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Divisibility

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Divisibility

by Troika » Mon May 28, 2012 2:40 am
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?

OA: 3, 6, 9 and 18

Source: MGMAT NP, pg. 134, q. 26
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by neelgandham » Mon May 28, 2012 3:23 am
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.e. 15/y = k + (y-3)/y , where k is a non negative integer.
i.e. 15 = ky + (y-3)
i.e. 15 = (k+1)y - 3
i.e. 18 = (k+1)*y = Positive Integer * Positive Integer

Let us try to represent the number 18 as a product of two different positive integers.
18 = 18*1 = (k+1)*y. So, k = 17 and y = 1.
18 = 9*2 = (k+1)*y. So, k = 8 and y = 2.
18 = 6*3 = (k+1)*y. So, k = 5 and y = 3.
18 = 3*6 = (k+1)*y. So, k = 2 and y = 6.
18 = 2*9 = (k+1)*y. So, k = 1 and y = 9.
18 = 1*18 = (k+1)*y. So, k = 0 and y = 18.

We, now, know that y can be 1,2,3,6,9,or 18. But y-3 is the remainder and the remainder is always non-negative. i.e. y-3>0. i.e y>3. y = 1,2 can be eliminated and we are left with 3,6,9,18
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by GMATGuruNY » Mon May 28, 2012 3:37 am
HG10 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?

OA: 3, 6, 9 and 18

Source: MGMAT NP, pg. 134, q. 26
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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