Doubt Clearance Thread: UPSC 2021
"When in doubt, observe and ask questions. When certain, observe at length and ask many more questions."
Created this thread as a one stop solution for all members so that all the doubts wherein any conceptual clarification is required can be solved here.
the_initiated,What is the remainder when 51 x 27 x 35 x 62 x 75 is divided by 100?
(a) 50 (b) 25 (c) 5 (d) 1
how to solve this?
There is a multiple of 2 and 5 so the last digit would be 0. Now the only possible remainder for 100 can be 50.
explaination is this-The expression is: 51 × 27 × 35 × 62 × 75 (51 × 27 × 35 × 62 × 75) / 100 = (51 × 27 × 35 × 62 × 3) / 4 = (51 × 27 × 35 × 31 × 3) / 2 So, we need to find the remainder when 51 × 27 × 35 × 31 × 3 is divided by 2. The value of 51 × 27 × 35 × 31 × 3 must be an odd number, as it is a multiplication of all odd numbers. Hence, when it is divided by 2, we will get 1 as remainder. But we cancelled out 50 earlier. So, the remainder when 51 × 27 × 35 × 62 × 75 is divided by 100 = 1 × 50 = 50 Hence, the correct answer is option (a).
not able to understand why 50 is coming in end
mehuifs,@Jurgen_klopp try to cancel 100 using numerators, then I think 2 would remain in denominator which would give answer 1.
Something wrong with this method.
@Jurgen_klopp you don't cancel out the common factors while taking remainders. Answer is 50 only
Woah, I did not know that.
Jurgen_klopp,
Rambo93,
I had issues with both of them back then and I struggle till date.Complex number, differentiation and integration is far easier.:P
Haha, that's one way to flex about mathematics😛.
UPSC released the official cut off for 2020 prelims.
Cutoff ka maatam toh kab ka mann gya, ab tehrvi ka samay aa gya.
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