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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.

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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