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N = a!×b!×c!×d!/e! where a, b, c, and d, are four distinct positive in 06 Oct 2023, 01:30
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\(N = \frac{a!×b!×c!×d!}{e!}\) 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



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gmatophobia
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N = a!×b!×c!×d!/e! where a, b, c, and d, are four distinct positive in 07 Oct 2023, 07:34
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Bunuel
\(N = \frac{a!×b!×c!×d!}{e!}\) 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
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"N is divisible by cube of product of the three smallest odd prime numbers"

N must be divisible by \(3^3 * 5^3 * 7^3\)

Therefore we can conclude that out of the four numbers in the numerator, three numbers should be 7, 8, and 9.

a, e are two consecutive numbers

\((a,e) = (2,3)\) or \((a,e) = (3,2)\)

To minimize the value of \(\frac{a+b+c+d}{e}\), we can have the greater number in the denominator and the smaller number in the numerator.

Hence, N = \(\frac{2!×7!×8!×9!}{3!}\)

Value of \(\frac{a+b+c+d}{e}\) = \(\frac{7+8+9+2}{3} = \frac{26}{3}\)

Option A
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N = a!×b!×c!×d!/e! where a, b, c, and d, are four distinct positive in 13 Oct 2023, 05:59
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Couldn't b, c, and d be 0, 15, and 7?

This would make the answer 24/3 --> 8
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DrAnkita91
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N = a!×b!×c!×d!/e! where a, b, c, and d, are four distinct positive in 13 Oct 2023, 06:15
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"Therefore we can conclude that out of the four numbers in the numerator, three numbers should be 7, 8, an 9"

Can I PLEASE know how did we derive these 7 , 8 AND 9?
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N = a!×b!×c!×d!/e! where a, b, c, and d, are four distinct positive in 13 Oct 2023, 15:00
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gmatophobia
Bunuel
\(N = \frac{a!×b!×c!×d!}{e!}\) 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

"N is divisible by cube of product of the three smallest odd prime numbers"

N must be divisible by \(3^3 * 5^3 * 7^3\)

Therefore we can conclude that out of the four numbers in the numerator, three numbers should be 7, 8, and 9.

a, e are two consecutive numbers

\((a,e) = (2,3)\) or \((a,e) = (3,2)\)

To minimize the value of \(\frac{a+b+c+d}{e}\), we can have the greater number in the denominator and the smaller number in the numerator.

Hence, N = \(\frac{2!×7!×8!×9!}{3!}\)

Value of \(\frac{a+b+c+d}{e}\) = \(\frac{7+8+9+2}{3} = \frac{26}{3}\)

Option A
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How did you come up with the 7, 8 ,9 ?
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rachana_shukla
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N = a!×b!×c!×d!/e! where a, b, c, and d, are four distinct positive in 15 Oct 2023, 05:58
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LucienH
gmatophobia
Bunuel
\(N = \frac{a!×b!×c!×d!}{e!}\) 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

"N is divisible by cube of product of the three smallest odd prime numbers"

N must be divisible by \(3^3 * 5^3 * 7^3\)

Therefore we can conclude that out of the four numbers in the numerator, three numbers should be 7, 8, and 9.

a, e are two consecutive numbers

\((a,e) = (2,3)\) or \((a,e) = (3,2)\)

To minimize the value of \(\frac{a+b+c+d}{e}\), we can have the greater number in the denominator and the smaller number in the numerator.

Hence, N = \(\frac{2!×7!×8!×9!}{3!}\)

Value of \(\frac{a+b+c+d}{e}\) = \(\frac{7+8+9+2}{3} = \frac{26}{3}\)

Option A


How did you come up with the 7, 8 ,9 ?
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Since the b!*c!*d! has to be divisible by \(3^3 * 5^3 * 7^3\) , this means that 7 has to occur in all the three number's factorial. Similarly 5 and 3 must also occur in all the three number's factorial.
Thats why the numbers should by 7,8,9
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N = a!×b!×c!×d!/e! where a, b, c, and d, are four distinct positive in 26 Dec 2024, 23:34
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