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A certain number when successively divided by 8 and 11 [#permalink]
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?

Code:

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 :)

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

_________________

ash
________________________
I'm crossing the bridge.........

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" _________________

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...

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.

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. _________________

ash
________________________
I'm crossing the bridge.........

Originally posted on MIT Sloan School of Management : We are busy putting the final touches on our application. We plan to have it go live by July 15...