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dhlee922
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remainder problem - NP ch. 10 q. 23

by dhlee922 Wed Apr 10, 2013 9:00 pm

you're given background info, integer x has a remainder of 5 when divided by 9, integer y has a remainder of 7 when divided by 9

question: what is the remainder when 5x - y is divided by 9?


i solved this by picking numbers, but i think that it's probably inefficient.

i did 2 examples and made sure i got the same number

taking info from above, x has remainder of 5 so:
x = 14, remainder is 5
5x = 70, remainder is 7

x = 23, remainder is 5
5x = 115, remainder is 7

i recall in chapter 10, it only mentions that if you multiply the remainders of x and y, it is equivalent to finding the remainder of xy, but i dont recall it saying that multiplying x by 5 is the same as multiplying Rx by 5 (remainder of x multiplied by 5)

could someone provide the explanation/proof/algebra of why this works?
RonPurewal
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Re: remainder problem - NP ch. 10 q. 23

by RonPurewal Fri Apr 12, 2013 6:04 am

dhlee922 wrote:you're given background info, integer x has a remainder of 5 when divided by 9, integer y has a remainder of 7 when divided by 9

question: what is the remainder when 5x - y is divided by 9?


i solved this by picking numbers, but i think that it's probably inefficient.


well, you're probably mistaken there.
on remainder problems, picking numbers is almost always easier and faster than algebra, which is often extremely difficult, and sometimes even impossible, on such problems.

there is an algebraic solution for this, but it's horrible and cumbersome:
if x/9 gives a remainder of 5, then x = 9(int1) + 5, where "int1" is some integer.
if y/9 gives a remainder of 7, then y = 9(int2) + 7, where "int2" is some other integer.
so, then, 5x - y is 45(int1) + 45 - 9(int2) - 7, or 36(int1) + 36.
that's 36(int1) + 36, or 9(4(int1) + 4). so, that's a multiple of 9, so the overall remainder on division by 9 is 0.

in my opinion, that's terrible, but plugging numbers is a piece of cake.
on the other hand, if you think the above is actually easier than plugging numbers, then by all means go for it.
THANU.KG
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Re: remainder problem - NP ch. 10 q. 23

by THANU.KG Wed May 08, 2013 4:37 pm

RonPurewal wrote:
dhlee922 wrote:you're given background info, integer x has a remainder of 5 when divided by 9, integer y has a remainder of 7 when divided by 9

question: what is the remainder when 5x - y is divided by 9?


i solved this by picking numbers, but i think that it's probably inefficient.


well, you're probably mistaken there.
on remainder problems, picking numbers is almost always easier and faster than algebra, which is often extremely difficult, and sometimes even impossible, on such problems.

there is an algebraic solution for this, but it's horrible and cumbersome:
if x/9 gives a remainder of 5, then x = 9(int1) + 5, where "int1" is some integer.
if y/9 gives a remainder of 7, then y = 9(int2) + 7, where "int2" is some other integer.
so, then, 5x - y is 45(int1) + 45 - 9(int2) - 7, or 36(int1) + 36.
that's 36(int1) + 36, or 9(4(int1) + 4). so, that's a multiple of 9, so the overall remainder on division by 9 is 0.

in my opinion, that's terrible, but plugging numbers is a piece of cake.
on the other hand, if you think the above is actually easier than plugging numbers, then by all means go for it.


Hi Ron,

Shouldn't it be 45(int1) + 25 - 9(int2) -7 ==> 45(int1) + 18 - 9(int2) ==> 9[5(int1)-9(int2)+2] which when divided by 9 gives a remainder of 0?

Am i wrong in my simplification?

Thanks
jlucero
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Re: remainder problem - NP ch. 10 q. 23

by jlucero Thu May 09, 2013 3:03 pm

You are correct, and the lengthy and convoluted (and in this case, mistaken) algebra, proves that picking numbers is probably going to be easier. Here's why picking numbers is going to be easier: the question asks "what is the remainder when 5x - y is divided by 9?" This means there is only ONE possible answer to this. Easiest numbers to pick? x=5, y=7

(yes, x can be 5 as 5 divided by 9 is zero with a remainder of 5. ditto for y=7)
Joe Lucero
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AnandT660
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Re: remainder problem - NP ch. 10 q. 23

by AnandT660 Tue Nov 01, 2016 12:20 am


Another way to look at it is is to say that y = x+2. 5x-y then becomes 4x-2, and when you plug different values of x in here, the remainder you get when dividing by 9 = 0
RonPurewal
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Re: remainder problem - NP ch. 10 q. 23

by RonPurewal Fri Nov 04, 2016 9:57 am

anandt660, do you have a question?

this discussion has been dormant for three and a half years, so there isn't much point in trying to "contribute" to it... lol
if you're asking a question, please clarify. thanks.