Prime Factory
a + b + c + d + a b c d = 2 0 1 8 a , b , c , d are primes(not necessarily distinct),and a ≤ b ≤ c ≤ d .How many solutions of ( a , b , c , d ) are there?
Try my set here
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3 solutions
Do you notice that a,b,c,d are not necessarily distinct,because a ≤ b ≤ c ≤ d ?
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Yup, totally missed that! Will re-think when I get a chance.
If there are 0,1,or 3 even numbers in a , b , c , d ,then a + b + c + d + a b c d will be odd.
If there are 4 even numbers, 2 + 2 + 2 + 2 + 1 6 = 2 4 = 2 0 1 8 .
So there are 2 even numbers,that means a = b = 2 , c + d + 4 c d = 2 0 1 4 .
2 0 1 4 ≡ 1 ( m o d 3 )
If c ≡ d ≡ 1 ( m o d 3 ) , c + d + 4 c d ≡ 0 ( m o d 3 ) .
If c ≡ d ≡ 2 , c + d + 4 c d ≡ 2 ( m o d 3 )
If c ≡ 2 , d ≡ 1 ( m o d 3 ) or c ≡ 1 , d ≡ 2 ( m o d 3 ) , c + d + 4 c d ≡ 2 ( m o d 3 ) .
So c = 3 ,solve for d and get d is not a prime number.
The only prime factors of 2018 are 2 and 1009. Easy to find out that it won't work :)
Did you factorize a + b + c + d + a b c d ?
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Hmmm I reread the problem. Thought it was about prime factors of 2018. I now realize that would be trivial haha. Never mind my solution.
A parity argument suffices here. If 2 is one of the primes, then a + b + c + d will be odd, but a b c d will be even; if 2 is not one of the primes, then a + b + c + d will be even, but a b c d will be odd. Neither possibility can give an even answer.