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@ satya
. 36 is the wrong answer because 36%9 != 2
. Let write some multiple
The multiple of 7 plus 1 are as follows
8,15,22,29,36,43,50,57,64,…….,337,344,351,…..
The multiple of 9 plus 2 are as follows 11,20,29,38,47,56,65,74,………,335,344,353…..
The multiple of 11 plus 3 are as follows
14,25,36,47,58,69,80,91,……..,333,344,355…

So, 344 is the number that is in common in all the three series so the number is 344.

there is no shortcut form for this question . the way to find the answer is the trail and error method. for that start with the 7 .check the 7+1 =8, 7*7+1=50, and 7*7*7+1=344. check the number 344 with 9 and 11 it gives the remainders 2 and 3 respectively. so 344 is the one of the required number . to calculate the remaining numbers the formula is lcm(7,9,11)*any real number + 344.
so the answer is the numbers 693*x+344. where x=0,1,2,……

I think trail and error method is the better solution for this problem. for that just start with 7. and consider the numbers which will give remainder 1 when divided with7. for that consider the powers of the 7. so 8,50,344,….
check with the 9 and 11. among them 344 is the one of the correct answers.
Thre are somany numbers which will give reminder 1,2,3 when divided with 7,9,and 11 respectively. to get all the numbers the formula is :
lcm(7,9,11)*x+344.
i.e 693*x+344, where x may be any positive integer.

Ans is : 344+k*693.
Point is that the LCM of the given 3 nos 7,9 and 11 is 693 , so in this case we can arrive at the no by observing all the common multiples upto 693 . Once we get that the series will repeat itself after every 693 times. For eg. next no after 344 will be 1037 which also satisfies this relation. In a nutshell all we have to do is write down multiples upto 693.

Submitted By: alex — October 7, 2008

+12 votes + - 344

Submitted By: Umang Rustagi — October 8, 2008

+12 votes + - 693*Q+344

where Q= any whole number

Submitted By: evg — October 8, 2008

-2 votes + - 344

Submitted By: Ratnam — October 11, 2008

+7 votes + - 344 is the number which leaves remainders as 1,2,3 when we divide 344 with 7,9 and 11 respectively

Submitted By: satya — October 11, 2008

-49 votes + - 36

Submitted By: Ammad Alam — November 7, 2008

+29 votes + - @ satya

. 36 is the wrong answer because 36%9 != 2

. Let write some multiple

The multiple of 7 plus 1 are as follows

8,15,22,29,36,43,50,57,64,…….,337,344,351,…..

The multiple of 9 plus 2 are as follows 11,20,29,38,47,56,65,74,………,335,344,353…..

The multiple of 11 plus 3 are as follows

14,25,36,47,58,69,80,91,……..,333,344,355…

So, 344 is the number that is in common in all the three series so the number is 344.

Submitted By: bharathi — November 29, 2008

+15 votes + - there is no shortcut form for this question . the way to find the answer is the trail and error method. for that start with the 7 .check the 7+1 =8, 7*7+1=50, and 7*7*7+1=344. check the number 344 with 9 and 11 it gives the remainders 2 and 3 respectively. so 344 is the one of the required number . to calculate the remaining numbers the formula is lcm(7,9,11)*any real number + 344.

so the answer is the numbers 693*x+344. where x=0,1,2,……

Submitted By: bharathi — December 1, 2008

+3 votes + - I think trail and error method is the better solution for this problem. for that just start with 7. and consider the numbers which will give remainder 1 when divided with7. for that consider the powers of the 7. so 8,50,344,….

check with the 9 and 11. among them 344 is the one of the correct answers.

Thre are somany numbers which will give reminder 1,2,3 when divided with 7,9,and 11 respectively. to get all the numbers the formula is :

lcm(7,9,11)*x+344.

i.e 693*x+344, where x may be any positive integer.

Submitted By: Abhinav — December 16, 2008

+5 votes + - Ans is : 344+k*693.

Point is that the LCM of the given 3 nos 7,9 and 11 is 693 , so in this case we can arrive at the no by observing all the common multiples upto 693 . Once we get that the series will repeat itself after every 693 times. For eg. next no after 344 will be 1037 which also satisfies this relation. In a nutshell all we have to do is write down multiples upto 693.

Submitted By: Muralidhar — December 21, 2008

-9 votes + - A simple mathematical formula.

N%7 = 1 ==> (7-1) is x

N%9 = 2 ==> (9-2) is y

N%11 = 3 ==> (11-3) is z

now N = GCD of (x,y,z) and it can be applied to any number of constraints.

Submitted By: Muralidhar — December 21, 2008

-9 votes + - Sorry its the GCM of x,y,z and not GCD.

Submitted By: tabinda — December 27, 2008

-4 votes + - N%7=1

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