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A positive integer n has the smallest 3 prime number [#permalink]
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28 Feb 2017, 04:20
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Q. A positive integer n has the smallest 3 prime numbers as its only prime factors. How many positive integers divide n completely? (1) The total number of times the prime factors of n occur in n is 5. (2) The product of the number of times each prime factor of n occurs in n is 4. Answer Choices : A. Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient to answer the question asked. B. Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient to answer the question asked. C. BOTH statements (1) and (2) TOGETHER are sufficient to answer the question asked, but NEITHER statement ALONE is sufficient to answer the question asked. D. EACH statement ALONE is sufficient to answer the question asked. E. Statements (1) and (2) TOGETHER are NOT sufficient to answer the question asked, and additional data specific to the problem are needed. Thanks, Saquib Quant Expert eGMATRegister for our Free Session on Number Properties (held every 3rd week) to solve exciting 700+ Level Questions in a classroom environment under the realtime guidance of our Experts
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A positive integer n has the smallest 3 prime number [#permalink]
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28 Feb 2017, 04:20
The official solution has been posted. Looking forward to a healthy discussion..
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Re: A positive integer n has the smallest 3 prime number [#permalink]
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07 Mar 2017, 07:24
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Given,
smallest three prime numbers are its ONLY prime factors, i.e, 2,3 & 5 . (Although we don't need to know what the factors are, just how many are there)
To find the number of integers that are factors of an integer n we need to know the power of prime factors.
Lets say n = 2^a * 3^b * 5^c
number of integers that are factors of n will be (a+1)(b+1)(c+1)
From 1st statement, a+b+c = 5
Since we know none of a,b,c can be zero, lets assume a = 1. This gives following values for a,b and c: 1. a=1, b=2, c=2 2. a=1, b=3, c=1
So the # of factors can be = 16 or 18
Not sufficient.
From statement 2, a*b*c = 4. Again fixing a=1 we get following values of a,b,c 1. a=1,b=2,c=2 2. a=1,b=4,c=1
Hence, # of factors = 18 or 20
Hence, insufficient.
From 1 and 2, we can get that the number of factors are 18. Hence, sufficient.
Answer is C.



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Re: A positive integer n has the smallest 3 prime number [#permalink]
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07 Mar 2017, 13:29
As per question stem, number N has the three smallest prime numbers as it's only prime factors which means N = 2^a * 3^b * 5^c
Question asks: How many positive integers divide n completely= which means, the total number of factors which N has. This can be denoted as = (a+1)*(b+1)*(c+1) So now we need to find out which statement(s) gives us the value of a, b & c.
Statement 1  The total number of times the prime factors of n occur in n is 5. This implies that, a+b+c = 5 Now we know (a,b,c) can take values such as (1,1,3) or (2,1,2) In the two cases, total number of factors is different. (1,1,3) => 2*2*4 = 16 (2,1,2) => 3*2*3 = 18
As we do not have a definite values, hence Statement 1 is not sufficient.
Statement 2  The product of the number of times each prime factor of n occurs in n is 4. This implies => a*b*c = 4 Now we know (a,b,c) can take values such as (1,1,4) or (2,1,2) In the two cases, total number of factors is different. (1,1,4) => 2*2*5 = 20 (2,1,2) => 3*2*3 = 18
As we do not have a definite values, hence Statement 2 is not sufficient.
Combining both Statements, we know
a+b+c =5 & a*b*c = 4 This means, that (a,b,c) can take any value (1,2,2) or (2,1,2) or (2,2,1), but the total number of factors will always remain as 18.
Therefore combining both Statement is sufficient.
Answer is C



