Post Reply 
 
Thread Rating:
  • 4 Votes - 1.5 Average
  • 1
  • 2
  • 3
  • 4
  • 5
REMAINDERS & DIVISIBILITY
08-24-2013, 08:29 PM (This post was last modified: 04-14-2014 12:37 AM by Maha Gupta.)
Post: #1
REMAINDERS & DIVISIBILITY
QUERY 1
25^25 is divided by 26, the remainder is?
  1. 24 
  2. 25
RONNIE BANSAL
Use remainder theorem. Here 25^25 is the polynomial and the divisor is 26.
We can write 26 = 25+1 = 25 - ( -1)
So the remainder is ( -1)^25= -1. But we dont take the remainder a negative term; so add it to the divisor.
So the remainder is 26 +( -1) = 25 (option '4')


QUERY 2
When (67^67 + 67) is divided by 68, the remainder is?
  1. 63 
  2. 66 
  3. 67
RONNIE BANSAL
The given expression is in the form x^67 + x ..........(a polynomial in x)
Now 68 = 67 + 1; means x +1

So according to the remainder theorem when a polynomial is divided by another of the form x + 1, the remainder is equal to p( -1) where p is the polynomial itself.
So the remainder is -1^67 + ( -1) = -1 + ( -1) = -2

But the remainder should not be described negative of a number; in such a situation it is added to the divisor to find the actual.
So the remainder is -2 + 68 = 66 (option '3')


QUERY 3
What is the remainder  when [(9^19) + 6] is divided by 8  i.e, [(9^19) + 6)/8. 

RONNIE BANSAL
The given expression is in the form (x^19) + c; where 'c' is a constant              ..........(a polynomial in x).
Now 8 = 9 - 1; means a polynomial in the form of x - 1

So according to the remainder theorem when a polynomial is divided by another of the form x - 1, the remainder is equal to p(1) where p is the polynomial itself. So using remainder theorem, the remainder is (1^19) + 6 = 1 + 6 = 7

Also see it and compare:
When (67^67 + 67) is divided by 68, the remainder is ___
  1. 1       
  2. 63       
  3. 66       
  4. 67 
RONNIE BANSAL
The given expression is in the form x^67 + x              ..........(a polynomial in x)Now 68 = 67 + 1; means a polynomial in the form of x +1 

So according to the remainder theorem when a polynomial is divided by another of the form x + 1, the remainder is equal to p( -1) where p is the polynomial itself.So using the remainder theorem, the remainder is (-1)^67 + ( -1) = -1 + ( -1) = -2 

But the remainder should not be described negative of a number; in such a situation it is added to the divisor to find the actual.So the remainder =  -2 + 68 = 66


QUERY 4
Find the remainders in :
1.     2^11/5
2.    7^7/2^4 


RONNIE BANSAL
1.    211/5
In questions like this we should avoid using the remainder theorem as it can really be difficult when the power of a number (greater than 1) which is derived from the remainder theorem is so high. Better convert the base in powers of such numbers which are easily divisible by the divisor, like:

211 = 24  x  2x  23= 16 x 16 x 8
On dividing 16 by 5 we get 1 as the remainder; and if 8 is divided by 5 we get 3
So the multiplication of all the remainders
= 1 x 1 x 3 = 3 which is our answer. 

2.     77/24      =7x 72 x 72 x 7/16      
= 49 x 49 x 49 x 7/16     
Now the remainder on dividing 49 by 16 =1     

The multiplication of all the remainders 1 x 1 x 1 x 7 = 7 which is the answer

Let's take another example:
Remainder when 2^31 divided by 5?

MAHA GUPTA
2^31 = [(2^4)^7]*2^3 
= (16^7)*8
On dividing 16 by 5 we get 1 as the remainder, so the remainder when 16^7 is divided by 5 = 1^7 = 1; and if 8 is divided by 5 we get 3 as the remainder.

So the remainder needed = 1*3 = 3

NOTE: If the divisor is 2, 5 or 10, one needs to know only the unit digit of the dividend and divide it by the divisor to find the remainder. In this case 5 is the divisor, so just find the unit digit in 2^31, which is 8 of course. So when 8 is divided by 5, the remainder is 3 (answer) 



QUERY 5
What is remainder when the product of 177, 414 & 837 divided by 12  

NIRMAL singh 
In such a question we divide each number by the divisor and keep dividing the product of all remainders thus found by the same divisor till we get a number smaller than that divisor, that number is our answer.

So the remainders of 177, 414 & 837 when divided by 12 are 9, 6 & 9 respectively
Now their product = 9 x 6 x 9
When divided by 12 we can write it as 54 x 9/12 
Again the remainders are 6 x 9
On division of it by 12 it will be 54/12

We see that the remainder is 6; and this is the answer.


