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When a number divided by 35, the remainder is 2;divided by

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Senior Manager
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When a number divided by 35, the remainder is 2;divided by [#permalink] New post 06 Sep 2006, 01:37
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When a number divided by 35, the remainder is 2;divided by 31, the remainder is 3. The number is:


*** Wanted to know the way to do this apart from plugging in numbers...
Director
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 [#permalink] New post 06 Sep 2006, 03:26
Construct the two equations and solve for X...

x/35=q+2
x/31=q+3.....

I probably made a mistake somewhere in the process and got the number 78.75....
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 [#permalink] New post 06 Sep 2006, 06:02
35m+2=31k+3

35m-1=31k

k= (35m-1)/31= (4m-1)/31 +m

Since k and m are an integers , 4m-1 must be a multiple of 31

m=8 is the smallest positive value that will do the trick- k would be 9

35(8)+2=282=31(9)+3

I hope this solve ur problem
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 [#permalink] New post 06 Sep 2006, 21:13
SimaQ wrote:
Construct the two equations and solve for X...

x/35=q+2
x/31=q+3.....

I probably made a mistake somewhere in the process and got the number 78.75....


you are assuming q to be same in two cases - which is not necessary..
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 [#permalink] New post 07 Sep 2006, 09:13
yezz wrote:
35m+2=31k+3

35m-1=31k

k= (35m-1)/31= (4m-1)/31 +m

Since k and m are an integers , 4m-1 must be a multiple of 31

m=8 is the smallest positive value that will do the trick- k would be 9

35(8)+2=282=31(9)+3

I hope this solve ur problem


Hi, remainder problems give me headaches. Can you explain how you get

(4m-1)/31 +m

Only thing I can assume is that 35-31 = 4

Thanks
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 [#permalink] New post 07 Sep 2006, 09:25
I did this way:

n = 35x+2
n = 31y + 3
35x+2 = 31y+3
x = (31y+1)/35

because x is an integer. 31y+1 must have last digit as 0 or 5.
Hence y must be an integer with unit digit 4 or 9.

When y = 4 then (31*4 + 1)/35 i.e 125/35 is not an integer.
when y = 9 then (31*9+1)/35 i.e 280/35 = 8 is an integer.
Hence n = 31 * 9 + 3 = 282
_________________

SAID BUSINESS SCHOOL, OXFORD - MBA CLASS OF 2008

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 [#permalink] New post 07 Sep 2006, 09:30
since

x= 35m+2
and x = 31k+3 where k,m are positive intigers

thus

35m+2 = 31k+3

thus k = 35m+2-3/31 = 35m-1/31

35m-1 could be expressed as 31m+4m-1

and since K is intiger

therfore

31m+(4m-1)/31 is intiger

31m/31 + 4m-1/31 and since 31m is completely divisable by 31

therfore 4m-1 must be divisible by 31

ie least possible value of 4m-1 = 31

therfore 4m =32 ie m = 8
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 [#permalink] New post 07 Sep 2006, 11:29
YES!!!! Now it makes perfect sense!!! thank you yezz & ps_dahiya

So I guess that's a standard technique for remainder problem. "That" being breaking up a variable such as 35m into 31m+4m to simplify into manageable components.

Thanks, very insightful.
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 [#permalink] New post 07 Sep 2006, 11:31
Let me tell you that remainder problems come in a wide range of representations

and there is multiple ways to approach each

so my suggestion is solve as much as u can

take care
  [#permalink] 07 Sep 2006, 11:31
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When a number divided by 35, the remainder is 2;divided by

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