Hello,
Question #38 asks: 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?
In the explanation given in page 51, it defines two ways to solve the problem but I do not understand very well the third paragraph that starts with: "The second thing to note is that y is at most 18." Then proceeds: "When y=18, the divisor (18) is 3 larger than 15, and the quotient is zero. As y increases, the remainder will not change. For example, if the divisor is 19, the remainder is still 15 and the quotient is still zero (the divisor is now 4 more than 15)" and gives another example when the divisor is 20 and says that that remainder is still 15.
Can you clarify on this please as I am having hard time trying to digest why the remainder is still 15 even when the divisor increases. Also I appreciate if you elaborate more thoroughly the two approaches to solve the question.

Thank you!
Renaud

hi renaud...

I am not so good in teaching but I was also stuck with this problem for sometime. I will try to help you though :D

just to make it short... the remainder is still 15 because 0 is a possible quotient. Thus, 15/18 = 0 remainder 15. The maximum remainder here (having quotient 0) is only 15.

15 as remainder satisfies y-3 since 15 is 3 less than y, note that y=18. Make y > 18 and you will never get a remainder which is y-3.

Hope that helps (somehow). :)

Hey MGMAT instructors? Do I make sense :D

Hello Rowie,
Yes you make sense, waiting for the teachers.

Thank you!
Renaud

Rowie's explanation is correct--thanks!

Renaud, between the two approaches (number listing and algebra), I strongly prefer number listing for remainder problems. These problems are infrequent enough that it is all too easy to set up a slightly wrong equation. Also note that the algebra solution produced values that had to be "plugged and checked" anyway: y = 1 and y = 2 were "solutions" but we eliminated them because remainders are positive by definition, so the constraint that remainder = y - 3 didn't work for those values.

The number listing method is "real," so you can be sure of your answer. It also allows you to see patterns, an easier way of recognizing the properties than memorizing them in words. On paper, I would do this:

15/1 = 15 + 0r, where "0r" is just shorthand for "remainder of 0."
15/2 = 7 + 1r
15/3 = 5 + 0r (r = 3-3 = 0, check)
15/4 = 3 + 3r
15/5 = 3 + 0r
15/6 = 2 + 3r (r = 6-3 = 3, check)
15/7 = 2 + 1r
15/8 = 1 + 7r
15/9 = 1 + 6r (r = 9-3 = 6, check)
15/10 = 1 + 5r

And here we see a pattern begin--for y = 8 through 15, the quotient is 1 with a remainder that decreases as we increase y. Thus, we can skip to:

15/16 = 0 + 15r
15/17 = 0 + 15r
15/18 = 0 + 15r ( r = 18-3 = 15, check)
15/19 = 0 + 15r
and so on...

This is the pattern Rowie described. Since it's quite reasonable to list these mathematically on paper, trying to reason through the problem verbally ("quotient...divisor...remainder...etc") is unnecessarily frustrating, in my opinion.

_________________
Emily Sledge
Instructor
ManhattanGMAT