Finding Remainders - MCQ Test UPSC Prelims Paper 2 CSAT MCQs & solutions

[7^{(4n + 3)}]6ⁿ
= 7⁴ⁿ x 7³ x 6ⁿ
= 49²ⁿ x 7³ x 6ⁿ
= When each factor is divided by 10 the remainders in each case = (-1)²ⁿ, 3, and 6                                                   (6 when raised to the power of any natural number is divided by 10 always gives remainder as 6 itself)

So, all the remainders thus found above are 1, 3 and 6
So their multiplication= 1*3*6= 18
So the remainder after 18 has been divided by 10 = 8 (option ‘C’)

One has to remember that each factor of the difference of two numbers gives the same remainder if those numbers are divided by it.

Now the difference here = 11284 – 7655 = 3629
Factors of 3629 are 1, 19, 191 and 3629
But we have to find the three digit number here, so 191 is the required number and their sum is 1+9+1 = 11 (option ‘A’)

We have three remainders, means the number comprising of the first digits i.e. 643 was divided first and we got 175 as the remainder.

Now according to DIVIDEND = DIVISOR x QUOTIENT + REMAINDER
=> DIVISOR x QUOTIENT = DIVIDEND – REMAINDER
=> DIVISOR x QUOTIENT = 643 – 175 = 468

We see that 468 is divisible by 234 only among all the answer options; so 234 (option ‘C’) is the divisor we need.

See carefully the difference between the divisor and remainder is having a certain trend. i.e.4 – 3 = 6 – 5 = 8 – 7=1

In such questions, take LCM of divisors and subtract the common difference from it.Now the LCM of 4, 6, 8 = 24Therefore the required number here = 24 – 1 = 23 (option ‘A’)

We have just seen above in TYPE-2 how to tackle the first part of the questionThus the number for the first part would be the [(LCM of 5, 6, 7) – (Common difference of divisors and their remainders)] i.e. 210 – 2 = 208

Here now, we have one more condition to satisfy i.e. remainder 1 when divided by 11

Here we should remember that if LCM of the divisors is added to a number; the corresponding remainders do not change i.e if we keep adding 210 to 208… the first 3 conditions for remainders will continue to be fulfilled.

Therefore now, let 208 + 210k be the number that will satisfy the 4th condition i.e. remainder 1 when (208 + 210k)/11

Now let’s see how
The expression (208 + 210k)/11 = 208/11 + 210k/11
Now the remainder when 208 is divided by 11 = 10
And remainder when 210k is divided by 11 = 1*k = k

Therefore the sum of both the remainders i.e. 10 + k should leave remainder 1 on division of the number by 11
Obviously k = 2
Hence the number = 208 + 210*2 = 628 (option ‘D’)