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A certain number when successively divided by 8 and 11

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A certain number when successively divided by 8 and 11 [#permalink] New post 06 Jul 2004, 15:48
A certain number when successively 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?

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EDIT by Praet: the correct term is successively, not successfully...boksana, i remember posting this problem myself. it was from some website.what does successfully divided mean anyway :)
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 [#permalink] New post 06 Jul 2004, 16:02
8-3 = 4
and
11-7=4

so we can get the number
using the formula i know


the least number which satisfies the conditions is LCM of 8,11 minus 4

88-4 = 84

84/88 leaves are remainder of 84
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 [#permalink] New post 06 Jul 2004, 16:09
No :( Try again
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Re: more remainders [#permalink] New post 06 Jul 2004, 16:17
Solved it by enumeration though not the best way to solve this problem

3 mins

X=8m+3=11n+7

or 8m=11n+4 (this has to be a multiple of 8, which is possible only if 11n is a mutiple of 4)

Hence we have 44 (+4 div by 8) 88 (+4 not div by 8) 132 (+4 div by 8) ....

Hence X can be 44+4=48 or 132+4=136 both divisible by 8

Dividing X by 88 leaves remainder of 48 in both cases

Anyone knows a better way to solve the problem???
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 [#permalink] New post 06 Jul 2004, 16:23
You are wrong, the OA is different
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 [#permalink] New post 06 Jul 2004, 16:30
boksana , u have big TYPO there ;-)

successfully should have been successively........gotcha :)

now heres my solution, A is the original number , B and C are intermediates.

A divided by 8 gives quotient B and leaves remainder 3 and B divided by 11 gives quotient C leaves remainder 7

8 | A |
_________
11 | B | 3
_________
| C | 7

let C = 1

B = 11C +7 = 18
A = 8.18 + 3 = 147

147/88 leaves remainder 59
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 [#permalink] New post 06 Jul 2004, 16:55
Sorry. But I didn't type I've copied from a source. :oops:
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 [#permalink] New post 06 Jul 2004, 19:48
I've got 51 as well. But OA is 59 :? Wiered :roll:
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 [#permalink] New post 06 Jul 2004, 20:06
59 is impossible. 51 must be it. Divide 59 by 11 and you get remainder 4.
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 [#permalink] New post 06 Jul 2004, 21:10
theere is a typo in the question......replace successfully by successively.....and then try to solve it.

boksana wrote:
A certain number when successfully 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?


heres my solution.....after i used the word "successively" and then i got the answer.


ashkg wrote:
boksana , u have big TYPO there ;-)

successfully should have been successively........gotcha :)

now heres my solution, A is the original number , B and C are intermediates.

A divided by 8 gives quotient B and leaves remainder 3 and B divided by 11 gives quotient C leaves remainder 7

8 | A |
_________
11 | B | 3
_________
| C | 7

let C = 1

B = 11C +7 = 18
A = 8.18 + 3 = 147

147/88 leaves remainder 59

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 [#permalink] New post 07 Jul 2004, 05:29
ashkg that was fantastic!
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 [#permalink] New post 08 Jul 2004, 00:16
Paul or Boksana
Can you please explain how did you get 51?

Question still holds good if we consider the word "Successfully". What would be the answer in that scenario?
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 [#permalink] New post 16 Jul 2004, 02:16
boksana wrote:
I've got 51 as well. But OA is 59 :? Wiered :roll:


Ans is indeed 59.

let X be the number

so X = 8 m + 3
but m = 11 n + 7 -> remember successively

So, X = 8(11n+7) + 3
X = 88n + 59

remainder when X is divided by product of 8 and 11(88) is 59

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 [#permalink] New post 16 Jul 2004, 03:40
I don't get someting:

This is how I solved it:
n=8k + 3.
Now, you have to divide the result you get after dividing n by 8, by 11:

The result from the first division is (k+3/8) and not k as Amith says.
For example, if you divide 27 by 8: the answer is 3+ 3/8. and not 3.

Sow now we divide again: K + 3/8 = 11J + 7.
So K=11J + (6+5/8)

Therefore, n=8*(11J + 6+5/8 ) + 3.
n = 88J + 56.

So the remainder is 56.

