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25 Aug 2008, 01:21
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The numbers from 1 to 37 are written in a line from left to right say
n1,n2,......n37 such that each number divides the sum of all its previous numbers. If n1=37 and n2=1 find the value of n3-n37.
1) 16 2)-16 3)17 4)15 5)None of these
n1,n2,......n37 such that each number divides the sum of all its previous numbers. If n1=37 and n2=1 find the value of n3-n37.
1) 16 2)-16 3)17 4)15 5)None of these
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25 Aug 2008, 02:39
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we know that the last number should be a factor all the numbers summed up i.e sum(1... 37) - n37
sum(1... 37) = (37*38)/2
= 37 * 19
= 703
we know that 19 is a factor of 703.
but if 19 is the last number, the sum of the previous numbers = 703-19 (which is still a multiple of 19), so the guess that 19 is a still a potential for n37.
note that 703 is not a multiple of 2 or 3, therefore, they cannot n37.
and no factors of 2 or 3 can be n37
similarly 5/7/11 cannot be n37 (at this point i stopped checking for prime numbers, but ideally should continue till 19)
we know n1 = 37
n2 = 1
therefore, n1 + n2 = 38
only factors of this are 2 and 19.
since 19 is already n37
=> n3 = 2
n3 - n37 = 2 - 19
= -17
Hence None of these
E)
OA plz
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