ARITHMETICMATHSRemainder & Divisibility

QUESTIONS ON REMAINDERS & DIVISIBILITY (PART-IV)

QUESTIONS ON REMAINDERS & DIVISIBILITY (PART-IV)

QUERY 31

When a number is divided by 7 the remainder is 4 and when the number is divided by 6, the remainder is 3. What will be the number?
A) 32
B) 46
C) 39
D) 53

MAHA GUPTA
This query is similar to the query No. 8 (TYPE-2) given above, where one needs to find the number when the differences between the divisors and the remainders are same.

In such questions, take LCM of divisors and subtract the common difference of divisors and remainders from it to find the required number.

Now the LCM of 7 and 6 = 42
And the common difference of divisors and remainders = 3
Therefore the required number here = 42 – 3 = 39 (option ‘C’)

NOTE: Better to do such question using answer options.

QUERY 32

When a number is divided by 24, the reminder is 16. The reminder when the same number is divided by 12 is?

A) 1
B) 2
C) 3
D) 4

MAHA GUPTA
We see that the second divisor 12 is a factor of the first divisor 24; when such is the case the given remainder is divided by the second divisor. The remainder thus got is the required remainder.

Hence the required remainder = the remainder of 16/12 = 4 (option ‘D’)

QUERY 33

When a number is successively divided by 9, 11 & 13 remainders left respectively are 8, 9 & 8. if the sequence of divisors is reversed; then what will be the remainders?

A) 10, 2, 5
B) 10, 1, 6
C) 7, 2, 9
D) 5, 3, 13

MAHA GUPTA
For such questions start doing from the rear end and better if the number to be found is taken as the smallest.

So, we want 8 as the remainder with 13 as the divisor; of course the smallest such number is 8 itself.

Obviously this 8 must have come as the quotient when the number was divided by 9 because the number was divided successively.
Therefore the number must have been 8*11 + 9 = 97 ……….(see 9 is the remainder with 11 as divisor)

It’s clear this 97 was the quotient when the number was divided by 9
Therefore the number must have been 97*9 + 8 = 881 ……….(see 8 is the remainder with 9 as divisor)

Hence the number is 881

Now we have to find the remainders when this number i.e. 881 is successively divided by reverse of 9, 11 & 13 i.e. 13, 11 & 9

The remainder when 881 is divided by 13 = 10; the quotient being 67

Now we have to find the remainder when 67 is divided by 11; obviously it’s 1; the quotient being 6

Now we have to find the remainder when 6 is divided by 9; obviously it’s 6 itself

So the required answer is 10, 1, 6 (options ‘B’)

QUERY 34

Find remainder if (3232)32 is divided by 7

A) 2
B) 3
C) 4
D) 1

MAHA GUPTA
Remainder obtained after dividing 32 by 7 is 4 (so when (432)32 is divided by 7 will give us the same remainder when (3232)32 is divided by 7

Now (432)32
= [(43)10)*42]32
= [(6410)*42]32

Remainder obtained after dividing 64 by 7 is 1 (so when [(110)*42]32 is divided by 7 will give us the same remainder when [(6410)*42]32 is divided by 7

Now [(110)*42]32
= (42)32
= 464
= (43)20)*44
= (6420)*44

Remainder obtained after dividing 64 by 7 is 1 (so when [(120)*44 is divided by 7 will give us the same remainder when (6420)*44 is divided by 7

Now (120)*44
= 256
You see when 256 is divided by 7 the remainder is 4 (option ‘C’)

TRICK
(3232)32
= 321024
= 2(5 × 1024)
= 25120
= 25118 × 2²
= 81706 × 2²
When 8 is divided by 7 remainder is 1
So total remainder when 81706 divided by 7 is 11706 = 1
And when 81706 × 2² divided by 7 remainder is 1 × 4 = 4 (option ‘C’)

QUERY 35

Find the least possible 5 digit number which when divided by 2, 4, 6, 8 leaves remainders 1, 3, 5, 7 respectively.

A) 10009
B) 10011
C) 10007
D) 10013

MAHA GUPTA
Here you see that the difference between the divisors and their remainders is same, you see it’s 1 in every case. Therefore first you should know how to find out the smallest number when the difference between the divisors and their remainders is same.

In such questions, take LCM of divisors and subtract the common difference from it.
So LCM of 2, 4, 6 and 8 = 24
Common difference = 1

So the smallest number satisfying the above condition = 24 – 1 = 23
But we have to find the smallest 5 digit such number.

Now the next higher number = 24x – 1; where x is any natural number greater than 1.
Now put the least possible value of x such that 420x – 1 greater than 10,000 or equal to 10,000
You can find it dividing 10000 by 24; hence it’s = 10,000/24 = greater than 416; so it’s 417

Now the required number =24x – 1 = 24*417 – 1 = 10,007 (option ‘C’)

TRICK 
Do it by answer options. Try every given answer and do, it’ll take seconds hardly.

QUERY 36

Find the number when it is divided by 9, 11, 13 leaves remainders 1, 2, 3 respectively.

