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by navdeepbajwa » Tue Oct 27, 2009 10:00 pm
If k is an integer, and 35^2-1/k is an integer, then k could be each of the following, EXCEPT

(A) 8(B) 9(C) 12(D) 16(E) 17
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by gopalkrishna.r » Tue Oct 27, 2009 10:59 pm
(D) 16 is the answer

How did I get this? 35^2 = 1225 (easy way is using vedic maths)..any number X5^2 would be X(X+1)25... so in this case, 3*4 25 which is 1225.

1225-1 = 1224

Now, actual problem is to check which of the given numbers is exactly divisible by 1224. From divisibility tests, 16 does not pass. 1224/16 = 72.5 and so it is the answer.
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by Ian Stewart » Wed Oct 28, 2009 5:19 am
Or you can factorize 35^2 - 1; it's a difference of squares:

35^2 - 1 = (35 + 1)(35 - 1)
= 36*34
= 2^2 * 3^2 * 2 * 17
= (2^3)(3^2)(17)

from which we can see that 2^3 = 8, 3^2 = 9, (2^2)(3) = 12 and 17 are all divisors of 35^2 - 1, whereas 2^4 = 16 is not.
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by mp2437 » Wed Oct 28, 2009 5:52 am
gopal, thanks for the little trick, never knew about that!

Nonetheless, still requires much work. I like Ian's style.
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by navdeepbajwa » Wed Oct 28, 2009 9:57 am
Thanks Ian
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