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# Question of the week - 29 (N = a! * b! * c! * d!/e! ................)

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Joined: 04 Jan 2015
Posts: 2942
Question of the week - 29 (N = a! * b! * c! * d!/e! ................)  [#permalink]

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Updated on: 27 Feb 2019, 21:43
00:00

Difficulty:

85% (hard)

Question Stats:

58% (03:19) correct 42% (02:55) wrong based on 69 sessions

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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. $$\frac{27}{2}$$

_________________

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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Re: Question of the week - 29 (N = a! * b! * c! * d!/e! ................)  [#permalink]

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28 Dec 2018, 05:13
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 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. $$\frac{26}{3}$$

B. 9

C. $$\frac{21}{2}$$

D. 13

E. $$\frac{27}{2}$$

_________________
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Re: Question of the week - 29 (N = a! * b! * c! * d!/e! ................)  [#permalink]

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02 Jan 2019, 02:07

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.

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Re: Question of the week - 29 (N = a! * b! * c! * d!/e! ................)  [#permalink]

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29 Jan 2019, 17:03
EgmatQuantExpert wrote:

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.

Hello EgmatQuantExpert !

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

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Re: Question of the week - 29 (N = a! * b! * c! * d!/e! ................)  [#permalink]

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29 Jan 2019, 22:17
jfranciscocuencag wrote:
EgmatQuantExpert wrote:

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.

Hello EgmatQuantExpert !

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

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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Re: Question of the week - 29 (N = a! * b! * c! * d!/e! ................)   [#permalink] 29 Jan 2019, 22:17
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