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# factors

Author Message
Manager
Joined: 13 Jul 2010
Posts: 162

Kudos [?]: 97 [0], given: 7

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09 Nov 2010, 12:33
when 777 is divided by positive integer n, the remainder is 77. How many possibilities are there for n?

a. 2
b. 3
c. 4
d. 5
e. 6

Let me your thought process on this. I know we have to put this in remainder format of n(y) + 77= 777 so n(y)=700 but from here I would find the number of factors for 700 to be 18. Is there a quicker way of figuring out the number of factors greater than 77?

Kudos [?]: 97 [0], given: 7

Math Expert
Joined: 02 Sep 2009
Posts: 42249

Kudos [?]: 132578 [0], given: 12326

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09 Nov 2010, 12:47
gettinit wrote:
when 777 is divided by positive integer n, the remainder is 77. How many possibilities are there for n?

a. 2
b. 3
c. 4
d. 5
e. 6

Let me your thought process on this. I know we have to put this in remainder format of n(y) + 77= 777 so n(y)=700 but from here I would find the number of factors for 700 to be 18. Is there a quicker way of figuring out the number of factors greater than 77?

Positive integer $$a$$ divided by positive integer $$d$$ yields a reminder of $$r$$ can always be expressed as $$a=qd+r$$, where $$q$$ is called a quotient and $$r$$ is called a remainder, note here that $$0\leq{r}<d$$ (remainder is non-negative integer and always less than divisor).

So we'd have: $$777=qn+77$$, where $$remainder=77<n=divisor$$ --> $$qn=700=2^2*5^2*7$$ --> as $$n$$ must be more than 77 then $$n$$ could take only 5 values: 100, 140, 175, 350, and 700 (the least factor more than 77 is $$2^2*5^2=100$$, now if you substitute 2, 2^2, and 5 by 7 you'll get the factors more than 100 plus if you include 7 you'll get one more factor, 700 itself, so total 1+3+1=5).

Discussed here: remainder-101074.html?hilit=possibilities#p782422

Hope it's clear.
_________________

Kudos [?]: 132578 [0], given: 12326

Manager
Joined: 13 Jul 2010
Posts: 162

Kudos [?]: 97 [0], given: 7

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09 Nov 2010, 14:47
Bunuel wrote:
gettinit wrote:
when 777 is divided by positive integer n, the remainder is 77. How many possibilities are there for n?

a. 2
b. 3
c. 4
d. 5
e. 6

Let me your thought process on this. I know we have to put this in remainder format of n(y) + 77= 777 so n(y)=700 but from here I would find the number of factors for 700 to be 18. Is there a quicker way of figuring out the number of factors greater than 77?

Positive integer $$a$$ divided by positive integer $$d$$ yields a reminder of $$r$$ can always be expressed as $$a=qd+r$$, where $$q$$ is called a quotient and $$r$$ is called a remainder, note here that $$0\leq{r}<d$$ (remainder is non-negative integer and always less than divisor).

So we'd have: $$777=qn+77$$, where $$remainder=77<n=divisor$$ --> $$qn=700=2^2*5^2*7$$ --> as $$n$$ must be more than 77 then $$n$$ could take only 5 values: 100, 140, 175, 350, and 700 (the least factor more than 77 is $$2^2*5^2=100$$, now if you substitute 2, 2^2, and 5 by 7 you'll get the factors more than 100 plus if you include 7 you'll get one more factor, 700 itself, so total 1+3+1=5).

Discussed here: remainder-101074.html?hilit=possibilities#p782422

Hope it's clear.

THanks for the reference Bunuel. So I am guessing its best to write out the factors in these types of questions? I didn't quite understand your methodology of replacing 2, 2^2, and 5 by 7 etc. I understand how it is applied here but if this were a different question I'd be lost. I think the key is just to mulitply the different factors to arrive at numbers larger than 77.

Kudos [?]: 97 [0], given: 7

Manager
Joined: 07 Jan 2010
Posts: 143

Kudos [?]: 96 [0], given: 57

Location: So. CA
WE 1: 2 IT
WE 2: 4 Software Analyst

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12 Nov 2010, 19:05
So qn=700=2^2*5^2*7 has 18 factors, on the gmat how do you quickly determine 5 out of the 18 factors are greater than 77 within 2-2.5 mins? i got kind of lost with the substitution method also.

Kudos [?]: 96 [0], given: 57

Re: factors   [#permalink] 12 Nov 2010, 19:05
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# factors

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