rahulp11 wrote:
Juliaz wrote:
Why is it wrong to transform 49*64/6 into 49*32/3 (divide numerator and denominator by 2)
Then remainder would be 2..
(48+1)(30+2)/3, 48 and 30 are multiples of 3, remainder=2
Why is it not possible to simplify the equation?
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BrentGMATPrepNow chetan2uPlease confirm if it's wrong to reduce the fraction for such remainder problems. Are we required to transform the numerator only while retaining the divisor of the question stem?
The logic used in the above solution is the same as if the denominator were to be 6 instead of 3, yet the remainders we get are different. Please help.
Yes, it would, and the remainder will actually get reduced by the divisor times. If divisor(common factor) is 5, then you will get actual remainder by multiplying the remainder you get in simplified form by 5.
Why does it happen?
Say, we are looking for remainder when 100 is divided by 15.
This will give us 10 as the remainder.
But if we reduce \(\frac{100}{15}\) to \(\frac{20}{3}\), that is we divide both by common term 5, the remainder is 2. The actual remainder is nothing but 2*5.
Now 100 is 15*6+10, so \(\frac{15*6+10}{15}=\frac{15*6}{15}+\frac{10}{15}\). So remainder is given by 5/15. But we are reducing it to simplified term 2/3, thereby reducing the remainder from 10 to 10/5 or 2