m = 20,25,35,40x + 14,19,29,34


so the least number with the resultant remainder will be when x =0

so answer = 14,19,29,34

Sorry but I am still unsure. Could you give a more detailed explanation

m/ 20,25,35,40 --> gives us a remainder of 14,19,29,34



thus we have to multiply 20,25,35,40 with a certain number x and add 14,19,29,34 to give us m


m = 20,25,35,40x + 14,19,29,34

Now that we have got the equation we will try to minimize m which will only happen when x = 0.

If i list down the equations

20x1 + 14 = m
25x2 + 19 = m
35x3 + 29 = m
40x4 + 34 = m

x will be different in all cases as it is merely the quotient which will differ and even if we do put all the x's as zero we wil get different values of m.

Please clarify

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

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.

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

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

yes, the answer is correct. One short-cut for this kind of problem :

Observe the difference between the divisor and the corresponding remainder

(20-14) = 6
(25-19) = 6
(35-29) = 6
(40-34) = 6

So, the least number satisfying the condition in question will be LCM(20,25,35,40) - 6 = 1400-6

=1394.

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?

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?

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.

the problems on HCF and LCM with common/same remainders are the only ones which can be done within the time limit and they may appear in the real test (though, the probability is quite less).

Thanks a bunch

E

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

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.