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

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CEO
Joined: 15 Aug 2003
Posts: 3469
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Kudos [?]: 702 [0], given: 781

stolyar , i am assuming that you did not see my post [#permalink]  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.
Manager
Joined: 26 Aug 2003
Posts: 233
Location: United States
Followers: 1

Kudos [?]: 2 [0], given: 0

Well the first number that satisfies the conditions is 51. But it's not a choice here so not sure now...
SVP
Joined: 03 Feb 2003
Posts: 1608
Followers: 6

Kudos [?]: 76 [0], given: 0

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
Senior Manager
Joined: 21 Aug 2003
Posts: 258
Location: Bangalore
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Kudos [?]: 4 [0], given: 0

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...?
SVP
Joined: 03 Feb 2003
Posts: 1608
Followers: 6

Kudos [?]: 76 [0], given: 0

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.
CEO
Joined: 15 Aug 2003
Posts: 3469
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Kudos [?]: 702 [0], given: 781

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.

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.
Manager
Joined: 26 Aug 2003
Posts: 233
Location: United States
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Kudos [?]: 2 [0], given: 0

But how can 59 be the answer, when:

59 mod 11 = 4
59 mod 88 = 59
?!?!
CEO
Joined: 15 Aug 2003
Posts: 3469
Followers: 61

Kudos [?]: 702 [0], given: 781

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"

thanks
praetorian
Manager
Joined: 26 Aug 2003
Posts: 233
Location: United States
Followers: 1

Kudos [?]: 2 [0], given: 0

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"

thanks
praetorian

Makes sense if it's successively.
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