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Divisibility: When the positive integer x is divided by 11,

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Divisibility: When the positive integer x is divided by 11,

by II » Tue May 20, 2008 3:12 pm
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

How would you approach this question ... would be great if you have multiple approaches to tackle this.

Thanks.
II
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by sxjain3 » Tue May 20, 2008 4:43 pm
Both 11 and 19 are Prime Numbers and hence their LCM = 11*19 = 209 is the closest number which can be looked at here.

Now when dividing this number the remainder is 3, so the number being divided i.e. x = 209+3 = 212.

So when 212 is divided by 11, the quotent y = 19 and remainder r = 3.

So when y = 19 is divided by 19, the remainder is 0.


The same will be true for any multiple of 209.
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by luvaduva » Tue May 20, 2008 6:08 pm
x/11 = y + 3 => x = 11y + 3
x/19 = z + 3 => x = 19z + 3

11y + 3 = 19z + 3

y =19 and z =11.

19/19 = 1 R 0
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by beeparoo » Tue May 20, 2008 7:00 pm
sxjain3 wrote:Both 11 and 19 are Prime Numbers and hence their LCM = 11*19 = 209 is the closest number which can be looked at here.
Hey sxjain3, curiously, how did you know to seek the LCM of 11 and 19. I too recognized that they were primes, but didn't know what to do with them.

Thanks in advance for your response.

Sandra
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by sxjain3 » Tue May 20, 2008 8:19 pm
Hi Sandra,

As both numbers were prime and were dividing the same number, their LCM was teh closest number they both could have divided, esp that they had the same remainder.

This is the reason, LCM will be a good starting point.
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by AleksandrM » Wed May 21, 2008 9:29 am
I used long divion to get

11y + 3 = 19z + 3

Then ended up with 11y = 19z

I then realized that since 11 and 19 are prime the only way that 11y can equate to 19z is if y = 11 and z = 19.

Hence y/19 = 19/19 = 1.

However, is it fair to say that y MUST be 11 and z MUST be 19????
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by chidcguy » Wed May 21, 2008 3:03 pm
AleksandrM wrote:I used long divion to get

11y + 3 = 19z + 3

Then ended up with 11y = 19z

I then realized that since 11 and 19 are prime the only way that 11y can equate to 19z is if y = 11 and z = 19.

Hence y/19 = 19/19 = 1.

However, is it fair to say that y MUST be 11 and z MUST be 19????
No Y can be 38 and Z can be 22 to make 11Y= 19Z

38X11=19X22 = 418

11y = 19z means 11y/19z =1 since 19 cannot divide 11 and vice versa, 19 must divide Y and 11 must divide z with out remainders to satisfy the condition. that means remainder is 0

also means 11y/19z=1
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by sandeep07 » Mon May 26, 2008 10:59 pm
In Continuation to what AleksandraM did...

11y = 19z. In this equation both y and z are quotients. So they are integers.

z = 11y/19. If z has to be an integer, then it must be that y is divisble by 19. If y is not divisible by 19, there is no way 11y/19 would become an integer.

Since y has to be divisble by 19, the remainder is Zero.
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