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# For any integer x greater than 1, x! denotes the product of all intege

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Manager
Joined: 28 May 2018
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Location: India
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For any integer x greater than 1, x! denotes the product of all intege  [#permalink]

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06 Apr 2019, 12:02
1
2
00:00

Difficulty:

75% (hard)

Question Stats:

38% (02:22) correct 63% (02:31) wrong based on 16 sessions

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For any integer x greater than 1, x! denotes the product of all integers from 1 to x, inclusive. n = a!*b!*c! where a, b, c are distinct positive integers > 1, What is the least value of a + b + c for which n is divisible by the square of the product of the three smallest prime numbers?

A) 8
B) 10
C) 12
D) 13
E) 15

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Senior Manager
Joined: 03 Mar 2017
Posts: 291
Re: For any integer x greater than 1, x! denotes the product of all intege  [#permalink]

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06 Apr 2019, 20:27
1
a,b,c are distinct positive integers and >1

First, three smallest prime numbers will be 2,3,5 and n = a!*b!*c!

As per question,n should be divisible 2^2*3^2*5^2.

The highest power of 5 is 2.

We have to find the values of a,b,and c such that we can get atleast 2 powers of 5.

Therefore one of a,b,and c must have a value of 10.

Option A and B can be discarded.

Option C is also incorrect as we cannot have same values for a,b,and c.

Therefore D is correct.

When we take value as 13, (a=10,b=2,c=1) we get all the highest powers of ( 2^2*3^2*5^2)

IMO D.
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Senior Manager
Joined: 25 Feb 2019
Posts: 287
Re: For any integer x greater than 1, x! denotes the product of all intege  [#permalink]

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06 Apr 2019, 20:41
1
three smallest prime numbers

2,3,5

so the number n should have two ,2s, 3s,5s

we see that only 6!*5! satisfy above mentioned condition

but third number can be 0! or 1! or 2! and so on

but from the give condition

each of a,b,c > 1

so we have least possible value as 2

so the sum.will be 13

IMO D

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Re: For any integer x greater than 1, x! denotes the product of all intege   [#permalink] 06 Apr 2019, 20:41
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