remainders

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remainders

by pathaniaus » Wed Jun 03, 2009 4:50 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?

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by abhinav85 » Wed Jun 03, 2009 6:43 pm
Answer is 9.

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by blackarrow » Wed Jun 03, 2009 8:47 pm
I am getting 1

only 6 is the integer value which can yield a remainder as 3, when 15 is divided

Abhinav, can u please explain how do u get 9?
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by dmateer25 » Thu Jun 04, 2009 2:54 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?


Try the possibilities:

y can't be 1 or 2 because y-3 would give negative remainders.
if y = 3 the remainder is 0. 3-3=0. So this is one possibility for y.
if y = 4 the remainder is 3. 4-3=1. Nope
if y = 5 the remainder is 0. 5-3=2. Nope
if y = 6 the remainder is 3. 6-3=3. This is a possibility for y
if y = 7 the remainder is 1. 7-3=4. Nope
if y = 8 the remainder is 7. 8-3=5. nope
if y = 9 the remainder is 6. 9-3=6. This is a possibility for y.

I'm not going to list all the rest of them but the other possibility for y is 18.
if y = 18 the remainder is 15. 18-3=15. This is a possibility for y.

so y can equal: 3, 6, 9, or 18.

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by avanishjoshi » Thu Jun 04, 2009 4:02 am
Say 15 = YN + Remainder = YN + Y-3

18 = Y(N + 1).

Now Y will be a integer for N = 1,2,5,8 and 17

so the corresponding values of Y are 9,6,3,2 and 17 .