# I want to make sure I did this problem correctly before I add it to the problems section

How many sets of a, b, c, d, e are possible if:
ac + ae + bd + bc + ad + be + 17 = 40
And if a, b, c, d, e are all integers greater than or equal to zero?

So subtracting 17 and factoring yields
(a+b)(c+d+e)=23
since 23 is prime, its only factors are 23 and 1
if a+b=1 then c+d+e=23
and if a+b=23 then c+d+e=1

the first case, a and b can add up to 1 in 2 ways, or 2C1 ways and c d and e can add up to 23 in 25C2 ways or 300 ways.
Therefore, there would be 2*300 or 600 ways for this to happen

the second case, a and b can add up to 23 in 24C1 or 24 ways and c d and e can add up to 1 in 3C1 ways or 3 ways.
Therefore, there would be 24*3 or 72 ways for this to happen

Can someone tell me whether this is done correctly? Thanks, Note by A Former Brilliant Member
1 year, 7 months ago

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Yup. Totally correct. Well done!

- 1 year, 7 months ago