QUERY 6
What is the remainder when [7^(4n + 3)]*6^n is divided by 10; where 'n' is a positive integer.
  1. 2       
  2. 4      
  3. 8        
  4. 6 
JAYANTCHARAN CHARAN 
[7^(4n + 3)]*6^n
= 7^4(n) (7^3) (6^n)
= [(49)^2(n)] (7^3) )6^n)
= 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 (answer) 

Harmeet Singh
The expression is [7^(4n + 3) ]* [6^n] 

To find the remainder when any expression is divided by 10 is SIMILAR to just finding the last digit, i.e. unit digit of the expression only. 

Now 7^(4n + 3) = (7^4n )* (7^3) 
Thus When 7^4n divided by 10 for all 'n' will give unit digit 1; and When 7^3 divided by 10 will give units digit 3 as 7^3 = 343  
6^n for all 'n' gives units digit 6 

So, their product is 1*3*6 = 18 
So the remainder after 18 has been divided by 10 = 8 (answer)


QUERY 7
Finding out smallest number which leaves same remainder with different divisors.  Type # 1. 

Find Smallest number other than 4, that leaves remainder 4 when divided by 6, 7, 8 or 9

JAYANTCHARAN CHARAN 
To solve such questions, take LCM of the divisors and add the remainder to it. 
Now, LCM of 6, 7, 8, 9 = 504
Therefore the required number = 504 + 4 = 508

QUERY 8
Finding out smallest number when the difference between the divisors and their remainders is same. Type # 2 

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

JAYANTCHARAN CHARAN 
Unlike the case mentioned at TYPE # 1, this time the remainder is not the same; but if u 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 = 24
Therefore the required number here = 24 - 1 = 23 


QUERY 9
Finding out smallest number when the difference between the divisors and their remainders is same; and leaves different remainder when divided by another number. Type # 3 

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

JAYANTCHARAN CHARAN 
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


QUERY 10
Finding out smallest number which leaves same remainder with different divisors;  and leaves different remainder when divided by another number. Type # 3............continued

Find smallest number that leaves remainder 2 when divided by 3, 4, 5;  and is fully divisible by 7 (means zero as remainder). 

JAYANTCHARAN CHARAN
Here the first part of the question is like TYPE # 1
Hence the desired number for that = LCM of 3, 4, 5 + the remainder = 60 + 2 = 62 

Now we have one more condition to satisfy i.e. the number is fully divisible by 7 (means zero as remainder); obviously we'll need to follow the same thing as has been described in TYPE-3 above
Therefore now, let 62 + 60k be the number that will satisfy the 4th condition i.e. remainder zero when (62 + 60k)/7

The  expression (62 + 60k)/7 = 62/7 + 60k/7
Now the remainder when 62 is divided by 7 = 6
And remainder when 60k is divided by 7 = 4*k = 4k

Therefore 6 + 4k should leave remainder zero on division of the number by 7 (the number is divisible by 7)
So k must be 2 Hence the number = 62 + 60*2 = 182 


QUERY 11
Finding out smallest number when there is no relation in the trends of divisors and remainders. Type # 4 

Find smallest number that leaves remainder 1 with 5, 4 with 7, 6 with 11 and 7 with 13. 

JAYANTCHARAN CHARAN 
We can see there's no relation among these divisors-remainders sets........neither the remainder nor the difference between divisor and remainder is having a trend. 

In such cases take 1 case and target another case...e.g. First take the case 7 with 13........and target 6 with 11. 

We see that when the number is divided by 13 the remainder is 7 (means 7 is the smallest number that we get as remainder. So all numbers of the form 7 + 13k will give 7 as remainder on division by 13. 

Now target 6 with 11............so we will have to divide 7 + 13k by 11
By doing so the remainder = 7 + 2k
Obviously for minimum value of k = 5
Therefore the number = 7 + 13*5 = 72 

Now that 2 conditions are fulfilled, lets target the third condition...say 4 with 7.
Now, if we add LCM of 11 and 13 i.e 143 to 72, 2 conditions already satisfied would continue being satisfied...
Hence the number is of the form 72 + 143 k.
Thus the remainder when 72 + 143k is divided by 7 = 2 + 3k
So the minimum value of k = 3
Therefore the number satisfying all 3 condition considered = 72 + 143*3 = 501 

Now, if we see carefully.....the 4th condition i.e. 'remainder 1 with 5' considering 501 as the number satisfies itself.
Hence 501 is the appropriate answer here. 

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. 

