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

### Explanatory Answer

Let us quickly recap tests of divisibility for the 5 numbers given in the question.

Test of divisibility for 2: The number should be even. The unit digit should be one of 0, 2, 4, 6, 8.

Test of divisibility for 3: The sum of the digits should be divisible by 3.

Test of divisibility for 4: The last two digits, the tens and unit place should be divisible by 4.

Test of divisibility for 6: The number should be divisible by both 2 and 3. i.e., it should be an even number and the sum of its digits should be divisible by 3.

Test of divisibility for 9: The sum of the digits of the number should be divisible by 9.

The given number N is 139,k50 where k takes values from 0 to 9, inclusive.

- The units digit is 0. Therefore, N is divisible by 2. Choice A is
**NOT**the answer - 1 + 3 + 9 + 5 + 0 + k = 18 + k. (18 + k) will be divisible by 3 when k = 0, 3, 6, and 9. So, N may be divisible by 3. Choice B is
**NOT**the answer. - The last two digits, the tens and the units place of the number are 5 and 0. i.e., the value is 50. 50 is not divisible by 4. So, 139,k50 will NOT be divisible by 4.
**Choice C is the answer.**

Srushti Patel says

Answer is 4.