How many 4 digit numbers greater than 4000 can be formed

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How many 4 digit numbers greater than 4000 can be formed using the digits from 0 to 8 such that the number is divisible by 4?

A. 508
B. 827
C. 828
D. 1034
E. 1035

The OA is D

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by SampathKp » Thu Jan 02, 2020 8:16 am
BTGmoderatorLU wrote:Source: e-GMAT

How many 4 digit numbers greater than 4000 can be formed using the digits from 0 to 8 such that the number is divisible by 4?

A. 508
B. 827
C. 828
D. 1034
E. 1035

The OA is D
Every 100 has 25 integers which are divisible by 4. 2 of the 25 i.e 92 and 96 contain 9 in them.

Hence every 100 has 23 integers which are divisible by 4 and do not contain 9 .

every thousand will have 23*9 = 207 integers that are divisible by 4 that do not contain a 9.

largest 4 digit number not containing 9 that is divisible by 4 is 8888.

We need to identify 4 digit numbers greater than 4000 and less than 8888 that divisible by 4 and do not contain a 9.

= (207*5) -1 = 1034

from 4004 to 5000 , we have 207 integers divisible by 4 that do not contain a 9.
Similarly , 5000's have 207, 6000's have 207, 7000's have 207, 8000's have 206 (1 less than others because 9000 cant be counted as it has 9)

So answer is D

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by Scott@TargetTestPrep » Sat Jan 04, 2020 7:20 pm
BTGmoderatorLU wrote:Source: e-GMAT

How many 4 digit numbers greater than 4000 can be formed using the digits from 0 to 8 such that the number is divisible by 4?

A. 508
B. 827
C. 828
D. 1034
E. 1035

The OA is D

Recall that, in order for a number to be divisible by 4, the last two digits of the number have to be divisible by 4. So the last two digits of the number must be one of the following:

00, 04, 08, 12, 16, ..., 80, 84, 88, 92, and 96.

However, since we can't use the digit 9, we have to omit 92 and 96. This still leaves us with
(88 - 0)/4 + 1 = 23 possibilities for the last two digits of the number.

For the first two digits of the number, it must be one of the following:

40, 41, 42, ..., 48, 50, 51, 52, ... , 58, ..., 80, 81, 82, ..., 88

We see that the first two digits of the numbers have 5 x 9 = 45 possibilities.

Therefore, there are 45 x 23 = 1035 such numbers. However, these 1035 numbers include the number 4000; the problem says the number must be greater than 4000. So we have to remove 4000 from the list and this leaves us with 1034 numbers.

Answer: D

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