Number Properties (Divisibility and Remainders)

[iil-london]

When the positive integer x is divided by 11, the quotient is y and the remainder 3. When x is divided by 19, the remainder is also 3. What is the remainder when y is divided by 19?

A) 0
B) 1
C) 2
D) 3
E) 4

What clues from the question would provide you with the indication on HOW to solve this question ? Please explain your thinking.
What would your primary strategy for this be?
Also, what would your secondary approach be ?

Thanks in advance.

[StaceyKoprince]

Please read (and follow!) the guidelines. Your subject should be the first 5 to 8 words of the question.

First, the question is talking about remainders, so I'd need to know how to represent this information algebraically.

The quotient times the divisor plus the remainder equals the dividend. In this case:
y * 11 + 3 = x
z*19 + 3 = x

Notice I introduced a new variable, z, because the second sentence didn't tell me the quotient. I also know some other things:
- x is a positive integer (because the problem told me)
- y and z are both integers (a quotient is always an integer)

I need to figure out what the remainder will be when y is divided by 19. I don't have enough info to actually solve for these variables, so it's going to have to depend on determining some characteristic of y.

I do notice that both equations equal x, so I can try setting them equal to each other to see whether that tells me anything about y.

11y + 3 = 19z + 3
11y = 19z
If 11*y is equal to 19*z, then the complete product of each must share all of the same factors. A similar example with smaller numbers:
4*6 = 8*3
The prime factors (equal for BOTH sides) are: 2,2,2,3
So the factors of 11*y must equal the factors of 19*z. 11 and 19 are both primes, so they both have to be among the prime factors, and those prime factors must hold for BOTH sides of the equation.

on the y side, we have 11*y. If I need a factor of 19 in there as well, that 19 must be a part of the y, since it isn't part of the 11. If 19 is a factor of y, then y is divisible by 19 - and, by definition, if y is divisible by 19, the remainder is zero. ("divisible by" means we get only an integer - that is, no remainder).

[iil-london]

Wow ... great explanation Stacy ! You guys at MGMAT are good huh !

[StaceyKoprince]

We do try! You're welcome. :)

[Guest]

Great explanation, thanks!

I have a question about what the answer would be if the numbers change. Let's say that that when x is divided by 19, the reminder is 5. How would we go about solving that, or there not enough information?

Thanks in advance.

[RonPurewal]

Great explanation, thanks!

I have a question about what the answer would be if the numbers change. Let's say that that when x is divided by 19, the reminder is 5. How would we go about solving that, or there not enough information?

Thanks in advance.

if both of the numbers are 5's, then the explanation works the same way it does in stacey's example.

if you say that x/11 gives a remainder of 3 and x/19 gives a remainder of 5: here's how i would handle that (which may not be the best, or most elegant, solution ... but it works)
this means that x is 3 more than a multiple of 11, and also 5 more than a multiple of 19.

1) first, use trial and error to figure out a value of x that happens to work
i would do this by counting up values that are 5 more than a multiple of 19, and seeing when you get one that's 3 more than a multiple of 11.
24: no
43: no
62: no
81: no
100: no
119: no
138: no
157: yes (= 14x11 + 3)
so y is 14, and the remainder is 14 (because 19 goes into 14 zero times, with a remainder of 14)

2) now consider the characteristics of all other x's that will leave the desired remainders
* you need x to leave the same remainder of 3 when it is divided by 11.
this means that any other x that works must differ from 157, the value we just found, by a multiple of 11; if it doesn't, it will leave a different remainder.
* you need x to leave the same remainder of 5 when it is divided by 19.
this means that any other x that works must differ from 157, the value we just found, by a multiple of 19; if it doesn't, it will leave a different remainder.
* consequence: x differs from 157 by a multiple of both 11 and 19.
* since 11 and 19 are prime numbers, this means that the difference between 157 and x is a multiple of 11x19.
* this means that x is one of the following numbers:
157
157 + 19(11)
157 + 38(11)
157 + 57(11)
etc.
if you divide these numbers by, you'll get the following quotients:
y = 14 (with a remainder of 3, as required)
y = 14 + 19 (with a remainder of 3, as required)
y = 14 + 38 (with a remainder of 3, as required)
etc.
the right hand terms won't contribute to the remainder when we divide y by 19, because they are all multiples of 19.
therefore the remainder always 14.

--

this sort of logic will always apply, so you'll always get a definite solution to the problem.

[viksnme]

Using the same principles as used by Stacey and Ron, we can find a solution to the original problem by taking numbers and playing with them.

If x is divisible by both 11 and 19 leaving a remainder of 3, surely x could be

x = 11*19 + 3 where y=19

or some multiple of 11*19 plus 3 i.e.

x = 11*19*2 + 3 , where y = 19*2 or
x = 11*19*3 +3 , where y = 19*3 etc

Hence, if we divide y by 19, the remainder would always be 0.
What say tutors ?

[RonPurewal]

Using the same principles as used by Stacey and Ron, we can find a solution to the original problem by taking numbers and playing with them.

If x is divisible by both 11 and 19 leaving a remainder of 3, surely x could be

x = 11*19 + 3 where y=19

or some multiple of 11*19 plus 3 i.e.

x = 11*19*2 + 3 , where y = 19*2 or
x = 11*19*3 +3 , where y = 19*3 etc

Hence, if we divide y by 19, the remainder would always be 0.
What say tutors ?

well played.