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Simple question about remaings

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Source: — Data Sufficiency |
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by liferocks » Sat Apr 10, 2010 7:08 pm
we can write the relation like this
n=mK+f(n)
now from 1,putting n=K+32 and simplyfying we get
km=k+24
or,K=24/(m-1)
since k is integer m-1 has to be a divisor of 24..ie one of 1,24,2,12,3,8,4 and 6
among these only m-1=24 satisfies relation km=k+24
so sufficient

proceeding similarly we get from condition2
km=K+36
or,k=36/(m-1)
factors of 36 are 1,2,3,4,6,9,12,18,36
here for m-1 we get two values(36 and 6 )which satisfies km=k+36
so insufficient

ans is A
now while checking for the factors ,its obvious that the first(24 for cond1 and 36 for 2) will satisfy the relation,so i have checked with only those factors which has values less than 10 to speed up.
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by dxgamez » Sat Apr 10, 2010 7:41 pm
Still not getting the formula for remainders.

Can the formula be used for all remainder qns?
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by liferocks » Sat Apr 10, 2010 7:52 pm
this is the general relation
dividend=divisor*quotient+reminder
you can use this for any reminder related problem.
does it clear the confusion?
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by dxgamez » Sat Apr 10, 2010 9:58 pm
Right. Much clearer. Thanks!
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