Question

What are the numbers which divide 364 leaving remainder as zero?

• A. 7
• B. 4
• C. 3
• D. 6

Answer 1

Answer: (A), (B)

The correct options are
A 7
B 4

For a three digit number abc,
100 × a + 10 × b + c can be written as
100a + 10b + c
= 98a + 7b + (2a + 3b + c)
Here, 7 is not a common factor.
Rewriting it in a way that 7 becomes a common factor,
7(14a + b) + (2a + 3b + c)
Now, to make it divisible by 7,
(2a + 3b + c) should be divisible by 7.

For, 364
Here a = 3, b = 6, c = 4,
(2a + 3b + c) = 2 × 3 + 3 × 6 + 4
= 6 + 18 + 4
= 28
Since, 28 is divisible by 7
$\therefore$ 364 is divisible by 7

364 is also divisible by 4 because the last two digit (i.e. 64) is divisible by 4.

364 is not divisible by 3 because
3 + 6 + 7 = 16 and 16 is not divisible by 3.

364 is not divisible by 6 because for a number to be divisible by 6 it is mandatory that the number is divisible by both 2 and 3.