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A password consists of four distinct digits such that their sum is 19 and such that exactly two of these digits are prime, for example 0397. The number of possibilities for the password is?

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    It seems like everyone just elaborated my answer. What is the $p$oint? The OP got the correct answer using my method before other answers a$p$peared. It is redundant to have duplicates.2012-11-15

4 Answers 4

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So primes here are 2, 3,5, 7.

Taking 2,3, you need 14 more to get a sum of 19. This can be done with 6,8 only.

Taking 2,5, you need 12 more to get a sum of 19. This can be done with 4,8 only.

Taking 2,7, you need 10 more to get a sum of 19. This can be done with 1,9 or 4,6 only.

Taking 3,5, you need 11 more to get a sum of 19. This cannot be done.

Taking 3,7, you need 9 more to get a sum of 19. This can be done with 1,8 or 0,9 only.

Taking 5,7, you need 7 more to get a sum of 19. This can be done 1,6 only.

That gives us 6 sets each represents $4!=24$ combinations, so there are 7*24=168 possible passwords.

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Here is a python snippet

primes = [2,3,5,7] others = [0,1,4,6,8,9] npasswords = 0 for i in primes:   for j in primes:     if i < j:       for k in others:         l = 19-i-j-k         if l < k and l >=0 and l not in primes:           print("%d %d %d %d" % (i,j,k,l))           npasswords += 24 # add !4 for all permutations print("found %d passwords" % npasswords) 

Here is the output

2 3 8 6 2 5 8 4 2 7 6 4 2 7 9 1 3 7 8 1 3 7 9 0 5 7 6 1 found 168 passwords 
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    Oh right, forgot one test, and edited code2012-11-15
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Well, we have 2, 3, 5, and 7 as our primes. Take 2 and 3. Now the problem reduces to finding $X$ and $Y$ in $23XY$ so that $2 + 3 + X + Y = 19$, $X \ne Y$, $X \ne 2$, $X \ne 3$, $Y \ne 2$, and $Y \ne 3$. You could do this by hand if you are systematic. Repeat the same process for 2 and 5, 2 and 7, etc. You could also write a Java program or ask for help at Programmers.

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    @JohnChang Yes.2012-11-15
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By hand

p1 p2 remaining other pairs 2  3     14         6,8 2  5     12         4,8  2  7     10      1,9  4,6       3  5     11          -    3  7      9      0,9   1,8   5  7      7         1,6 

So $4! \times 7 = 168$ possibilities