later.

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Post Tue Sep 26, 2017 10:32 pm
pannalal wrote:
(33*32*31*30 * 28!) mod 29 is not equal to (-4 * -3 * -2 * -1 * 28!) mod 29 =>

Rather, it should be (4*3*2*1*28!) mod 29.
Ah, but that isn't a mistake! It'd be a mistake to say that 33 = -4 mod 29, but that isn't what I said Wink

Of course, I could see why it would seem that way from what I typed, but hey, I'm wriggling out of this one.

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pannalal Junior | Next Rank: 30 Posts Default Avatar
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Post Tue Sep 26, 2017 11:24 pm
Matt@VeritasPrep wrote:
pannalal wrote:
(33*32*31*30 * 28!) mod 29 is not equal to (-4 * -3 * -2 * -1 * 28!) mod 29 =>

Rather, it should be (4*3*2*1*28!) mod 29.
Ah, but that isn't a mistake! It'd be a mistake to say that 33 = -4 mod 29, but that isn't what I said Wink

Of course, I could see why it would seem that way from what I typed, but hey, I'm wriggling out of this one.
Matt, somehow, I still feel that there is a mistake. If you write:
(33*32*31*30 * 28!) mod 29 = (-4 * -3 * -2 * -1 * 28!) mod 29

How do you prove that you are right. The final answer is correct because 4*3*2*1 = 24 and (-4)*(-3)*(-2)*(-1) = 24. Somehow I am not ready to accept that it is not a mistake. In any case, please state how do you arrive (33*32*31*30) mod 29 = (-4 * -3 * -2 * -1) mod 29.

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Post Tue Sep 26, 2017 10:32 pm
pannalal wrote:
(33*32*31*30 * 28!) mod 29 is not equal to (-4 * -3 * -2 * -1 * 28!) mod 29 =>

Rather, it should be (4*3*2*1*28!) mod 29.
Ah, but that isn't a mistake! It'd be a mistake to say that 33 = -4 mod 29, but that isn't what I said Wink

Of course, I could see why it would seem that way from what I typed, but hey, I'm wriggling out of this one.

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pannalal Junior | Next Rank: 30 Posts Default Avatar
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Post Tue Sep 26, 2017 11:24 pm
Matt@VeritasPrep wrote:
pannalal wrote:
(33*32*31*30 * 28!) mod 29 is not equal to (-4 * -3 * -2 * -1 * 28!) mod 29 =>

Rather, it should be (4*3*2*1*28!) mod 29.
Ah, but that isn't a mistake! It'd be a mistake to say that 33 = -4 mod 29, but that isn't what I said Wink

Of course, I could see why it would seem that way from what I typed, but hey, I'm wriggling out of this one.
Matt, somehow, I still feel that there is a mistake. If you write:
(33*32*31*30 * 28!) mod 29 = (-4 * -3 * -2 * -1 * 28!) mod 29

How do you prove that you are right. The final answer is correct because 4*3*2*1 = 24 and (-4)*(-3)*(-2)*(-1) = 24. Somehow I am not ready to accept that it is not a mistake. In any case, please state how do you arrive (33*32*31*30) mod 29 = (-4 * -3 * -2 * -1) mod 29.

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pannalal Junior | Next Rank: 30 Posts Default Avatar
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Post Tue Sep 26, 2017 9:16 pm
Matt@VeritasPrep wrote:
Every term other than 33!/29 will still have a 29 in its numerator, so every term other than 33!/29 is divisible by 29.

That means that problem can be stated as

Quote:
What is the remainder when 33!/29 is divided by 29?
(33*32*31*30*28*27*...*1) mod 29 =>

(33*32*31*30 * 28!) mod 29 =>

(-4 * -3 * -2 * -1 * 28!) mod 29 =>

(24 * 28!) mod 29

From here, I'd use Wilson's Theorem, which is not at all appropriate for the GMAT (but then again, neither is this question). Since 29 is prime, we know that 28! mod 29 = -1. That means the answer we seek is (24 * -1) mod 29, or 5.
Superb, Matt. You are wonderful and your answer is correct except one small mistake:
(33*32*31*30 * 28!) mod 29 is not equal to (-4 * -3 * -2 * -1 * 28!) mod 29 =>

Rather, it should be (4*3*2*1*28!) mod 29.

Rest is all right.

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