How many five-digit numbers can be formed from the digits 0,

A number to be divisible by 4 its last two digit must be divisible by 4 (similarly a number to be divisible by 2 its last digit must be divisible by 2; to be divisible by 8, last three digits must be divisible by 8 and so on).

Thus the last two digit must be 04, 12, 20, 24, 32, 40, or 52.

If the last two digits are 04, 20, or 40, the first three digits can take 4*3*2= 24 values.
Total for this case: 24*3 = 72.

If the last two digits are 12, 24, 32, or 52, the first three digits can take 3*3*2= 18 values (that's because the first digit in this case cannot be 0, thus we are left only with 3 options for it not 4, as in previous case).
Total for this case: 18*4 = 72.

Grand total 72 +72 =144.

Answer: E.
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My approach:
Total of 600 possible number: 5x5x4x3x2x1 (zero can't be the 1st digit)
Of those 600 number, 300 are even, because you have 3 odd and 3 even numbers. Of those 300 even numbers, about half are multiple of 4. So answer choice E.

Similar question to practice: how-many-five-digit-numbers-can-be-formed-using-digits-91597.html
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Dear Bunnel,

i dont understand the highlighted part... when the last 2 digits are 04, 20 or 40 then we have only 3 digits left for the 1st 3 to choose from... as 0,2 or 4 are gone???

same for the next highlighted part.

please explain.

Yes, we are told that no digits can repeat. So, if the last two digits are 04, 20, or 40, we are left with 4 digits for the first, 3 digits for the second and 2 digits for the third one. The same applies to the second case.

Hope it's clear.
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no what i mean is that we have 0, 1, 2, 3, 4, 5 to choose from. so, if the last two digits are 04, 20, or 40... then 3 of the 6 digits to choose from are gone. we will be left with 2,3,5??? how 4?

For EACH case of 04, 20, or 40 we used 2 digits and we are left with 4. Isn't it?
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To be divisible by four the number needs to end in 04, 40, 20, 12, 32 or 52.

Now then, we have 4 numbers remaining and 3 slots. 4C3 * 3! ways to order them.
We do this for each combination so: 4C3*3!*6=24*6=144

Answer: E

Hope this helps

You have 6 digits and you need a 5 digit number.

In how many ways can the number be divisible by 4?
If it ends with 04 or 12 or 20 or 24 or 32 or 40 or 52, it will be divisible by 4.

All 3 combinations that have a 0 as one of the last two digits can be formed by using basic counting principle.
4 * 3 * 2= 24. (First digit in 4 ways, second in3 ways and third in 2 ways)
Since there are 3 such combinations, you get 24*3 = 72

The other 4 combinations which do not have a 0 in the last two digits can be formed by 3 * 3 * 2 = 18
(First digit in 3 ways (no 0), second in 3 ways (leftover 2 digits and 0) and third in 2 ways.
Since there are 4 such combinations, you get 18*4 = 72

Total number of ways = 72 + 72 = 144
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Sir what i dont understand is how can the 3 numbers be arranged in 4 ways in case 2? (highlighted text)

Thanks

In that case, the last two digit can take 4 different values: 12, 24, 32, or 52. The first three digits can take 18 values. Total = 4*18.
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To be divisible by 4, the last two digits of the number must be divisible by 4. Therefore, they can be 04, 12, 20, 24, 32, 40, and 52. We can split these into two groups: 1) 04, 20, 40, and 2) 12, 24, 32, 52

Group 1:

If the last two digits are 04, then there are 4 choices for the first (or ten-thousands) digit, 3 choices for the second (or thousands) digit, and 2 choices for the third (or hundreds) digit. So we have 4 x 3 x 2 = 24 such numbers if the last two digits are 04. Also there should be 24 numbers if the last two digits are 20 or 40. So we have 24 x 3 = 72 numbers in this group.

Group 2:

If the last two digits are 12, then there are 3 choices for the first (or ten-thousands) digit (since it can’t be 0), 3 choices for the second (or thousands) digit, and 2 choices for the third (or hundreds) digit. So we have 3 x 3 x 2 = 18 such numbers if the last two digits are 12. Also there should be 18 numbers if the last two digits are 24, 32 or 52. So we have 18 x 4 = 72 numbers in this group also.

Therefore, there are a total of 72 + 72 = 144 numbers.

Answer: E
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By far the BEST solution I have seen, given the time limit GMAT has, other solutions are not doable.

OE:
The digits given are 0,1,2,3,4, and 5.

The divisibility rule of 4 says, if the last two digits are divisible by 4, the number is divisible by 4. So, the last two digits can be 04, 12, 20, 24, 32, 40, 52.

If the last two digits are 04 or 20 or 40, then the number of 5-digit numbers possible for each condition is = 4 * 3 * 2 = 24, since all four remaining digits are possible for the first number.

Since this applies for the cases 04, 20, and 40, the total number of 5 digit numbers possible is 3 * 24 = 72 numbers.

Looking at the rest of the possibilities, we cannot use 0 as the first digit, so we will treat these numbers differently. If the last two digits are 12 or 24 or 32 or 52, then the number of 5 digits number possible for each condition is = 3 * 3 * 2 = 18.

Hence the total number of 5 digit number possible = 18*4+72 =144. Thus, the correct answer is E.

How many five-digit numbers can be formed from the digits 0, 1, 2, 3, 4, and 5, if no digits can repeat and the number must be divisible by 4?

For a no. to be divisible by 4, last two digits must be divisible by 4
We will take one by the pairs such that those are last two digits of the five digit number as required and are divisible by 4

1) 04 Total no.s = 4*3*2 = 24
2) 12 Total no.s = 2*2*2*3 = 24
3) 20 Total no.s = 4*3*2 = 24
4) 32 Total no.s = 2*2*2*3 = 24
5) 40 Total no.s = 4*3*2 = 24
6) 52 Total no.s = 2*2*2*3 = 24

Answer = 144

Hence E