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There are 3 ways to make the number 12 using products of two positive

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There are 3 ways to make the number 12 using products of two positive  [#permalink]

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New post 24 Jul 2018, 00:26
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[Math Revolution GMAT math practice question]

There are 3 ways to make the number 12 using products of two positive integers. These are 1*12, 2*6, and 3*4. In how many ways can 2700 be written as the product of two positive integers?

A. 6
B. 12
C. 15
D. 18
E. 36

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There are 3 ways to make the number 12 using products of two positive  [#permalink]

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New post 24 Jul 2018, 01:24
1
MathRevolution wrote:
[Math Revolution GMAT math practice question]

There are 3 ways to make the number 12 using products of two positive integers. These are 1*12, 2*6, and 3*4. In how many ways can 2700 be written as the product of two positive integers?

A. 6
B. 12
C. 15
D. 18
E. 36


2700 = (2^2 × 3^3 × 5^2)
total number of factors are (2+1)(3+1)(2+1)=36 including 1 and itself
no of ways 2700 can be written as product of two positive int = 36/2=18
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Re: There are 3 ways to make the number 12 using products of two positive  [#permalink]

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New post 24 Jul 2018, 02:57
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1
MathRevolution wrote:
[Math Revolution GMAT math practice question]

There are 3 ways to make the number 12 using products of two positive integers. These are 1*12, 2*6, and 3*4. In how many ways can 2700 be written as the product of two positive integers?

A. 6
B. 12
C. 15
D. 18
E. 36


To count the factors of a positive integer:
1. Prime-factorize the integer
2. Write the prime-factorization in the form \((a^p)(b^q)(c^r)\)...
3. The number of factors = \((p+1)(q+1)(r+1)\)...

\(2700 = 2^2 * 3^3 * 5^2\)

Adding 1 to each exponent and multiplying, we get:
Total number of factors = \((2+1)(3+1)(2+1) = 36\)

These 36 factors can be used to form 18 FACTOR PAIRS, as follows:
1*2700
2*1350
3*300
And so on.


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There are 3 ways to make the number 12 using products of two positive  [#permalink]

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New post 24 Jul 2018, 03:28
2
1
MathRevolution wrote:
[Math Revolution GMAT math practice question]

There are 3 ways to make the number 12 using products of two positive integers. These are 1*12, 2*6, and 3*4. In how many ways can 2700 be written as the product of two positive integers?

A. 6
B. 12
C. 15
D. 18
E. 36



Most of us by now know how to find factors. But even after knowing this, you can go wrong if you miss out on "pair of factors" and mark "number of factors"

1) Prime factorise 2700..
\(2700=3^3*10^2=3^3*2^2*5^2\)
2) number of factors ..
(3+1)(2+1)(2+1)=4*3*3=36..

But we are looking for pairs so 35/2=18 as product of two factors say 1*2700=10*270=2*1350...

Some extra points
If a number is a square, the number of factors will be ODD
For example say it was 8100..
\(8100=3^4*2^2*5^2.....5*3*3=45\)
So now 45/2 is not an integer...so what does one do..
Two cases..
1) if question asks you pair of factors then add 1 as \(90*90 =8100\)
So \(\frac{(45+1)}{2}=23\)
2) but if question asks you pair of different factors then subtract one as 90*90 will not be included so \((\frac{45-1)}{2}=22\)
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1) Absolute modulus : http://gmatclub.com/forum/absolute-modulus-a-better-understanding-210849.html#p1622372
2)Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html
3) effects of arithmetic operations : https://gmatclub.com/forum/effects-of-arithmetic-operations-on-fractions-269413.html


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Re: There are 3 ways to make the number 12 using products of two positive  [#permalink]

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New post 24 Jul 2018, 09:08
chetan2u wrote:
MathRevolution wrote:
[Math Revolution GMAT math practice question]

There are 3 ways to make the number 12 using products of two positive integers. These are 1*12, 2*6, and 3*4. In how many ways can 2700 be written as the product of two positive integers?

