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When 777 is divided by positive integer n, the remainder is [#permalink]
 03 Apr 2008, 04:42
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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
Please, suggest how to solve the above problem quickly ..
Thank you
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Re: PS: division with remainder 777 [#permalink]
04 Apr 2008, 04:02
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chica wrote: Instead of counting the factors of 700 that are greater than 77, let’s count the ones that are less than or equal to 700/77.
As 700 = 50 × 2 × 7, we can see that there are 5 factors of 700 that are less than or equal to 7: 1 , 2 , 4 , 5 , and 7.
Thus there are 5 possible values of n (i.e. factors of 700) greater than 77.
We don’t need to know these values, but for the curious, they are 700, 350, 175, 140 and 100. Excellent! 700=(factor >77)*(factor <700/77) OE say that for each factor >70 we have one factor <700/77 in order to have 700 as a product of those factors. So, we can count only all factors <700/77 700/77~8.9 --> 1,2,4,5,7
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Re: PS: division with remainder 777 [#permalink]
03 Apr 2008, 05:03
I suggest just picking numbers: 100, 175, 350, 700. Thus, C. 100 and 700 are the most obvoius, then goes 350 (half of 700) and 175 (half of 350). I can't think of any other way to solve this kind of problems. What do others think?
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Re: PS: division with remainder 777 [#permalink]
03 Apr 2008, 05:11
777 = n*k+77 (n>77)
n*k=700=7*4*25
try combination for n>77
without 7 -> 4*25
with 7 -> 7*2*25,7*4*5,7*4*25,7*25
answer D
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Re: PS: division with remainder 777 [#permalink]
03 Apr 2008, 05:20
I agree, Albert. I forgot to include 140. D.
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Re: PS: division with remainder 777 [#permalink]
03 Apr 2008, 05:26
I get D as well..
here is how..
777=n*K+77
where N and K are positive integers..
N*K=can be 700 in the first case..so lets look at the prime factors of 700=7*5^2*2^2
number of factors of 700 are 2+2+1=5
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Re: PS: division with remainder 777 [#permalink]
03 Apr 2008, 05:30
I actually believe that the question is really asking us how many factors of 700 are there..
in that case the actual answer should be (1+1)(2+1)(2+1)=18 there are 18 factors of 700 and either one of those could be a possible value of N..
If we had 18..i would have really gone with 18..
what do the Math geniuses here think?
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Re: PS: division with remainder 777 [#permalink]
03 Apr 2008, 05:52
I'm no math genius, but I'll point out that if you take factors of 700, which are less than 77, you'll end up with remainder being less than 77. That's why we take only the factors greater than 77.
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Re: PS: division with remainder 777 [#permalink]
03 Apr 2008, 06:18
you are totally correct..all factors greater than 77... wow..i could have made this problem really hard by just throwing in 18 as an answer option..then this would have been a 700 level question
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Re: PS: division with remainder 777 [#permalink]
03 Apr 2008, 06:28
Thank you OA is indeed D
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Re: PS: division with remainder 777 [#permalink]
03 Apr 2008, 12:55
AlbertNTN wrote: 777 = n*k+77 (n>77)
n*k=700=7*4*25
try combination for n>77
without 7 -> 4*25
with 7 -> 7*2*25,7*4*5,7*4*25,7*25
answer D is there an approach we can use without randomly guessing?
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Re: PS: division with remainder 777 [#permalink]
03 Apr 2008, 16:42
I think AlbertNTN approach is not guessing. 777 = n*k+77 (n>77, k - an integer) n*k=700=2^2*5^2*7the next step is to choose all possibilities for which n>77: one prime number: 2, 4, 5, 25, 7- 0 possibilities for n two prime numbers: 2*5, 2*25, 4*5, 4*25, 2*7, 4*7, 5*7, 25*7 - 2 possibilities for n three prime numbers: 2*5*7, 4*5*7, 2*25*7, 4*25*7- 3 possibilities for n So, we have 5 possible values of n: 4*25, 25*7, 4*5*7, 2*25*7, 4*25*7
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Re: PS: division with remainder 777 [#permalink]
04 Apr 2008, 03:32
walker wrote: I think AlbertNTN approach is not guessing.
777 = n*k+77 (n>77, k - an integer)
n*k=700=2^2*5^2*7
the next step is to choose all possibilities for which n>77:
one prime number: 2, 4, 5, 25, 7- 0 possibilities for n
two prime numbers: 2*5, 2*25, 4*5, 4*25, 2*7, 4*7, 5*7, 25*7 - 2 possibilities for n
three prime numbers: 2*5*7, 4*5*7, 2*25*7, 4*25*7- 3 possibilities for n
So, we have 5 possible values of n: 4*25, 25*7, 4*5*7, 2*25*7, 4*25*7 I meant to post the actual explanation that came with the question.. and which is still confusing to me.. If anybody can evaluate the latter, it will be great (bolded part..): If the remainder is 77, then n must logically be greater than 77. Also, there must be a positive integer q such that 777= nq + 77. i.e. nq = 700. Therefore, the factors of 700 greater than 77 comprise the possible values of n. Instead of counting the factors of 700 that are greater than 77, let’s count the ones that are less than or equal to 700/77.
As 700 = 50 × 2 × 7, we can see that there are 5 factors of 700 that are less than or equal to 7: 1 , 2 , 4 , 5 , and 7.
Thus there are 5 possible values of n (i.e. factors of 700) greater than 77.We don’t need to know these values, but for the curious, they are 700, 350, 175, 140 and 100.
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Re: PS: division with remainder 777 [#permalink]
04 Apr 2008, 04:21
walker wrote: chica wrote: Instead of counting the factors of 700 that are greater than 77, let’s count the ones that are less than or equal to 700/77.
As 700 = 50 × 2 × 7, we can see that there are 5 factors of 700 that are less than or equal to 7: 1 , 2 , 4 , 5 , and 7.
Thus there are 5 possible values of n (i.e. factors of 700) greater than 77.
We don’t need to know these values, but for the curious, they are 700, 350, 175, 140 and 100. Excellent! 700=(factor >77)*(factor <700/77) OE say that for each factor >70 we have one factor <700/77 in order to have 700 as a product of those factors. So, we can count only all factors <700/77 700/77~8.9 --> 1,2,4,5,7 Many thanks, walker!!!!!!!!!! I got it now
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Re: PS: division with remainder 777
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04 Apr 2008, 04:21
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