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Re: How many integers from 1 to 100 (both inclusive) have odd number of [#permalink]
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ggurface wrote:
Engr2012 wrote:
GMATinsight wrote:
How many integers from 1 to 100 (both inclusive) have odd number of factors?

A) 7
B) 8
C) 9
D) 10
E) Greater than 10

Kudo for the correct solution.


The question could have been a bit more specific that by factors it means factors including 1 and n.

The catch here is that the integers having odd number of factors MUST be perfect squares (all other integers will have even number of factors). There are 10 perfect squares from 1 to 100 and thus D is the correct answer.

FYI, total number of factors: \(n = a^p*b^q*c^r\) with a,b,c,p,q,r \(\geq 1\) = (p+1)(q+1)(r+1) (this includes 1 and n as the factors). Now, for perfect squares, p, q , r will be even and thus p+1, q+1, r+1 will all be ODD.


All perfect squares have odd # of factors. Squaring integers 1-10 produces 1, 4, 9, 16, ... 100 which all have odd # of factors. Answer is D

Question. Do ONLY perfect squares have odd # of factors then?


YES, ONLY perfect squares have odd # of factors.
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Re: How many integers from 1 to 100 (both inclusive) have odd number of [#permalink]
How many integers from 1 to 100 (both inclusive) have odd number of factors?

so, we are talking about the series:1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16......100

4=2^2 --> # of factors (2+1)=3 odd
3=3^2 --> # of factors (2+1)=3 odd
16=2^4 --> # of factors (4+1)=5 odd

how can we find out the number of perfect squares from 1 to 100 quickly...as looking at each number is very time consuming?

regards
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Re: How many integers from 1 to 100 (both inclusive) have odd number of [#permalink]
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GMATinsight wrote:
How many integers from 1 to 100 (both inclusive) have odd number of factors?

A) 7
B) 8
C) 9
D) 10
E) Greater than 10

Kudo for the correct solution.


The detailed explanation of the problem is here.

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Re: How many integers from 1 to 100 (both inclusive) have odd number of [#permalink]
How many integers from 1 to 100 (both inclusive) have odd number of factors?

A) 7
B) 8
C) 9
D) 10 --> correct: square number has a odd number of factors. For example, 4= 2^2 has 3 factors. so 1<=n^2<=100 => 1<=n<=10, so total integers with odd number of factors are 10
E) Greater than 10
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Re: How many integers from 1 to 100 (both inclusive) have odd number of [#permalink]
Mansoor50 wrote:
How many integers from 1 to 100 (both inclusive) have odd number of factors?

so, we are talking about the series:1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16......100

4=2^2 --> # of factors (2+1)=3 odd
3=3^2 --> # of factors (2+1)=3 odd
16=2^4 --> # of factors (4+1)=5 odd

how can we find out the number of perfect squares from 1 to 100 quickly...as looking at each number is very time consuming?

regards


As it is rightly mentioned above " The integers having the odd number of factors MUST be perfect squares (all other integers will have even number of factors)."

Here we don't have to find a square of 100 no.s but we have to find perfect squares between 1 to 100.

Each no. between 1 to 10 have their perfect square lying between 1 to 100, after 10 the further no.s have square out of the range- 1 to 100. eg 11^2= 121, 12^2=144...

so there would be only 10 perfect square & only that integers have the odd no. of factors.

1^2= 1 (inclusive)
2^2= 4
3^2=9
4^2=16
5^2=25
6^2=36
7^2=49
8^2=64
9^2=81
10^2=100 (inclusive)

Answer is D
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Re: How many integers from 1 to 100 (both inclusive) have odd number of [#permalink]
36=4*9= 2^2 * 3^2 thus factors of 36= (2+1)*(2+1)=3*3=9=odd so the integral values should have 1,2,3,4,5,6,7,8,9,10 and numbers such as 36..thus option E more than 10...why is only 10 correct?
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Re: How many integers from 1 to 100 (both inclusive) have odd number of [#permalink]
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saumyagupta1602 wrote:
36=4*9= 2^2 * 3^2 thus factors of 36= (2+1)*(2+1)=3*3=9=odd so the integral values should have 1,2,3,4,5,6,7,8,9,10 and numbers such as 36..thus option E more than 10...why is only 10 correct?


Hi saumyagupta1602

You can watch the video I have posted above to understand in detail.

The bottom line is "Only Perfect squares have odd number of factors"

From 1 to 100 (both inclusive) we have 10 perfect squares
{1, 4, 9, 16, 25, 36, 49, 64, 81, 100}

So there are 10 numbers from 1 to 100 which have odd number of factors

Please subscribe my YouTube

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How many integers from 1 to 100 (both inclusive) have odd number of [#permalink]
Expert Reply
GMATinsight wrote:
How many integers from 1 to 100 (both inclusive) have odd number of factors?

A) 7
B) 8
C) 9
D) 10
E) Greater than 10

Kudo for the correct solution.


An integer with an odd number of factors must be a square. If we take any normal number like 24, we can see we can break it down into 2 factors numerous ways:

\(24 = 1 * 24 = 2 * 12 = 3 * 8 = 4 * 6\) (8 factors).

These are all factors of 24 and they usually come in pairs. Then the only way to get an odd number of factors is if we have the same factor repeating itself in this breakdown, for example, \(16 = 1 * 16 = 2 * 8 = 4 * 4\) has 5 factors.

The 4 only counts as one factor, hence only squares can have an odd number of factors.

Hence we only need to count squares within this range, which is 1 to 100 or the square of 1 through square of 10, which is 10 integers.

Ans: D
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Re: How many integers from 1 to 100 (both inclusive) have odd number of [#permalink]
Only perfect square has odd number for factors
So we need to findthe perfect square from 1 to 100
Which will include 1 4 9 16 25 36 49 64 81 100

Answer option = D
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Re: How many integers from 1 to 100 (both inclusive) have odd number of [#permalink]
Expert Reply
Mansoor50 wrote:
how can we find out the number of perfect squares from 1 to 100 quickly...as looking at each number is very time consuming?


\(1 ≤ x^2 ≤ 100\)
Since x must be a positive, we can take the positive square root of each value, as follows:
\(1 ≤ x ≤ 10\)
Implication:
x can be any positive integer between 1 and 10, inclusive -- a total of 10 options.
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