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655-705 Level|   Probability|      
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For any integer P greater than 1, P! denotes the product of all the in 01 Dec 2018, 10:21
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B
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53% (01:56) correct 47% (02:08) wrong based on 244 sessions

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GMATbuster's Weekly Quant Quiz#11 Ques #7


For Questions from earlier quizzes: Click Here

For any integer P greater than 1, P! denotes the product of all the integers from 1 to P, inclusive. If one integer is selected at random between 5! And 5! + 10, inclusive, what is the probability that the chosen number will have only two factors?

A)0
B)1/11
C)1/10
D)2/11
E)2/10
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B
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For any integer P greater than 1, P! denotes the product of all the in 01 Dec 2018, 10:41
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5! = 120
5! + 10 = 130
From 120 to 130 (total N = 11)
only 127 is prime number
Probability = 1/11
hence choice B
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For any integer P greater than 1, P! denotes the product of all the in 01 Dec 2018, 10:56
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the number with only 2 factors are called prime numbers. firs find the range of numbers given by question from 5! (120)to 5!+10(130) inclusive 130-120+1=11
primer number in only one 127
1/11
Answer choice B

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For any integer P greater than 1, P! denotes the product of all the in 01 Dec 2018, 11:14
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GMATbuster's Weekly Quant Quiz#11 Ques #7


For Questions from earlier quizzes: Click Here

For any integer P greater than 1, P! denotes the product of all the integers from 1 to P, inclusive. If one integer is selected at random between 5! And 5! + 10, inclusive, what is the probability that the chosen number will have only two factors?

A)0
B)1/11
C)1/10
D)2/11
E)2/10
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IMO-D


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For any integer P greater than 1, P! denotes the product of all the in 20 Jan 2019, 19:09
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workout

GMATbuster's Weekly Quant Quiz#11 Ques #7


For Questions from earlier quizzes: Click Here

For any integer P greater than 1, P! denotes the product of all the integers from 1 to P, inclusive. If one integer is selected at random between 5! And 5! + 10, inclusive, what is the probability that the chosen number will have only two factors?

A)0
B)1/11
C)1/10
D)2/11
E)2/10
Show more

We are looking at numbers between 5! and 5!+10 or between 120 and 130, inclusive
So, 130-120+1 = 11 numbers. We can eliminate C and E

The only numbers with 2 factors are prime numbers.
We can eliminate 120, 121, 122, 123, 124, 125, 126, 128, 129, 130 (divisible by 2,3,5 and 11).
So we're left with 127, which it turns out is a prime (tested 7 to make sure), so B.

For reference 101–200 has 21 prime numbers (101, 103, 107, 109, 113, 127, 131, 137, 139, 149, 151, 157, 163, 167, 173, 179, 183, 193, 197, 199)
It could've easily been D if the range had been slightly different (e.g. 130-140 inclusive)
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For any integer P greater than 1, P! denotes the product of all the in 22 Nov 2024, 22:50
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5! - 5! + 10 has 10 numbers i.e., 5! , 5! + 1, ... , 5! +10

5! = 120
5!+1 = 120 + 1
.
.
.
.
5! + 10 = 130

So, the set is 120 - 130 inclusive

Only prime number have 2 factors
Only 127 is prime number. So, 1/10
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