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02 Jan 2025, 01:30
Kudos
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D
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
95%
(hard)
Question Stats:
45% (02:27) correct
55%
(02:25)
wrong
based on 152
sessions
History
Date
Time
Result
Not Attempted Yet
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
A. 8
B. 10
C. 12
D. 13
E. 15
02 Jan 2025, 03:36
Kudos
Bookmarks
Divisor= 5^2 * 3^2 * 2^2
We need at least two 5, 3 and 2 in n.
A = 5! = 5*4*3*2*1 (got one five, one 3 and 2^3 three 2's)
Now we need one more 5 and one more 3
B = 6! Will have both
C = 2! For least no and > 1
A+B+C= 6+5+2= 13
We need at least two 5, 3 and 2 in n.
A = 5! = 5*4*3*2*1 (got one five, one 3 and 2^3 three 2's)
Now we need one more 5 and one more 3
B = 6! Will have both
C = 2! For least no and > 1
A+B+C= 6+5+2= 13
02 Jan 2025, 02:17
Kudos
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Bunuel
Product of three smallest prime numbers = 2*3*5
Square of this product = \(2^2*3^2*5^2\)
n = a! * b! * c! where a, b, c > 1
For n to be divisible by \(2^2*3^2*5^2\), it needs to have atleast 2 powers of 5, which is only possible if
1. 2 numbers out of a, b, c are greater than or equal to 5
2. 1 number out of a, b, c is greater than or equal to 10
In either case if you observe, a+ b + c > 10, so eliminate option choice A and B.
For first possibility, smallest triplet possible is (2,5,6) => Sum = 13
For second possibility, smallest triplet would be (2,3,10) => Sum = 15
So smallest sum possible should be 13.
IMO: D
