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

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Math Revolution GMAT Instructor
Joined: 16 Aug 2015
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There are 3 ways to make the number 12 using products of two positive  [#permalink]

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24 Jul 2018, 01:26
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85% (hard)

Question Stats:

37% (01:16) correct 63% (01:53) wrong based on 100 sessions

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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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"Only $99 for 3 month Online Course" "Free Resources-30 day online access & Diagnostic Test" "Unlimited Access to over 120 free video lessons - try it yourself"  Math Revolution Discount Codes Economist GMAT Tutor Discount Codes Optimus Prep Discount Codes Director Joined: 20 Feb 2015 Posts: 546 Concentration: Strategy, General Management There are 3 ways to make the number 12 using products of two positive [#permalink] ### Show Tags 24 Jul 2018, 02:24 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 Manager Joined: 04 Aug 2010 Posts: 243 Schools: Dartmouth College Re: There are 3 ways to make the number 12 using products of two positive [#permalink] ### Show Tags 24 Jul 2018, 03:57 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 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. _________________ GMAT and GRE Tutor Over 1800 followers Click here to learn more GMATGuruNY@gmail.com New York, NY If you find one of my posts helpful, please take a moment to click on the "Kudos" icon. Available for tutoring in NYC and long-distance. For more information, please email me at GMATGuruNY@gmail.com. Math Expert Joined: 02 Aug 2009 Posts: 6561 There are 3 ways to make the number 12 using products of two positive [#permalink] ### Show Tags 24 Jul 2018, 04:28 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 $$(\frac{45-1)}{2}=22$$ _________________ 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 GMAT online Tutor VP Status: Learning Joined: 20 Dec 2015 Posts: 1218 Location: India Concentration: Operations, Marketing GMAT 1: 670 Q48 V36 GRE 1: Q157 V157 GPA: 3.4 WE: Engineering (Manufacturing) Re: There are 3 ways to make the number 12 using products of two positive [#permalink] ### Show Tags 24 Jul 2018, 10:08 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? _________________ Please give kudos if you found my answers useful Math Revolution GMAT Instructor Joined: 16 Aug 2015 Posts: 6046 GMAT 1: 760 Q51 V42 GPA: 3.82 Re: There are 3 ways to make the number 12 using products of two positive [#permalink] ### Show Tags 26 Jul 2018, 01:45 => $$2700 = 2^2*3^3*5^2$$ 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 _________________ MathRevolution: Finish GMAT Quant Section with 10 minutes to spare The one-and-only World’s First Variable Approach for DS and IVY Approach for PS with ease, speed and accuracy. "Only$99 for 3 month Online Course"
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Math Expert
Joined: 02 Aug 2009
Posts: 6561
Re: There are 3 ways to make the number 12 using products of two positive  [#permalink]

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27 Jul 2018, 23:37
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,

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

GMAT online Tutor

Re: There are 3 ways to make the number 12 using products of two positive &nbs [#permalink] 27 Jul 2018, 23:37
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