# How many four digit numbers can formed using digits 0 to 7 which is divisible by 4 without repetition?

How many four digit numbers can formed using digits 0 to 7 which is divisible by 4 without repetition?

We can use digits from 0 to 7only. Following are the possibilities.

04, 12, 16, 20, 24, 32, 36, 40, 52, 56, 60, 62, 64 - total 13

A number can’t start with zero, therefore cases above having a 0, will have to be dealt separately.

04,20,40,60 - 4possibilities

For first 2 digits we can use product rule to get 6 × 5=30. Because, out of 8digits 2 have been already used up. So, in all, there are 30×4=120possibilities.

For the remaining 9 cases, the first 2digits can be chosen in 5×5=25 Because, out of remaining 6digits, we can’t choose 0 as the starting digit. So, there are 25 ×9=225

So, combining the above cases, we get the answer to be 120+225=345

i dont understand what i am unable to get here. help me to sort out this.

• What is your question really? – RGS Nov 29 '16 at 17:48
• i am asking wheather i am wrong or right? – Riyas Randhava Nov 29 '16 at 17:50
• Note that $62$ is not divisible by $4$. – Joffan Nov 29 '16 at 17:56
• Um... those are two digit numbers; not 4. 62 is not divisible by 4. 20,60, 80 do not start with 0. xx04 doesn't start with 0 either. – fleablood Nov 29 '16 at 18:00
• 62 is out and 72 and 76 are in. 14. So the answer is 4x30 + 10*25 = 370. – fleablood Nov 29 '16 at 18:23

There are 4 even number and 4 odd. 2 are divisible by 4 (0 and 4) and 2 are even not divisible by 4 (2 and 6).

100 is divisible by4 so $xxAB$ is divisible by 4 iff $AB$ is divisible by 4.

$AB$ is divisible by $4$ if i)$B$ is divisible by 4 and $A$ is even. $2*(4-1) = 6$ possibilities for this. or ii)$B$ is even but not divisible by 4 and $A$ is odd. $2*4 = 8$ possibilities for this.

So there are $14$ possibilities of $AB$. $cdAB$ may not have $c = 0$. If $AB$ are both not $0$ then there are $5$ choices for $c$ and $5$ for $d$ so $25$ choices. If $A$ or $B$ is $0$ there are six choices for $c$ and $5$ for $d$ so $30$ choices.

If $B = 0$ there are $3$ even numbers that $A$ can be. If $A=0$ then $B$ is one number divisible by $4$ that $B$ can be. So there are $4$ choices for $AB$ with a zero and $10$ choices where neither are zero.

So there are $4*30+10*25 = 370$ total choices.

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The only thing you did wrong was i) put 62 in your list ii) leave 72 and 76 out of the list.