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Given the number ABCD, how many numbers are there given that A+B+C+D is a multiple of 10?

I was thinking of using mutually exclusive cases:

ABCD are all the same number

then its X+X+X+X

so that means there are 9 combinations

ABCD have two repeating letters

This is where it kind of got tricky for me because while even if I figure out the combinations of it, the permutations are different

But if I was to continue this process it would just be Brute Forcing

I was wondering if there was a more effective way


The Failed Brute Force Method I Used

list 1-9 for the letter A

1

     There are 90 combinations for A=1

     A|B|C|D-> 1|9|x| all values of x works, same when B is 8 and below

.

.

.

9

For A=1 I found that all values of B works

For A=2, I found that all values of B also works and on and on but I know that there are not 810 combinations that exists so this didn't work

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  • $\begingroup$ ......Somewhat OT ..... Is that pic a Shiba Inu $\endgroup$ – Peter May 16 '17 at 12:53
  • $\begingroup$ How do you get $90$ combinations for $A=1$. Did you forget $x=0$? In particular, for $B=9$, $1900$, $1919$, $1928$, $1937$, $1946$, $1955$, $1964$, $1973$, $1982$, and $1991$. $\endgroup$ – Michael Burr May 16 '17 at 14:02
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Hints:

  • If you choose $A$, $B$, and $C$, then there is a unique choice for $D$ that makes $A+B+C+D$ divisible by $10$.

  • Therefore, since there are $9$ choices for $A$ (assuming no leading digit of $0$), $10$ choices for $B$, $10$ choices for $C$, and $D$ is uniquely determined, the answer appears to be

    900

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  • $\begingroup$ but wouldn't that mean you would have to brute force by listing 1-9 for A and list 10 potential numbers for B and same with C? Or is there a better method? $\endgroup$ – John Rawls May 16 '17 at 12:40
  • $\begingroup$ No, you can count the number without listing them. You know what the possibilities for $A$, $B$, $C$, and $D$ are (but you don't need to write down the $ABCD$'s to know the possibilities.) $\endgroup$ – Michael Burr May 16 '17 at 13:01
  • $\begingroup$ In many cases, you can count items faster than you can list them. It's often a trick in discrete math/computer science that you can compute the number of objects through a simple computation, even when there are a lot of them. For example, how many positive numbers are less than or equal to $1,000,000$. You can calculate this directly (it's $1,000,000$) without listing all of them. $\endgroup$ – Michael Burr May 16 '17 at 14:07

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