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## Remainder from Sum containing Factorial

This topic has 2 expert replies and 3 member replies
pannalal Junior | Next Rank: 30 Posts
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#### Remainder from Sum containing Factorial

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.

pannalal Junior | Next Rank: 30 Posts
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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

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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Matt@VeritasPrep GMAT Instructor
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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

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
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Sun Sep 24, 2017 6:54 pm
As nobody has posted any solution, I am giving first hint.

Hint 1: Please note that S = (33!/1 + 33!/2 + 33!/3 + ..... + 33!/32 + 33!/33).
When you divide each term by 29, you will find that every term is divisible by 29 except 33!/29. For example 33!/1/29 is divisible by 29. Similarly, 33!/2/29 is also divisible by 29. The only term which is not divisible is 33!/29 because it has already 29 as denominator. So, practically speaking, you need to find remainder when (33!/29) is divided by 29. So, now, instead of 33 terms, you are dealing with only one term.

Note: I hope, many viewers will be able to find solution with the above hint. If I don't get solution in another 12 hours, I shall post second hint.

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Matt@VeritasPrep GMAT Instructor
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Tue Sep 26, 2017 8:15 pm
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.

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pannalal Junior | Next Rank: 30 Posts
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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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