4-digit numbers divisible by 4

[euro]

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]

[kvcpk]

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!!

[euro]

Great work. thanks for the detailed explanation. It is really helpful.

[sixpointer]

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!!

number of last two digit possible 22

A can be filled in 7( as it cant take zero)

B can take value of zero So, 7 ways

7*7*22=1078

I got this where Iam doing mistake?

[shovan85]

number of last two digit possible 22

A can be filled in 7( as it cant take zero)

B can take value of zero So, 7 ways

7*7*22=1078

I got this where Iam doing mistake?

See the 22 has 6 options which include zero (04,08,20,40,60,80). So when you have last two digits out of these 6 choices total number of numbers = 6*7*8 = 336
Rest 22-6 = 16 option which do not include 0 total number of numbers = 16*7*7 = 784

Thus total = 784+336 = 1120.

You cannot take only 7*7*22 as 22 has two parts: one with zeros (6 options) and other without zero (16 options)

[kvcpk]

number of last two digit possible 22

A can be filled in 7( as it cant take zero)

B can take value of zero So, 7 ways

7*7*22=1078

I got this where Iam doing mistake?

We cant approach it that ways because, 0 plays a tricky role here. We need to make sure that 0 is not repeated.

Let me put this in more clear terms:
Two main cases Arise:
Zero included in CD = 6 cases
A can be filled in 8 ways and B can be filled in 7 ways. = 56
Zero not included in CD = 16 cases
B is Zero
A can be filled in 7 ways = 7
B is not Zero
A can be filled in 7 ways and B can be filled in 6 ways, bacause B cant be 0,A,C,D = 42
6*56 + 16*7 + 16*42
= 336 + 112 + 672
= 1120

Hope this helps!!