+0

+1
1498
3
+1713

How many different 4-digit numbers can be formed using the digits 2, 4, 5, 6, and 7 such that no digits repeat and the number is divisible by four?

I got the answer 36, but how would you do it quickly - I wrote them all out...

-\(tommarvoloriddle\)

Edit: I mean, how would you do it without writing it out? It took me,  let's say, around 45 minutes One hour One hour and a half A VERY LONG TIME to solve this trying to write everything out.

I wrote all the 4-digit numbers out.

I typed them out... :\

Jul 22, 2019
edited by tommarvoloriddle  Jul 22, 2019
edited by tommarvoloriddle  Jul 22, 2019
edited by tommarvoloriddle  Jul 22, 2019
edited by tommarvoloriddle  Aug 14, 2019

#1
+3

That is right! That is what I got by having the the computer arrange them in order and also make sure that all are divisible by 4. You have to have the ability to program the computer in some language such as Python, Java, C.......etc. Here is the list:

(2456, 2476, 2564, 2576, 2756, 2764, 4256, 4276, 4572, 4576, 4652, 4672, 4752, 4756, 5264, 5276, 5472, 5476, 5624, 5672, 5724, 5764, 6452, 6472, 6524, 6572, 6724, 6752, 7256, 7264, 7452, 7456, 7524, 7564, 7624, 7652) = 36 - numbers and all are divisible by 4.

Jul 22, 2019
#2
+2

No, it shouldn't have taken you that long! If you had made a list of the 2 ending digits that are divisible by 4, you should have been able to arrange them by hand in a matter of a few minutes. Here are the 2-digit endings divisible by 4: 24, 52, 56, 64, 72, 76. As you can see, there are only 6 of them. Each one of these 6 has 6 combinations with the remaining digits, so that: 6 x 6 = 36

Jul 22, 2019
#3
+37057
+1

If the number ends in 24   44  52  56  64  72  76   it is divisible by 4  (throw out 44 ....can only use number onece) so 6 choices

Then the first number can be one of the other 5    i.e.  3 choices

the second number then can be    one of 2   choices

6 x 3 x 2 = 36 possibles

Jul 22, 2019