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factors

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Manager
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factors [#permalink] New post 09 Nov 2010, 11:33
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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?
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Re: factors [#permalink] New post 09 Nov 2010, 11:47
Expert's post
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).

Answer: D.

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

Hope it's clear.
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Re: factors [#permalink] New post 09 Nov 2010, 13:47
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).

Answer: D.

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.
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Re: factors [#permalink] New post 12 Nov 2010, 18:05
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.
Re: factors   [#permalink] 12 Nov 2010, 18:05
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