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How many different positive integers d are there such that [#permalink]
 16 Apr 2008, 13:52
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How many different positive integers d are there such that 48 divided by d yields a remainder of d - 4?
Two
Three
Four
Five
Six
Last edited by kevincan on 16 Apr 2008, 14:01, edited 2 times in total.
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Re: PS. Remainders [#permalink]
16 Apr 2008, 14:10
Answer is B - Three
4 is fairly obvious. 48/4 gives remainder of 0 and d-4 = 0
26 is also fairly obvious since we need a number d such that
d+d-4 = 48 which gives d = 26
Since 26 and 48 both have a common factor (2), there is another number, 13 which should satisfy the condition as 13 does.
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Re: PS. Remainders [#permalink]
16 Apr 2008, 14:22
the way i did this..
48=DN+r where r=d-4
48=D*N+D-4
52=D(N+1)
i quickly look for prime factors of 52..they are 13*2^2...
number of possible ways of writting 52..13*4, 26*2, 52*1 i.e 3 ...
Last edited by FN on 16 Apr 2008, 14:32, edited 1 time in total.
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Re: PS. Remainders [#permalink]
16 Apr 2008, 14:32
fresinha12 wrote: the way i did this..
48=DN+r where r=d-4
48=D*N+D-4
52=D(N+1)
i quickly look for prime factors of 52..they are 13*2^2...
number of possible divisors of 52..13*4, 26*2, 52*1 i.e 3 ...
can you explain why u prime factorized?
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Re: PS. Remainders [#permalink]
16 Apr 2008, 14:33
i shouldnt use the word divisors..i meant how many ways can your get 52 using integers? 3 ways
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Re: PS. Remainders [#permalink]
16 Apr 2008, 15:53
kevincan wrote: How many different positive integers d are there such that 48 divided by d yields a remainder of d - 4?
Two
Three
Four
Five
Six Three. 48 = d*n + (d-4) 52 = d*(n+1) We know that 52 = 2*2*13 = 2^2 * 13^1, which yields 3*2 = 6 factors. The 6 factors are: 1, 2, 4, 13, 26, 52 This means d can only equal to these 6 numbers List them out (d, remainder of 48/d, d-4) (1, 0, -3) <-No (2, 0, -2) <-No (4, 0, 0) <-Yes (13, 9, 9) <-Yes (26, 22, 22) <-Yes 52 cannot yield remainder. d can be 4, 13, 26
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Re: PS. Remainders [#permalink]
16 Apr 2008, 16:00
what if i changed the question and asked how many integers are there such that when 48 divided by d gives remainder d-3?
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Re: PS. Remainders [#permalink]
16 Apr 2008, 16:06
fresinha12 wrote: what if i changed the question and asked how many integers are there such that when 48 divided by d gives remainder d-3? 51 = d*(n+1) 3*17 = 51 There are 4 factors: 1, 3, 17, 51 Same way (d, remainder, d-3) (1, 0, -2) (3, 0, 0) (17, 14, 14) 51 won't give it. Two I got.
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Re: PS. Remainders [#permalink]
16 Apr 2008, 16:21
exactly..ways to write 51, is 13*7 and 51*1, i.e 2 ways.. there are 2 such numbers.. bkk145 wrote: fresinha12 wrote: what if i changed the question and asked how many integers are there such that when 48 divided by d gives remainder d-3? 51 = d*(n+1) 3*17 = 51 There are 4 factors: 1, 3, 17, 51 Same way (d, remainder, d-3) (1, 0, -2) (3, 0, 0) (17, 14, 14) 51 won't give it. Two I got. |
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Re: PS. Remainders [#permalink]
16 Apr 2008, 16:48
I'm thinking FOUR...3,17,1,51
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Re: PS. Remainders [#permalink]
16 Apr 2008, 21:34
fresinha12 wrote: the way i did this..
48=DN+r where r=d-4
48=D*N+D-4
52=D(N+1)
And my reasoning: D is different possitive interger, factor of 52 52=2^2 *13. So D = (2+1)*(1+1)=6 E
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Re: PS. Remainders [#permalink]
16 Apr 2008, 21:57
bkk145 wrote: kevincan wrote: How many different positive integers d are there such that 48 divided by d yields a remainder of d - 4?
Two
Three
Four
Five
Six Three. 48 = d*n + (d-4) 52 = d*(n+1) We know that 52 = 2*2*13 = 2^2 * 13^1, which yields 3*2 = 6 factors. The 6 factors are: 1, 2, 4, 13, 26, 52 This means d can only equal to these 6 numbers List them out (d, remainder of 48/d, d-4) (1, 0, -3) <-No (2, 0, -2) <-No (4, 0, 0) <-Yes (13, 9, 9) <-Yes (26, 22, 22) <-Yes 52 cannot yield remainder. d can be 4, 13, 26 beautiful work but, imo, we can add 52 as well. 48 divided by 52 has a reminder of 48 which is equal to 52 - 4. so possible values for d = 4, 13, 26 and 52.
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Re: PS. Remainders [#permalink]
16 Apr 2008, 23:41
Ans 4.
Did it just like fresinha
plus added 4 as fourth integer.
So numbers are 4,13,26,52.
Lilke your questions Kevincan. Makes me think.
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Re: PS. Remainders [#permalink]
17 Apr 2008, 06:27
Dreaming here again...agree with 52
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Re: PS. Remainders [#permalink]
17 Apr 2008, 07:11
how do we quickly see that 48/52 will give us a remainder of 48?
i totally missed 52..
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Re: PS. Remainders [#permalink]
17 Apr 2008, 08:07
fresinha12 wrote: how do we quickly see that 48/52 will give us a remainder of 48?
i totally missed 52.. This is from wiki: "When dividing 3 by 10, 3 is the remainder as we always take the front number as the remainder when the second number is of higher value." http://en.wikipedia.org/wiki/Remainder
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Re: PS. Remainders [#permalink]
17 Apr 2008, 08:25
Yea.. Answer should be 4.
I didn't consider 52 myself.
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Re: PS. Remainders
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17 Apr 2008, 08:25
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