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09 Mar 2016, 12:32
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C
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Difficulty:
15%
(low)
Question Stats:
76% (01:08) correct
24%
(01:21)
wrong
based on 168
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If the integer N = 139,k50 where k is the hundreds digit, N can NEVER be divisible by which of the following?
A. 2
B. 3
C. 4
D. 6
E. 9
A. 2
B. 3
C. 4
D. 6
E. 9
09 Mar 2016, 23:46
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It is a problem of divisibility rules.Here is the sol.
It is divisible by 2 irrespective of k value as unit digit's value is even(0).
It can be divisible by 3 for k values 2,5,8 as sum will be divisible by 3
It can be divisible by 6 if divisible by 2 and 3.
It can be divisible by 9 for k value 2 as sum will be divisible by 9.
Therefore answer is 4 as number is divisible if last two digits are divisible.In this case 50(No)
Hence answer C
It is divisible by 2 irrespective of k value as unit digit's value is even(0).
It can be divisible by 3 for k values 2,5,8 as sum will be divisible by 3
It can be divisible by 6 if divisible by 2 and 3.
It can be divisible by 9 for k value 2 as sum will be divisible by 9.
Therefore answer is 4 as number is divisible if last two digits are divisible.In this case 50(No)
Hence answer C
14 Mar 2016, 05:28
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Apply the rule here that for a number to be divisible by 4 the last two digits must be divisible by 4
hence C is correct
hence C is correct
