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stolyar , i am assuming that you did not see my post

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stolyar , i am assuming that you did not see my post [#permalink] New post 29 Sep 2003, 20:24
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stolyar ,

i am assuming that you did not see my post about "successively". The problem had a typo.

A certain number when[b] successively [/b]divided by 8 and 11 leaves remainders of 3 and 7 respectively. What will be remainder when the number is divided by the product of 8 and 11, viz 88?

(1) 3
(2) 21
(3) 59
(4) 68

lets say the integer is N

1. divide by 8
(N-3)/8 = M (some integer )

2. divide M by 11 (successive division according to stem)

[ M - 7]/11
= [((N -3)/8 ) - 7] / 11

= [ (N - 59) / (8 *11)

= (N -59)/88

now lets put it in standard from

(N - 59)/88 = K ( some integer)

N = 88 K + 59

Thus , 59 is the remainder...

agree now? :)

thanks
praetorian

Last edited by Praetorian on 07 Oct 2003, 23:59, edited 3 times in total.
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 [#permalink] New post 30 Sep 2003, 05:12
Well the first number that satisfies the conditions is 51. But it's not a choice here so not sure now...
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 [#permalink] New post 04 Oct 2003, 02:25
N=8n+3
N=11k+7
------------
N=88z+R, in which 0<=R<=87
88z is divisible by 8 and 11; therefore, the remainders of divisions of N by 8 and 11 are equal to the remainders of divisions of R by 8 and 11.

{R=8p+3
{R=11q+7
{0<=R<=87

R=51
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 [#permalink] New post 04 Oct 2003, 02:32
stolyar how do u get R=51 from the equations below:
{R=8p+3
{R=11q+7
{0<=R<=87

One method is trial and error(write all mutiples of 8 & 11 side by side and check), is there any other way...?
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 [#permalink] New post 04 Oct 2003, 06:47
I employed this method, but usually there should be a correct answer among options, so we just have to check five numbers, no more, maybe less.
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 [#permalink] New post 07 Oct 2003, 15:25
stolyar wrote:
I employed this method, but usually there should be a correct answer among options, so we just have to check five numbers, no more, maybe less.


official answer : 59

When a number is successively divided by two divisors d1 and d2 and two remainders r1 and r2 are obtained, the remainder that will be obtained by the product of d1 and d2 is given by the relation

d1r2 + r1.

Where d1 and d2 are in ascending order respectively and r1 and r2 are their respective remainders when they divide the number.

In this case, the d1 = 8 and d2 = 11. And r1 = 3 and r2 = 7
Therefore, d1r2 + r1 = 8*7 + 3 = 59.


thanks
praetorian

Last edited by Praetorian on 07 Oct 2003, 18:17, edited 1 time in total.
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 [#permalink] New post 07 Oct 2003, 17:42
But how can 59 be the answer, when:

59 mod 11 = 4
59 mod 88 = 59
?!?!
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 [#permalink] New post 07 Oct 2003, 18:20
wonder_gmat wrote:
But how can 59 be the answer, when:

59 mod 11 = 4
59 mod 88 = 59
?!?!


A brilliant guy pointed out that the "successfully " should be "successively"

The answer is indeed 59

thanks
praetorian
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 [#permalink] New post 08 Oct 2003, 04:47
praetorian123 wrote:
wonder_gmat wrote:
But how can 59 be the answer, when:

59 mod 11 = 4
59 mod 88 = 59
?!?!


A brilliant guy pointed out that the "successfully " should be "successively"

The answer is indeed 59

thanks
praetorian


Makes sense if it's successively.
  [#permalink] 08 Oct 2003, 04:47
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stolyar , i am assuming that you did not see my post

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