Finding Remainders - MCQ Test


5 Questions MCQ Test UPSC Prelims Paper 2 CSAT - Quant, Verbal & Decision Making | Finding Remainders - MCQ Test


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QUESTION: 1

What is the remainder when [7(4n + 3)]*6n is divided by 10; where ‘n’ is a positive integer.

Solution:

[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’)

QUESTION: 2

Numbers 11284 and 7655, when divided by a certain number of three digits, leave the same remainder. Find that number of 3 digits and their sum.

Solution:

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’)

QUESTION: 3

64329 is divided by a certain number. While dividing, the numbers 175, 114 and 213 appear as three successive remainders, the divisor is?

Solution:

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.

QUESTION: 4

Find smallest number that leaves remainder 3, 5, 7 when divided by 4, 6, 8 respectively.

Solution:

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’)

QUESTION: 5

Find smallest number that leaves remainder 3, 4, 5 when divided by 5, 6, 7 respectively and leaves remainder 1 when divided by 11.

Solution:

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’)

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