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Known the digits $a,b,c$, and $d$. If multiplied all ($abcd$), the results will be 960. If all of the digits have a different value from each other, find the value of $a + b + c+d$

I am practicing for a contest and I need some help solving this question. Any hints or solutions will certainly help me. Thank you!

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    You could look at the prime decomposition of 960 and look what possibilities this gives you for $a,b,c,d$... – 2017-02-04
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    I wish that downvoter leave a comment out.. – 2017-02-04
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    Probably since you did not show what you have tried... – 2017-02-04
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    @Student Should I post what I have guessed? This is not one of the types where you have to show your work. – 2017-02-04
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    I guess you did try something? And its better if you include/ give a general idea of what you did. This makes it easier to help you or point out some of your mistakes. In this case and in general when you have number questions whit relatively small numbers and you don't see where to start, prime number decompositions can be helpfull and are not hard to compute :) – 2017-02-04
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    However, IMHO people tend to downvote pretty quick, and I Found this question interesting myself, so +1 – 2017-02-04
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    @Student Ok, and thanks from the reply! – 2017-02-04

1 Answers 1

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$960 = 2^6 \cdot 3 \cdot 5$

We know that $a,b,c,d$ are digits, i.e. they are all $<10$. So one of the digits has to be $5$ (we need a multiple of $5$ to make the product of 960, but any multiple other than 5 itself is larger than 10)

Another digit has to be either 3 or 6 (same reasoning). If it was 3, then we are left with $2^6$, which has to be split up to make 2 numbers smaller than 10 that are not equal. This is not possible (neither can be $2^4=16$ or higher since that is not smaller than 10, meaning that they both would have to be equal to $2^3$, which is not allowed since they have to be different). Thus the second digit has to be a 6.

We are left with $2^5$ out of which we have to make two numbers smaller than 10. The only way to do so is to set one equal to $2^2=4$ and one equal to $2^3=8$.

Our digits are thus 4, 5, 6 and 8, the sum of which equals 23.

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    Nice explanation +1 and accepted! – 2017-02-04