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Hard Number system problem:

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Hard Number system problem:

by \'manpreet singh » Sun Nov 04, 2012 8:37 pm
When 15 is divided by y, the remainder is y - 3. If Y must be an integer, what are all the
possible values of y?

I was able to do this problem by algebraic method, but was confused when i saw the solution given using the remainder property. Need to a explained solution using latter method.
Thanks

Manpreet
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by GMATGuruNY » Sun Nov 04, 2012 9:02 pm
'manpreet singh wrote:When 15 is divided by y, the remainder is y - 3. If Y must be an integer, what are all the
possible values of y?

I was able to do this problem by algebraic method, but was confused when i saw the solution given using the remainder property. Need to a explained solution using latter method.
Thanks

Manpreet
Since a remainder must be nonnegative:
y-3≥0
y≥3.

When 15 is divided by y, the remainder is y - 3.
In other words, 15 is a multiple of y plus (y-3) more:
15 = ky + (y-3)
18 = ky + y
18 = y(k+1)

The equation above implies that y is a factor of 18.
Thus, y can be any factor of 18 that is greater than or equal to 3:
3, 6, 9, 18.
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by eaakbari » Mon Nov 05, 2012 6:58 am
GMATGuruNY wrote:
'manpreet singh wrote:When 15 is divided by y, the remainder is y - 3. If Y must be an integer, what are all the
possible values of y?

I was able to do this problem by algebraic method, but was confused when i saw the solution given using the remainder property. Need to a explained solution using latter method.
Thanks

Manpreet
Since a remainder must be nonnegative:
y-3≥0
y≥3.

When 15 is divided by y, the remainder is y - 3.
In other words, 15 is a multiple of y plus (y-3) more:
15 = ky + (y-3)
18 = ky + y
18 = y(k+1)

The equation above implies that y is a factor of 18.
Thus, y can be any factor of 18 that is greater than or equal to 3:
3, 6, 9, 18.
GMATGuruNY,

Do correct me if I am wrong but I got 5 possible values of y

y could be 18,9,6,3,1

Substituting in y = qx + r

15 = qy + (y-3)

y= 18/(q+1)

For y to be integer, q can have values 0,1,2,5,17.

Hence y can be 18,9,6,3,1
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by Brent@GMATPrepNow » Mon Nov 05, 2012 8:20 am
'manpreet singh wrote:When 15 is divided by y, the remainder is y - 3. If Y must be an integer, what are all the
possible values of y?
There's a nice rule that say, "If N divided by D equals Q with remainder R, then N = DQ + R"
For example, since 17 divided by 5 equals 3 with remainder 2, then we can write 17 = (5)(3) + 2

We're told that 15 divided by y equals some integer (say k) with remainder y-3.
So, we can conclude that 15 = ky + (y-3), for some integer k
Simplify: 18 = ky + y
Factor: 18 = y(k+1)

Since y and k+1 are positive integers, we can see that y could equal all of the factors of 18. That is, y could equal 1, 2, 3, 6, 9 or 18

IMPORTANT: Remainders cannot be negative. For example, we'd never say that 17 divided by 5 equals 4 with remainder -3.

So, for this question, we need to ensure that the remainder, (y-3), is never negative.
Since y-3 will be negative when y=1 and when y=2, we should remove 1 and 2 from the list of possible y-values.

This leaves us with 3, 6, 9 and 18 as possible y-values.

Cheers,
Brent
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