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When the positive integer x is divided by 11, the quotient

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When the positive integer x is divided by 11, the quotient [#permalink] New post 31 May 2007, 04:19
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
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 [#permalink] New post 31 May 2007, 04:42
(A) for me :)

We have:
o x = 11*y + 3
and
o x = 19*v + 3 where v is an integer

So,
11*y = 19*v
<=> y = 19/11*v = 19 * w + 0 where w = v / 11
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 [#permalink] New post 31 May 2007, 05:51
Fig wrote:
(A) for me :)

We have:
o x = 11*y + 3
and
o x = 19*v + 3 where v is an integer

So,
11*y = 19*v
<=> y = 19/11*v = 19 * w + 0 where w = v / 11


(A)

That's most appealing to me.

I arrived at 11*y = 19*v like Fig's first few lines, but couldn't work out logically how to conclude that the remainder should be Zero.

Anything about the prime numbers 11 and 19?

Can anyone make it clear?
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 [#permalink] New post 31 May 2007, 06:10
solidcolor wrote:
Fig wrote:
(A) for me :)

We have:
o x = 11*y + 3
and
o x = 19*v + 3 where v is an integer

So,
11*y = 19*v
<=> y = 19/11*v = 19 * w + 0 where w = v / 11


(A)

That's most appealing to me.

I arrived at 11*y = 19*v like Fig's first few lines, but couldn't work out logically how to conclude that the remainder should be Zero.

Anything about the prime numbers 11 and 19?

Can anyone make it clear?


Actually, we know that v and y are integers. And, yes, as 11 and 19 are prime numbers, v must be formed like this : v = 11*j where j is an integer. :)

Hope that helps :)
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 [#permalink] New post 31 May 2007, 08:58
OA is 0.

These simple looking ones really throw you off track taking a test.

Carrying on from Fig's

11*y = 19*v
This means that 11y is a multiple of 19.
So 11*y/19 is an integer.
Obviously 19 is not a factor of 11. So 19 must be a factor of y.
And we have 0 as the remainder
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 [#permalink] New post 31 May 2007, 08:59
Right. I'll memorize this. Thanks Fig.

:wink:
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 [#permalink] New post 31 May 2007, 09:07
Of the same track, let me introduce another problem about Factors and Prime numbers:

OG 11, PS, Q209 (p.181)

How many different positive integers are factors of 441?
A. 4
B. 6
C. 7
D. 9
E. 11
















:wink: :wink: :wink: :wink: :wink: :wink:








Nice hint from the Explanation:

441 = 9 * 49 = [3^2] * [7^2]

So the set of 441's factors should be 3, 3^2, 3*7, (3^2)*7, 3*(7^2), 7^2, 7, and don't forget 1 and 441. Count these factors, you'll say 9 is the answer.
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 [#permalink] New post 31 May 2007, 13:51
from the stem we know that "when the positive integer x is divided by 11, the quotient is y and the remainder 3" this can also be rewritten as:

x = 11*y+3

the same way that 30/7 = 4 and a remiander of 2

7*4+2 = 30

after we arrived to x = 11*y+3 we are told that x will also give a remiander of 3 when divided in 19, so:

x/19 = (11*y+3)/19 = will give a remainder of 3

we are asked what will be the remainder when y/19 , we don't know, but we can try to solve for (11*y+3)/19

one not so good way to solve is to plug numbers ! I don't like this way , but we will use it later on for check.

we are still stuck with (11*y+3)/19 , now if (11*y+3)/19 gives a remiander of 3 , we can say that 11*y/19 will give us a remainder of 0 ! (since we took out the 3, if we will add it the remainder will be once again 3)

but we were asked about y/19, not 11*y/19.

Since 19 is a prime number it will be very easy to find y that will give us 11*y/19 = an integer (no remiander).

y=19

11*19/19 = 11

so 19/19 will give you yet again a remiander of 0.

The answer is (A)

:-D
  [#permalink] 31 May 2007, 13:51
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