A. 6
B. 12
C. 15
D. 18
E. 36



Most of us by now know how to find factors. But even after knowing this, you can go wrong if you miss out on "pair of factors" and mark "number of factors"

1) Prime factorise 2700..
\(2700=3^3*10^2=3^3*2^2*5^2\)
2) number of factors ..
(3+1)(2+1)(2+1)=4*3*3=36..

But we are looking for pairs so 35/2=18 as product of two factors say 1*2700=10*270=2*1350...

Some extra points
If a number is a square, the number of factors will be ODD
For example say it was 8100..
8100=3^4*2^2*5^2.....5*3*3=45
So now 45/2 is not an integer...so what does one do..
Two cases..
1) if question asks you pair of factors then add 1 as 900*900 =8100
So (45+1)/2=23
2) but if question asks you pair of different factors then subtract one as 900*900 will not be included so (45-1)/2=22


Hi chetan2u,
I the question asks about the sum of the factors then how to go about that?
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Re: There are 3 ways to make the number 12 using products of two positive  [#permalink]

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New post 26 Jul 2018, 00:45
=>

\(2700 = 2^2*3^3*5^2\)
The number of distinct factors of \(2700\) is \((2+1)(3+1)(2+1) = 36.\)
Since the order of multiplication does not matter (i.e. \(30 * 90 = 90*30\)), the number of pairs of positive integers that multiply to give \(2700\) is \(\frac{36}{2} = 18.\)

Therefore, the answer is D.
Answer: D
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Re: There are 3 ways to make the number 12 using products of two positive  [#permalink]

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New post 27 Jul 2018, 22:37
arvind910619 wrote:
chetan2u wrote:
MathRevolution wrote:
[Math Revolution GMAT math practice question]

There are 3 ways to make the number 12 using products of two positive integers. These are 1*12, 2*6, and 3*4. In how many ways can 2700 be written as the product of two positive integers?

A. 6
B. 12
C. 15
D. 18
E. 36



Most of us by now know how to find factors. But even after knowing this, you can go wrong if you miss out on "pair of factors" and mark "number of factors"

1) Prime factorise 2700..
\(2700=3^3*10^2=3^3*2^2*5^2\)
2) number of factors ..
(3+1)(2+1)(2+1)=4*3*3=36..

But we are looking for pairs so 35/2=18 as product of two factors say 1*2700=10*270=2*1350...

Some extra points
If a number is a square, the number of factors will be ODD
For example say it was 8100..
8100=3^4*2^2*5^2.....5*3*3=45
So now 45/2 is not an integer...so what does one do..
Two cases..
1) if question asks you pair of factors then add 1 as 900*900 =8100
So (45+1)/2=23
2) but if question asks you pair of different factors then subtract one as 900*900 will not be included so (45-1)/2=22


Hi chetan2u,
I the question asks about the sum of the factors then how to go about that?


hi..
not likely you will such a question in actual, it will limit to number of factors and not sum of factors ..
but formula for it is ..
1) Prime factorise 2700..
\(2700=3^3*10^2=3^3*2^2*5^2\)
    a) number of factors ..
    \((3+1)(2+1)(2+1)=4*3*3=36..\)
    b) sum of factors ..
    for \(a^x*b^y*c^z\), it is \(\frac{a^{(x+1)}-1}{a-1}*\frac{b^{(y+1)}-1}{b-1}*\frac{c^{(z+1)}-1}{c-1}\)..
    \(\frac{2^{(2+1)}-1}{2-1}*\frac{3^{(3+1)}-1}{3-1}*\frac{5^{(2+1)}-1}{5-1}=7*\frac{3^4-1}{2}*\frac{5^3-1}{4}=7*40*31=8680\)

_________________

1) Absolute modulus : http://gmatclub.com/forum/absolute-modulus-a-better-understanding-210849.html#p1622372
2)Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html
3) effects of arithmetic operations : https://gmatclub.com/forum/effects-of-arithmetic-operations-on-fractions-269413.html


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Re: There are 3 ways to make the number 12 using products of two positive &nbs [#permalink] 27 Jul 2018, 22:37
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