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eaakbari
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Dividing 26 Mar 2010, 05:13
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Find least number which when divided by 20,25,35,40 leaves remainders 14,19,29,34.


I dont have OE or OA. Someone do tell me how to approach this



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eaakbari
Find least number which when divided by 20,25,35,40 leaves remainders 14,19,29,34.


I dont have OE or OA. Someone do tell me how to approach this
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I got the answer as 1394. Will post the strategy shortly. it took me more than 2 mins, still looking for a shorter approach.

-Ravi
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Dividing Updated on: 26 Mar 2010, 20:13
Originally posted by crack700 on 26 Mar 2010, 17:49.
Last edited by crack700 on 26 Mar 2010, 20:13, edited 1 time in total.
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eaakbari
Find least number which when divided by 20,25,35,40 leaves remainders 14,19,29,34.


I dont have OE or OA. Someone do tell me how to approach this
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If a number divided by 20 leaves a remainder of 14 it can be expressed as 20a-6 or 20a+14.
Similarly if the number gives remainders of 19,29,34 when divided by 25,35,40 respectively it can be expressed as 25b-6 or 25b+19 , 35c-6 or 35c+29 , 40d-6 or 40d+34.

Let's call the number N then N= 20a-6 = 25b-6 = 35c-6 = 40d-6 ( Took the -6 form since it's common across the 4 divisors)

The least value for N would be the (least common multiple of 20 ,25 ,35 ,40) - 6 = 1400-6 = 1394

Thanks
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Dividing 26 Mar 2010, 18:45
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crack700
eaakbari
Find least number which when divided by 20,25,35,40 leaves remainders 14,19,29,34.


I dont have OE or OA. Someone do tell me how to approach this

If a number divided by 20 leaves a remainder of 14 it can be expressed as 20a-6 or 20a+14.
Similarly if the number gives remainders of 19,29,34 when divided by 25,35,40 respectively it can be expressed as 25b-6 or 25b+19 , 35c-6 or 35c+29 , 40d-6 or 40d+34.

Let's call the number N then N= 20a-6 = 25b-6 = 35b-6 = 40b-6 ( Took the -6 form since it's common across the 4 divisors)

The least value for N would be the (least common multiple of 20 ,25 ,35 ,40) - 6 = 1490-6 = 1394

Thanks
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That's nice, Kudosss

Could get to the answer but took little over 2 mins, but this is the right approach. Thanks
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Dividing 26 Mar 2010, 20:12
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eaakbari
Find least number which when divided by 20,25,35,40 leaves remainders 14,19,29,34.


I dont have OE or OA. Someone do tell me how to approach this

If a number divided by 20 leaves a remainder of 14 it can be expressed as 20a-6 or 20a+14.
Similarly if the number gives remainders of 19,29,34 when divided by 25,35,40 respectively it can be expressed as 25b-6 or 25b+19 , 35c-6 or 35c+29 , 40d-6 or 40d+34.

Let's call the number N then N= 20a-6 = 25b-6 = 35c-6 = 40d-6 ( Took the -6 form since it's common across the 4 divisors)

The least value for N would be the (least common multiple of 20 ,25 ,35 ,40) - 6 = 1400-6 = 1394

Thanks

That's nice, Kudosss

Could get to the answer but took little over 2 mins, but this is the right approach. Thanks
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Thank you. With multiple choices in the exam substitution might be a better approach to do it faster
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eaakbari
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Dividing 27 Mar 2010, 02:12
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Wow thats a good answer but could you tell me how you would solve it if they didnt have 6 as common, it would be much more complex then


Thanks
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Dividing 10 Sep 2010, 03:26
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Hi,

Any number N when divided by d1,d2,d3 etc leaving remainders r1,r2,r3 etc respectively such that d1-r1=d2-r2=d3-r3 = "v" (say) then the number "N=K-v" where K=LCM(d1,d2,d3,....)

In this case divisors are (20,25,35,40) and remainders (14,19,29,34)
K = LCM (20,25,35,40) = 1400
v= 20-14 = 25-19 = 35-29 = 40-34 = 6

So our N = 1400 - 6 = 1394

This approach took me less than 1 min.

Hope it helps
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Ya, I would rather go by choices and eliminate answers
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Dividing 13 Sep 2010, 03:13
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This one is simple LCM problem... i know a bunch of this problem and how to solve it... i will post those later on
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