Bunuel wrote:
\(N = \frac{a!×b!×c!×d!}{e!}\) where a, b, c, and d, are four distinct positive integers, which are greater than 1, and a, e are two consecutive numbers, which are prime. What is the least possible value of \(\frac{a+b+c+d}{e}\), if N is divisible by cube of product of the three smallest odd prime numbers?
A.26/3
B. 9
C. 21/2
D. 13
E. 27/2
Breaking Down the Info:>
a, e are two consecutive numbers, which are primeThe only consecutive primes are 2 and 3, thus a and e are 2 and 3 (we don't know which is which, however).
>
N is divisible by cube of product of the three smallest odd prime numbersThe three smallest odd primes are 3, 5, and 7. The product of 3, 5, and 7 is 105. The cube of this product is \(3^3*5^3*7^3\).
We really only need to care about the \(7^3\), however, as the rest of the lower factors are automatically in 7!.
We need three multiples of 7, and since the letters must be distinct, we can use \(7!*8!*9!\) for three multiples of 7. (Another interesting solution is 7!*14!*3! which we'll talk about later)
> Finally, we should let a = 2 and e = 3 to minimize the fraction.
Then \(N = \frac{2! * 7! * 8! * 9!}{3!}\).
\(\frac{a+b+c+d}{e} = \frac{26 }{ 3}\)
Answer: AThere is actually another solution to this, which is to let \(N = \frac{2!*7!*14!*3!}{3!}\). For the d in the numerator, we must have a positive integer greater than 1, and 2 is taken so our minimal value for d in this scenario is 3. We get the same answer either way, but with different conditions, we would be able to get a smaller fraction with this method.
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Jerry | Dream Score Tutor
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