What am i doing wrong ?
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 [#permalink] New post 16 Jul 2004, 04:04
Well, successfully changes the meaning of the question. Successfully means that when X/8, you get a remainder of 3 and when the same X is divided by 11, you get a remainder of 7.
I just laid out the different possibilities until I reached 51

multiples of 8: 8-16-24-32-40-48
But remainder is 3 when divided by 8 so you have to add 3 to the previous line: 11-19-27-35-43-51
multiples of 11: 11-22-33-44
But remainder is 7 when divided by 11 so you have to add 7 to the previous line: 18-29-40-51

51 would have been my answer for "successfully"
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 [#permalink] New post 16 Jul 2004, 05:22
Paul wrote:
Well, successfully changes the meaning of the question. Successfully means that when X/8, you get a remainder of 3 and when the same X is divided by 11, you get a remainder of 7.
I just laid out the different possibilities until I reached 51

multiples of 8: 8-16-24-32-40-48
But remainder is 3 when divided by 8 so you have to add 3 to the previous line: 11-19-27-35-43-51
multiples of 11: 11-22-33-44
But remainder is 7 when divided by 11 so you have to add 7 to the previous line: 18-29-40-51

51 would have been my answer for "successfully"


Paul, Sorry if i am wrong here...

Here is my proof which will contradict your answer.

According to you,
X = 8 m + 3
and X = 11n + 7

So, 8m+3 = 11n + 7
=> 8m = 11n + 4

since 11 is prime, one can see that for each n = 4,8,12,16
m will be integer. So we will get many numbers which willl satisfy the given requirement ( remainder of 3 when devided by 8 and remainder of 7 when devided by 11)..

i.e I get X = { 51, 95 , 139, 183 ... } which will satisfy the given requirements. But when devided by 88(product of 8 and 11 as asked in the question), I get 2 different remainders {51, 7, 51, 7 ..} respectively for the above numbers....So with your logic(of considering 'successfully' instead of 'successively'), we can get either 7 or 51 as remainder. So...

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.. [#permalink] New post 16 Jul 2004, 05:25
that was my post(forgot to login before replying) :-)
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 [#permalink] New post 16 Jul 2004, 06:41
agree with 4 and 12 but 8 and 16 do not yield an integer. Hence, only possible remainder is 51 and X is 51 - 139 - 227, etc
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 [#permalink] New post 16 Jul 2004, 06:42
amithmp is right here. It is 59, and his method is right on.

But for the record, if it were successfully, I'd stay away from all the formulas and just find multiples of 11 that give a remainder of 7 until hitting the one that also give a remainder of 3 when divided by 8.

All these formulas on this forum must be overwhelming to the struggling student. There are so many times when problems can be broken down into easy english.

With remainders, once a number works for both, most of the time, you've got it.

So numbers that have remainder 7 with 11 are 7, 18, 29, 40, 51. And 51 works for 8, too. Divide 51 by 88, and the remainder is 51. The next one would HAVE TO BE 51 + 88.

Amithmp, 95 doesn't work here, because when dividing by 8, you get remainder 7.

By the way, I started with 11 because it's larger, so there are less multiples to work through. You could start with 8 also, but that'll take more time.
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 [#permalink] New post 16 Jul 2004, 09:36
I don't think that any formula is being used here by anyone above. They are all simple maths concepts and I strongly feel that a student should concentrate more on understanding the concepts than just getting the answer by picking numbers.

Picking numbers may work for a good number of times, but if a student learns both ways then nothing can stop him from solving the question.

Here's my solution again. It doesn't use any formula but the actual concept of division.

One should understand the quesiton first that you divide the original number(say A) by 8 and get a remainder 3 and a quotient B.
Now divide B ( that is what is meant by successively) by 11 and you get a quotient ( say C) and a remainder 7.

now we know that

dividend = quotient*divisor + remainder

so

A = 8B + 3
B = 11C + 7

now there can be many possibilities for getting a set of numbers satisfying the equations. We shall go for the first set. This is by taking C=1

so B = 18
use this value in first equation.
A = 8.18 + 3 = 147

So remainder = 147/88 = 59

If one understands this then this method is faster than picking numbers.
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  [#permalink] 16 Jul 2004, 09:36
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