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The number must have 4 in its unit's place and that it must be an odd multiple of 25 or 35 when 19 or 29 is subtracted from it, respectively. The LCM of 20, 25, 35, and 40 is 1400, which is not an odd multiple of 25 or 35. Hence, for me, this is an impossible question since no multiple of 20 or 40 can be an odd multiple of 25 or 35. Moreover, GMAT hardly ever tests remainders so profoundly. Besides, you can still repair your question to get my (not so useful for GMAT) explanation to this great kind of remainders question, generally tested on the Combined Admission Test in India.eaakbari wrote: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
yes, the answer is correct. One short-cut for this kind of problem :eaakbari wrote:I did get the answer after some more time.
IMO its right. Please give me your views
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
But this is a very complex method of solving and no way i could do this in 2 minutes. Please do tell me your method of solving for the CAT or whatever.
Thanks
in that case, the problem would be too difficult and would not be tested on the GMAT. So, the pragmatic way will be not to worry much about them.eaakbari wrote:Thanks Harsha but I have 1 more question
What would happen if they didnt have the common difference of 6. How would we go about it then?
Oh yes! Didn't think the other way round. The numbers are correct, and my (CAT) explanation is just same as yours, but we further minimize the efforts on CAT just as what harshavardhanc did.eaakbari wrote:I did get the answer after some more time.
IMO its right. Please give me your views
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
But this is a very complex method of solving and no way i could do this in 2 minutes. Please do tell me your method of solving for the CAT or whatever.
Thanks
