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# How many different positive integers d are there such that

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How many different positive integers d are there such that [#permalink]  16 Apr 2008, 12:52
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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, 13:01, edited 2 times in total.
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Re: PS. Remainders [#permalink]  16 Apr 2008, 13:10

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, 13:22
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, 13:32, edited 1 time in total.
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Re: PS. Remainders [#permalink]  16 Apr 2008, 13:32
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, 13:33
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, 14:53
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, 15:00
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, 15:06
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, 15:21
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, 15:48
I'm thinking FOUR...3,17,1,51
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Re: PS. Remainders [#permalink]  16 Apr 2008, 20:34
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, 20:57
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, 22:41
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, 05:27
Dreaming here again...agree with 52
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Re: PS. Remainders [#permalink]  17 Apr 2008, 06:11
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, 07:07
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, 07:25

I didn't consider 52 myself.
Re: PS. Remainders   [#permalink] 17 Apr 2008, 07:25
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