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How many integers less than 100
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15 May 2020, 06:47
15 May 2020, 06:47
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Question Stats:
43% (01:29) correct
56% (01:54) wrong based on 32 sessions
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How many integers less than 100 have exactly 4 odd factors but no even factors?
A) 11
B) 13
C) 15
D) 17
E) 19
May someone explain the steps of solving and strategy in tackling this question?
A) 11
B) 13
C) 15
D) 17
E) 19
May someone explain the steps of solving and strategy in tackling this question?
Re: How many integers less than 100
[#permalink]
15 May 2020, 10:52
15 May 2020, 10:52
2
Expert Reply
Two cases are possible:
\(x^1y^1=xy=odd<100\), where \(x\) and \(y\) are odd prime numbers, then # of factors will be \((1+1)(1+1)=4\);
If x=3, then y can be: 5, 7, 11, 13, 17, 19, 23, 29, 31 - 9 numbers.
If x=5, then y can be: 7, 11, 13, 17, 19 - 5 numbers.
If x=7, then y can be: 11, 13 - 2 numbers.
OR:
\(x^3=odd<100\), where \(x\) is odd prime number, then # of factors will be \((3+1)=4\).
x can be only 3 - 1 number.
Total \(9+5+2+1=17\).
Answer: D.
P.S. Finding the Number of Factors of an Integer:
First make prime factorization of an integer \(n=a^p*b^q*c^r\), where \(a\), \(b\), and \(c\) are prime factors of \(n\) and \(p\), \(q\), and \(r\) are their powers.
The number of factors of \(n\) will be expressed by the formula \((p+1)(q+1)(r+1)\). NOTE: this will include 1 and n itself.
Example: Finding the number of all factors of 450: \(450=2^1*3^2*5^2\)
Total number of factors of 450 including 1 and 450 itself is \((1+1)*(2+1)*(2+1)=2*3*3=18\) factors.
\(x^1y^1=xy=odd<100\), where \(x\) and \(y\) are odd prime numbers, then # of factors will be \((1+1)(1+1)=4\);
If x=3, then y can be: 5, 7, 11, 13, 17, 19, 23, 29, 31 - 9 numbers.
If x=5, then y can be: 7, 11, 13, 17, 19 - 5 numbers.
If x=7, then y can be: 11, 13 - 2 numbers.
OR:
\(x^3=odd<100\), where \(x\) is odd prime number, then # of factors will be \((3+1)=4\).
x can be only 3 - 1 number.
Total \(9+5+2+1=17\).
Answer: D.
P.S. Finding the Number of Factors of an Integer:
First make prime factorization of an integer \(n=a^p*b^q*c^r\), where \(a\), \(b\), and \(c\) are prime factors of \(n\) and \(p\), \(q\), and \(r\) are their powers.
The number of factors of \(n\) will be expressed by the formula \((p+1)(q+1)(r+1)\). NOTE: this will include 1 and n itself.
Example: Finding the number of all factors of 450: \(450=2^1*3^2*5^2\)
Total number of factors of 450 including 1 and 450 itself is \((1+1)*(2+1)*(2+1)=2*3*3=18\) factors.
Re: How many integers less than 100
[#permalink]
15 May 2020, 10:53
15 May 2020, 10:53
Expert Reply
I did not find it in the GRE Big Book as source
