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

a)168                            b)176                             c)180                            d)216

Mar 13, 2018

#1
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I don't have an answer to this exact question but at this website - http://newb.kettering.edu/wp/matholympiad/past-problems-solutions/ you may find other problems together with solutions.
This essay writing service can answer your math problems.

Mar 13, 2018
edited by EstelJordan  Mar 13, 2018
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You'd think a math Olympiad wouldn't be cheating more or less on line to look for answers. Not my place to say but it is kinda funny

Mar 13, 2018
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Here is my guess!!

Since there are 4 prime numbers between 0 - 9, therefore, there are: 4C2 =6 combinations:

23, 25, 27, 35, 37, 57 for a total sum of =5, 7, 9, 8, 10, 12. So, for the first 23, you need two of the numbers from the remaining six {0, 1, 4, 6, 8, 9} that will add to 14. And the only 2 numbers that will give you that are 6, 8. I count 28 such combinations. So: 6 x 28 =168.

And that is my guess!!.

Mar 13, 2018