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Re: A positive integer n has the smallest 3 prime number [#permalink]
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27 Mar 2017, 05:41
Solution Steps 1 & 2: Understand Question and Draw InferencesThe question tells us about a positive integer n which has smallest 3 prime numbers as its only prime factors. The questions asks us to find the number of positive integers that divide n completely • Let’s first analyze the information given in the question statement. The prime factors of n are the 3 smallest prime numbers i.e. 2, 3 and 5. • So, we can write \(n = 2^a * 3^b * 5^c\) where a, b, c are positive integers • We are asked to find the number of integers that divide n completely i.e. we are asked to find the number of factors of n. • To find the number of factors of n, we need to know the following:
a. Number of prime factors of n à this information is known to us b. Number of times the prime factors of n occur in nà this information is unknown to us. • Number of factors of n = (a+1)(b+1)(c+1) • So, we need to find the information about the number of times a prime factor of n occurs in n to find out the number of factors of n i.e. we need to find the values of a, b and c which give us a unique value of (a+1)(b+1)(c+1). • With the above understanding, let’s see the information provided in the statements. Step 3: Analyze Statement 1 independently1. The total number of times the prime factors of n occur in n is 5 That is, a + b + c = 5. As a, b, c are positive integers, their minimum values can be 1. Thus the possible values of {a, b, c} in any order can be:
• Case 1  {2, 2, 1}
o In this case, Total number of factors = (2+1)*(2+1)*(1+1) = 3*3*2 = 18 • Case 2  {3, 1, 1}
o In this case, Total number of factors = (3+1)*(1+1)*(1+1) = 4*2*2 = 16 As we do not have a unique answer, statement1 is insufficient to answer the question.Step 4: Analyze Statement 2 independently2. The product of the number of times each prime factor of n occurs in n is 4. That is a*b*c = 4. As a, b, c are positive integers, their minimum values can be 1. Thus the possible values of {a, b, c} in any order can be:
• Case 1  {4, 1, 1}
o In this case, Total number of factors = (4+1)*(1+1)*(1+1) = 5*2*2 = 20 • Case 2  {2, 2, 1}
o In this case, Total number of factors = (2+1)*(2+1)*(1+1) = 3*3*2 = 18 As we do not have a unique answer, statement2 is insufficient to answer the questionStep 5: Analyze Both Statements Together (if needed)1. From Statement 1, we inferred that {a, b, c} = {2, 2, 1} or {3, 1, 1} in any order 2. From Statement 2, we inferred that {a, b, c} = {4, 1, 1} or {2, 2, 1} in any order The only set of values that satisfy both the statements is {a, b, c} = {2, 2, 1} in any order. Therefore, we can now find a unique value of product of a+1, b+1 and c+1 Sufficient to answerAnswer: CThanks, Saquib Quant Expert eGMATRegister for our Free Session on Number Properties (held every 3rd week) to solve exciting 700+ Level Questions in a classroom environment under the realtime guidance of our Experts
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A positive integer n has the smallest 3 prime number [#permalink]
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18 Aug 2017, 02:34
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EgmatQuantExpert wrote: Q. A positive integer n has the smallest 3 prime numbers as its only prime factors. How many positive integers divide n completely?
n = \(2^a\)*\(3^b\)*\(5^c\). (a+1)(b+1)(c+1) = ?? We need the values for a, b, c. Quote: (1) The total number of times the prime factors of n occur in n is 5. (2) The product of the number of times each prime factor of n occurs in n is 4. Let's solve Statement 2 first, as Statement 1 was convoluted to begin with, for me at least. I had to look at Statement 2 to see that Statement 1 meant Sum. 2) a*b*c = 4 a=1, b=2, c=2 a=4, b=1, c=1 Insufficient. 1) S2 had product, so this statement will have the sum a + b + c = 5 Insufficient. 1+2) a*b*c=4 a+b+c=5 => the values have to be 2, 2, 1 => (a+1)(b+1)(c+1) = 3*3*2 = 18 Sufficient. C is the answer.
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Re: A positive integer n has the smallest 3 prime number [#permalink]
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18 Aug 2017, 11:18
I did not think 1 was a prime number...



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Re: A positive integer n has the smallest 3 prime number [#permalink]
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18 Aug 2017, 11:21




Re: A positive integer n has the smallest 3 prime number
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