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EgmatQuantExpert
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Question of the week - 29 (N = a! * b! * c! * d!/e! ................) Updated on: 27 Feb 2019, 21:43
Originally posted by EgmatQuantExpert on 28 Dec 2018, 04:02.
Last edited by EgmatQuantExpert on 27 Feb 2019, 21:43, edited 1 time in total.
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A
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!

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75% (hard)

Question Stats:

63% (03:19) correct 37% (03:11) wrong based on 495 sessions

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e-GMAT Question of the Week #29


\(N = \frac{a! * b! * c! * d!}{e!}\) 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. \(\frac{26}{3}\)

    B. 9

    C. \(\frac{21}{2}\)

    D. 13

    E. \(\frac{27}{2}\)

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Question of the week - 29 (N = a! * b! * c! * d!/e! ................) 02 Jan 2019, 02:07
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Solution


Given:
    • \(N = \frac{a! * b! * c! * d!}{e!}\)
    • a, b, c and d are four distinct positive integers, which are greater than 1
    • a, e are two consecutive numbers, which are prime
    • N is divisible by cube of product of the three smallest odd primes

To find:
    • The least possible value of \(\frac{(a + b + c + d)}{e}\)

Approach and Working:
    • (a, e) = (2. 3). Since, (2, 3) is the only pair of numbers, which are prime

And, we are given that N is divisible by \(3^3 * 5^3 * 7^3\)
    • Thus, b, c and d must contain at least one 7, one 5 and one 3 each

So, the minimum value that (b, c, d) can take is (7, 8 , 9), since, 7! 8! and 9! Covers the powers f all prime numbers
    • Therefore, the minimum value of \(\frac{(a + b + c + d)}{e}\) will be when a = 2 and e = 3
      o \(\frac{(2 + 7 + 8 + 9)}{3} = \frac{26}{3}\)

Hence the correct answer is Option A.

Answer: A

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Question of the week - 29 (N = a! * b! * c! * d!/e! ................) 28 Dec 2018, 05:13
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given a * e are two consecutive prime integers which can only be 2 & 3 let a = 2 and e= 3

since we need to find least value of \(\frac{(a + b + c + d)}{e}\) , if N is divisible by cube of product of three smallest odd prime no. which would be 3,5,7

so
\(N = \frac{a! * b! * c! * d!}{e! *(3*5*7)^3}\) should be divisible

for which value of b=7!, c=8!, d=9! would suffice the relation

so least value of \(\frac{(a + b + c + d)}{e}\)
(2+7+8+9)/3 = 26/3 IMO A


EgmatQuantExpert
\(N = \frac{a! * b! * c! * d!}{e!}\) 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. \(\frac{26}{3}\)

    B. 9

    C. \(\frac{21}{2}\)

    D. 13

    E. \(\frac{27}{2}\)

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Question of the week - 29 (N = a! * b! * c! * d!/e! ................) 29 Jan 2019, 17:03
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Solution


Given:
    • \(N = \frac{a! * b! * c! * d!}{e!}\)
    • a, b, c and d are four distinct positive integers, which are greater than 1
    • a, e are two consecutive numbers, which are prime
    • N is divisible by cube of product of the three smallest odd primes

To find:
    • The least possible value of \(\frac{(a + b + c + d)}{e}\)

Approach and Working:
    • (a, e) = (2. 3). Since, (2, 3) is the only pair of numbers, which are prime

And, we are given that N is divisible by \(3^3 * 5^3 * 7^3\)
    • Thus, b, c and d must contain at least one 7, one 5 and one 3 each

So, the minimum value that (b, c, d) can take is (7, 8 , 9), since, 7! 8! and 9! Covers the powers f all prime numbers
    • Therefore, the minimum value of \(\frac{(a + b + c + d)}{e}\) will be when a = 2 and e = 3
      o \(\frac{(2 + 7 + 8 + 9)}{3} = \frac{26}{3}\)

Hence the correct answer is Option A.

Answer: A

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Hello EgmatQuantExpert !

Could you please explain to me why are we choosing 7, 8 and 9, respectively?

Thank you in advance!
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Question of the week - 29 (N = a! * b! * c! * d!/e! ................) 29 Jan 2019, 22:17
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Solution


Given:
    • \(N = \frac{a! * b! * c! * d!}{e!}\)
    • a, b, c and d are four distinct positive integers, which are greater than 1
    • a, e are two consecutive numbers, which are prime
    • N is divisible by cube of product of the three smallest odd primes

To find:
    • The least possible value of \(\frac{(a + b + c + d)}{e}\)

Approach and Working:
    • (a, e) = (2. 3). Since, (2, 3) is the only pair of numbers, which are prime

And, we are given that N is divisible by \(3^3 * 5^3 * 7^3\)
    • Thus, b, c and d must contain at least one 7, one 5 and one 3 each

So, the minimum value that (b, c, d) can take is (7, 8 , 9), since, 7! 8! and 9! Covers the powers f all prime numbers
    • Therefore, the minimum value of \(\frac{(a + b + c + d)}{e}\) will be when a = 2 and e = 3
      o \(\frac{(2 + 7 + 8 + 9)}{3} = \frac{26}{3}\)

Hence the correct answer is Option A.

Answer: A


Hello EgmatQuantExpert !

Could you please explain to me why are we choosing 7, 8 and 9, respectively?

Thank you in advance!
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Because for \(N\) to be divisible by \(3^3*5^3*7^3\), the numerator \(a!*b!*c!*d!\) has to be divisible by \(3^3*5^3*7^3\) and for this to happen the numerator should have atleast three instances of 3, 5 and 7 each. Therefore, we considered the numbers 7, 8 and 9 because they are the least numbers that when in factorial form will give you three instances of 7 and the instances of 3 and 5 will follow accordingly.


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Question of the week - 29 (N = a! * b! * c! * d!/e! ................) 21 Apr 2020, 11:31
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OMG. Too difficult question. I've thought that N is divisible by (3*5*7)ˆ3 => N must include 21!
Therefore, I can't find any choice matched though I've carefully checked my calculation.
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Question of the week - 29 (N = a! * b! * c! * d!/e! ................) 21 Apr 2020, 19:45
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OMG. Too difficult question. I've thought that N is divisible by (3*5*7)ˆ3 => N must include 21!
Therefore, I can't find any choice matched though I've carefully checked my calculation.
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In 8! Also there is one 7 and in 9! Also there is one 7 so taking 7! ,8! and 9! Is sufficient to have three 7's. You might have missed the factorial in a number so you have to go to 21.

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Question of the week - 29 (N = a! * b! * c! * d!/e! ................) 16 Dec 2025, 13:44
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