euro wrote:How many 4-digit numbers divisible by 4 can be formed by using the digits 0-9, so that no two digits are repeated?
(A) 336
(B) 784
(C) 1120
(D) 1804
(E) 1936
[spoiler]Don't know the OA. I am getting (C)? [/spoiler]
A 4 digit number is divisible by 4, only if the last 2 digits are dvisible by 4, or when the last 2 digits are zeroes.
But, since the digits cant repeat, last 2 digits cant be zeroes.
Hence last 2 digits should be divisible by 4.
ABCD
CD should be divisible by 4. Hence there will be 100/4 = 24 possible multiples of 4, excluding 100. OF these 44,88 have digits repeated.
Hence 22 possible combinations for CD.
A can be fille din 8 ways [including 0]
B can be filled in 7 ways.
Hence possibilities = 22*8*7 = 1232
These contian numbers that include zeroes in the beginning.
So the answer will be slightly smaller than this count.
pick C.
To perform the calculation, we need to write out the terms and see how many have zero included in the last 2 places.
04,08,20,40,60,80 = 6 terms
So there are 16 endings that do not end in zero.
There are 7 possible combinations for B, without 0.
Hence, 16*7 = 112 combinations must be reduced from the calculated value.
1232 - 112 = 1120.
pick C.
Hope this helps!!
