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Remainder from Sum containing Factorial

This topic has 2 expert replies and 3 member replies

Remainder from Sum containing Factorial

Post Sun Sep 24, 2017 5:50 am
Elapsed Time: 00:00
  • Lap #[LAPCOUNT] ([LAPTIME])
    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 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.

    GMAT/MBA Expert

    Post 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.

    Thanked by: pannalal
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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.

    GMAT/MBA Expert

    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.

    Thanked by: pannalal
    Enroll in a Veritas Prep GMAT class completely for FREE. Wondering if a GMAT course is right for you? Attend the first class session of an actual GMAT course, either in-person or live online, and see for yourself why so many students choose to work with Veritas Prep. Find a class now!
    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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