Digit problem?

Question

If the digits 2,3,7 and 6 are arranged randomly so as to form a four-digit number,then how many numbers will be formed that will be divisible by 4 (repetition of digit is not allowed)?
a)4
b)6
c)8
d)12

which one is correct ?

Answer

Let “abcd” be the four digit number where, a,b,c,d are just some permutations of the digits 2,3,7,6.

Obviously, 3 and 7 CANNOT be the last digit for if it were so, the resulting four digit number ending in 3 or 7 will be odd and thus NOT divisible by 4. This means that d is either 2 or 4 only, d = 2 or 4.

Note, writing in base 10,

“abcd” = 1000a + 100b + 10c + d

“abcd” = 4 x ( 250a + 25b ) + 10c + d

This shows, for a four digit number to be divisible by 4, It suffices for the LAST TWO DIGITS BE DIVISIBLE BY FOUR!

That is we just need to find the number of “cd” such that “cd” is divisible by 4.

Note, d = 2 or 4 only.

If d = 2, we need “c2” that are divisible by 4.
Possibilities for “c2” = 32, 72, 62. Among these 32 and 72 are divisible by four. So, if d=2, the four digit number is either “ab32” or “ab72” and is divisible by four regardless of the digits a and b. In both cases, permutation for “ab” is just 2! = 2 ways, for “ab” are chosen from {6,7} for “ab32” and {3,6} for “ab72”. So the total number of four digit numbers in this case that are divisible by four is just
= 2 x 2 = 4 numbers.

If d = 6, we need “c6” that are divisible by 4.
Possibilities for “c6” = 36, 76, 26. Among these 36 and 76 are divisible by four. So, if d=6, the four digit number is either “ab36” or “ab76” and is divisible by four regardless of the digits a and b. In both cases, permutation for “ab” is just 2! = 2 ways, for “ab” are chose from {2,7} for “ab36” and {2,3} for “ab76”. So the total number of four digit numbers in this case that are divisible by four is just = 2 x 2 = 4 numbers.

Hence, total numbers that you seek is 4 + 4 = 8.

DONE.

Remark.
I wouldn’t encourage you “finding” by brute force all the possibilities and then deciding which are divisible by four because I suggest a more “theoretical” way. Anyway, in this case, the number of possibilities is small, thus here are the 8 four digit numbers of your problem that is divisible by four 2376, 3276, 2736, 7236, 6732, 7632, 2672, 6272.