Hard quant

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Hard quant

by jsl » Mon Oct 27, 2008 2:59 pm
If N is an integer, What is the units' digit of (N-1)! + N! + (N+1)! +2N*3N+1?

(1) N is greater than 1
(2) N = 4K + 2, where K is an integer equal to or greater than 1

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by cramya » Mon Oct 27, 2008 3:54 pm
Is the answer B)

OA?

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by cramya » Mon Oct 27, 2008 4:09 pm
Sorry, I think I was wrong. Even if I was right I dont have a generic approach.

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Re: Hard quant

by Ian Stewart » Mon Oct 27, 2008 4:29 pm
jsl wrote:If N is an integer, What is the units' digit of (N-1)! + N! + (N+1)! +2N*3N+1?

(1) N is greater than 1
(2) N = 4K + 2, where K is an integer equal to or greater than 1
It's not strictly necessary, but I'd start by factoring out (N-1)! from the factorials:

(N-1)! + N! + (N+1)! +2N*3N+1 = (N-1)!*(1 + N + N*(N+1)) + 2N*3N + 1

Now, if N > 5, (N-1)! will be at least 5!, so will be divisible by 5 and by 2, and so will end in 0. That means, if we know N > 5, the units digit of (N-1)! is 0, so the units digit of (N-1)!*(1 + N + N*(N+1)) will also be 0 (the units digit of a product comes from the product of the units digits). If N > 5, we then only care about the 2N*3N + 1 = 6N^2 + 1, since the units digit of the sum of the other terms must be zero. However, if N = 6, this will end in 7, while if N = 10, this will end in 1. Since even using both statements together, N could be 6 and could be 10, the statements are insufficient together, and the answer is E.

I'd add that at first, reading the question, I wondered whether it was meant to read 2N*(3N + 1) at the end, but it makes little difference to the solution in either case.
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by cramya » Mon Oct 27, 2008 5:20 pm
Ian,
Thanks!

I was getting 0 as the units for all factorials individually and collectively as a sum but different units digit for 6,10 and 14 (4k+2) for the expression 6*N^2+1. I intitially said B because I made a calculation mistake and then decided E.



Thanks for the confirmation! Your explanations are always good!

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by deepamohn » Mon Oct 27, 2008 6:31 pm
Is the answer E?

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by cramya » Mon Oct 27, 2008 6:32 pm
Yes

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by cramya » Mon Oct 27, 2008 6:35 pm
Yes; it should be per Ian.

JSL,
In the future if u could post the OA using the SPOILER function it would be very much appreciated.

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by daphyy » Wed Nov 12, 2008 6:43 am
Actually I think the question might have been 2^N*3^(N+1) towards the end. So I got 3*(6^N), whose units' digit is always gonna be 8 so the answer should be B. :)