(A) $299$
(B) $296$
(C) $298$
(D) $297$
This kind of sums are too problematic.
Please provide a method which could give the correct answer in about a minute. :)
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5xum
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Harshal Gajjar
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No, the answer is $(A)$ but it takes me quite a few minutes to figure that out :-( – achille hui Mar 27 '14 at 09:42
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2@DiffeoR, sure it can. by your logic $3$ doesn't divide $6$... – achille hui Mar 27 '14 at 09:49
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Oh right thanks ! even number will not divide odd, but not vice versa. :D – DiffeoR Mar 27 '14 at 09:55
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Is this really under a minute answer type question ? :P – DiffeoR Mar 27 '14 at 10:53
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Probably yes. No. of questions is almost equal to the no. of minutes proved. (..and the other questions were not so easy :P) – Harshal Gajjar Mar 27 '14 at 10:58
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1I think this is doable in a minute only for people trained to factorize small numbers quickly. If one know $299 = 13\times 23$, it is not hard to use this to show the number is divisible by both $13$ and $23$. Of course, if one do this systematically, one will probably try the deal with the factor $2^{?}$ in $296 = 2^3\times 37$ and $298 = 2\times149$ and the factor $3$ in $297 = 3^3\times 11$ first. Those two checks themselves probably cost one a great fraction of a minute and one doesn't have too many seconds left. – achille hui Mar 27 '14 at 13:23