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Remainder

by ramannjit » Wed Oct 06, 2010 5:15 pm
When the positive integer x is divided by 11, the quotient is y and the remainder is 3. When x is divided by 19, the remainder is also 3. What is the remainder when y is divided by 19?

Help me with this problem.
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by krazy800 » Wed Oct 06, 2010 7:30 pm
ramannjit wrote:When the positive integer x is divided by 11, the quotient is y and the remainder is 3. When x is divided by 19, the remainder is also 3. What is the remainder when y is divided by 19?

Help me with this problem.
I am getting "o". what is the OA. herez my approach

x=11Y +3 from statement 'positive integer x is divided by 11, the quotient is y and the remainder is 3' ---(1)

X=19z+3 from When x is divided by 19, the remainder is also 3.----(2)

from (1) & (2)

11y=19Z ---(3)

where y should be 19 or multiples of 19 for equation 3 to be satisfied.

therefore y/19 leaves no or 0 remainder.

hope I am right!!!
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by Alex Lbn » Wed Oct 06, 2010 7:34 pm
ramannjit wrote:When the positive integer x is divided by 11, the quotient is y and the remainder is 3. When x is divided by 19, the remainder is also 3. What is the remainder when y is divided by 19?

Help me with this problem.
x/11=y+3
x/19=z+3
y/19=?


(x-3)/11=y
(x-3)/19=z
We assume that y and z are integers. (x-3) is evenly divisible by both 11 and 19. Therefore, (x-3) must be at least the Least Common Multiple of 11 and 19,since 11 and 19 have no prime factors in common. So, x-3>=209 and x>=212. When x=212, y=19.
The next value of x that satisfies the conditions above will be 2*LCM(11&19)+3=421, in this case y=38.
Whatever the value of x we would to take, x-3 must be divisible by both 11 and 19, so x-3 will always be LCM(11*19)*some integer n, where n[1:infinity]. Thus, when we divide x by 11 y will always be a multiple of 19. Therefore, the remainder of y/19 will be zero.

Let's try to put it down the other way.
11 and 19 are prime numbers. Therefore their least common multiple will be 209. In other words, their LCM will be just a product of 11*19=209. In addition, there will be no such positive integer x, in the range of 0<x<209, which after division by both 11 and 19 will leave a remainder of 3.
There will be no integer x between 0 and 209 that we can take and divide it first by 11 and get a remainder of 3, and then the same integer x divide by 19 and also get a remainder of 3. In a given range, in order to get a remainder of 3 after division by both 11 and 19, integer x must always be different.
So the minimum possible value of integer x that satisfies given conditions is LCM of 11 and 19 +3, or (11*19)+3.
Therefore, minimum possible x is 212 and y in this case is 19. The next possible value of x will be 421, or 11*2*19+3. In this case y will be 38. We can conclude that in order to satisfy given conditions x-3 must always have 19 and 11 as factors, so the quotient y, when x-3 is divided by 11 will always be a multiple of 19. Therefore, the remainder of y/19 will always be 0.
I'm sure someone from the BTG community will offer more concise explanation.

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by GMATGuruNY » Wed Oct 06, 2010 7:50 pm
ramannjit wrote:When the positive integer x is divided by 11, the quotient is y and the remainder is 3. When x is divided by 19, the remainder is also 3. What is the remainder when y is divided by 19?

Help me with this problem.
x is a multiple of 11 plus 3 more and a multiple of 19 plus 3 more.

Plug in x = 11*19 + 3 = 212
212/11 = 19 R3, so y = 19.
y/19 = 19/19 = 1 R0.
So the remainder is 0.
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by ramannjit » Wed Oct 06, 2010 8:14 pm
OA is indeed "0"

Thanks everyone for the approaches of explainations to help me understand.
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by pzazz12 » Thu Oct 07, 2010 5:15 am
GMATGuruNY wrote:
ramannjit wrote:When the positive integer x is divided by 11, the quotient is y and the remainder is 3. When x is divided by 19, the remainder is also 3. What is the remainder when y is divided by 19?

Help me with this problem.
x is a multiple of 11 plus 3 more and a multiple of 19 plus 3 more.

Plug in x = 11*19 + 3 = 212
212/11 = 19 R3, so y = 19.
y/19 = 19/19 = 1 R0.
So the remainder is 0.

thank you.... simply great.......