A) 640
B) 620
C) 649
D) 658

MAHA GUPTA
We can see there’s no relation among these divisors-remainders sets

In such a case we take one of the cases and target another case…e.g. First take the case 3 with 13……..and target 2 with 11.

We see that when the number is divided by 13 the remainder is 3 (means 3 is the smallest number that we get as remainder. So all numbers of the form 3 + 13k will give 3 as remainder on division by 13.

Now target 2 with 11…………so we will have to divide 3 + 13k by 11
By doing so the remainder = 3 + 2k
Obviously for minimum value of k is 5; so that we get 2 as remainder on dividing 3 + 2k by 11.
Therefore the number = 3 + 13*5 = 68

Now that 2 conditions are fulfilled, lets target the third condition, i.e. 1 with 9.
Now, if we add LCM of 11 and 13 i.e 143 to 68, 2 conditions that were already satisfied would continue being satisfied…

Hence the number is of the form 68 + 143 k.
Thus the remainder when 68 + 143k is divided by 9 = 5 + 8k
So the minimum value of k = 4
Therefore the number satisfying all 3 condition considered = 68 + 143*4 = 640 (option ‘A’)

NOTE-I: For ease of calculation, start from biggest divisor and gradually move to smaller ones…you’ll always see that last 1-2 conditions will be satisfied automatically.

NOTE-II: There are theorems for solving above type of questions…viz Chinese theorem etc…but I find this approach very practical and easy. And I believe I can tackle any twist in the question devised by Cat Makers through this method.

TRICK 
Do it by answer options. Try every given answer and do, it’ll take seconds hardly.

QUERY 37

If the number 23583ab is completely divisible by 80, then the value of a – b is

A) 2
B) 3
C) 4
D) 0

MAHA GUPTA
Any number must have ‘0’ as its unit digit to be divisible by a number ending in ‘0’. So b = 0.

Now we have to find the value of ‘a’ so that 23583a is divisible by 8.
We know a number is divisible by 8 only if the number formed by the last three digits of this number is either zeroes or multiple of 8.

Hence 83a must be multiple of 8 here; so a = 2

Now a – b = 2 – 0 = 2 (option ‘A’)

QUERY 38

Find the smallest number when is divide by 8, 5 and 11 leaves remainders 5, 2 and 6 respectively.

A) 237
B) 242
C) 245
D) 337

MAHA GUPTA
We can see there’s no relation among these divisors-remainders sets

In such a case we take one of the cases and target another case…e.g. first take the case 6 with 11……..and target 5 with 8.

We see that when the number is divided by 11 the remainder is 6 (means 6 is the smallest number that we get as remainder. So all numbers of the form 6 + 11k will give 6 as remainder on division by 11.

Now target 5 with 8…………so we will have to divide 6 + 11k by 8
By doing so the remainder = 6 + 3k
Obviously for minimum value of k is 5; so that we get 5 as remainder on dividing 6 + 3k by 8.
Therefore the number = 6 + 11*5 = 61

Now that 2 conditions are fulfilled, lets target the third condition, i.e. 2 with 5.
Now, if we add LCM of 8 and 11 i.e 88 to 61, 2 conditions that were already satisfied would continue being satisfied…

Hence the number is of the form 61 + 88 k.
Thus the remainder when 61 + 88k is divided by 5 = 6 + 3k
So the minimum value of k = 2 so that remainder is 2 when the number is divided by 5
Therefore the number satisfying all 3 condition considered = 61 + 88*2 = 237 (option ‘A’)

NOTE-I: For ease of calculation, start from biggest divisor and gradually move to smaller ones…you’ll always see that last 1-2 conditions will be satisfied automatically.

NOTE-II: At times such a question can be solved by just seeing the answer options.

TRICK
Try every given answer option, it’d hardly take seconds.

QUERY 39

Find the smallest six digit number which is exactly divisible by 15, 20, 25 & 30.

A) 100235
B) 100200
C) 101115
D) 101215

MAHA GUPTA
The smallest 6 digit number is 1,00,000
When a number is the smallest in any number of digits and divisible by some numbers, divide it by the LCM of those divisors. On division you will get a number in a fraction. Just make it rounded off to the nearest higher number and multiply it by that LCM.

Here LCM of 15, 20, 25 and 30 = 300
1,00,000/300 = 333.33
The next rounded off number to it = 334

So the required number = 334*300 = 1,00,200 (option ‘B’)

QUERY 40

165 + 215 is divisible by?

A) 31
B) 13
C) 27
D) 33

MAHA GUPTA
165 = (24)5 = 220 = 25
x 215 = 32(215)

Thus, 165 + 215 = 32(215) + 215 = 215 x (32 + 1) = 215 x 33

Hence, 33 is a factor of the given expression; means divides it fully (option ‘D’)

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Maha Gupta

Maha Gupta

Founder of www.examscomp.com and guiding aspirants on SSC exam affairs since 2010 when objective pattern of exams was introduced first in SSC. Also the author of the following books:

1. Maha English Grammar (for Competitive Exams)
2. Maha English Practice Sets (for Competitive Exams)