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


QUERY 12
Finding out smallest number when there is no relation in the trends of divisors and remainders. Type # 4...........Continued

Find smallest number that must be subtracted from 1000 so that the resultant number leaves remainder 1, 3, 4, 8 with divisors 2, 6, 5, 13 respectively.

JAYANTCHARAN CHARAN 
From TYPE 4 we can find out the smallest number satisfying all 4 conditions is 99. 
Obviously if we add LCM of 2, 6, 5, 13 i.e. 390 to 99 the remainders will remain unchanged.

So we need such value of 99 + 390k so that the resultant number is just below 1000.
Easily we can see that for k = 2, we get that such value which is 99 + 390*2 = 879.

Hence the required number = 1000 - 879 = 121


QUERY 13
Finding out smallest number which leaves specific remainders when SUCCESSIVELY divided by two or three numbers. Type # 5

Find the smallest number that leaves remainders 3, 2, 4 when successively divided by 5, 6, 7 respectively.

JAYANTCHARAN CHARAN 
For such questions start doing from the rear end.

So, we want 4 as the remainder with 7 as the divisor; of course the smallest such number is 4 itself.

Obviously this 4 must have come as the quotient when the number was divided by 6
Therefore the number must have been 4*6 + 2 = 26                                 ..........(see 2 is the remainder with 6 as divisor)

It's clear this 26 was the quotient when the number was divided by 5
Therefore the number must have been 26*5 + 3 = 133                             ..........(see 3 is the remainder with 5 as divisor)

Hence the required number is 133


QUERY 14
Finding out smallest number when divided by the LCM of the earlier divisors. Type # 6

A number leaves remainder 3 when divided by 5 and remainder 8 when successively divided by 11. What is the remainder when this number is divided by 55?

JAYANTCHARAN CHARAN 
If looked carefully we'll find that 55 is the LCM of earlier divisors 11 and 5.

It would be easier if we start solving such a question from the rear end
Now take the smallest number that leaves remainder 8 with 11 as the divisor. Obviously it's 8.

Thus according to the question this 8 will be the quotient when the main number is divided by 5. We can see that it also leaves remainder 3 as given in the question.

Hence, the main number = 8*5 + 3 = 43

Now dividing 43 by 55 we see that the remainder is 43 itself.
Hence the answer 43

So 43 is the answer.


QUERY 15
Finding out the largest number that leaves the same remainder when it divides two different numbers. Type # 7

Find the largest number that leaves same remainder when it divides 3398 and 6578.

JAYANTCHARAN CHARAN 
This concept is very much simple to understand. Hence it could  easily be understood that the difference between two dividends must be divisible by the divisor.

Therefore 6578 - 3398 = 3180 should be divisible by the divisor to leave same remainders.

You can see the largest number that divides 3180 is 3180 itself.

Hence, the answer is 3180.


QUERY 16
Finding out the largest number that leaves the same remainder when it divides more than two different numbers. Type # 7..........Continued

Find the largest number that leaves same remainder when it divides 16009, 9009, 7509 and 14009.

JAYANTCHARAN CHARAN 
The approach is same as is in the above type of question
Take difference of the numbers in ascending or descending order as below
16009 - 14009 = 2000, 
14009 - 9009 = 5000,
9009 - 7509 = 1500.

Now to leave same remainder, each of the intervals must be divisible by the divisor.

Obviously we'll need the HCF of these numbers 2000, 5000 and 1500, and that is 500

Hence, the answer is 500.


QUERY 17
Finding out the remainder when a multiple of a number is divided by a different number. Type # 8

If a number is divided by 15, it leaves the remainder 7, if thrice the number is divided by 5; then what is the remainder?
  1. 1       
  2. 2        
  3. 3        
  4. 4        
  5. 0
JAYANTCHARAN CHARAN 
Such questions are difficult to frame as one has to find a pattern between divisors n remainders; but actually they are easy and can be cracked very comfortably. Why I'm putting it here because I have a very short...practical approach for solving such questions.

Now choose a number that leaves 7 as the remainder with 15 as divisor
Let's take it 7 only.

So thrice of 7 = 21

Now, according to the question we have to divide it by 5 and find the remainder. Of course it's 1
Hence the answer is 1

NOTE: Since the number should be giving same result for all values that give 7 as remainder with 15 as divisor, therefore its always better to take some value and solve it instead of taking an algebraic approach.


QUERY 18
Which two digit number when divided by 3 gives remainder of 1; when divided by 4 gives remainder of 2; when divided by 5 gives remainder of 3; and when divided by 6 gives remainder of 4? [Type # 2, see query 8]

RONNIE BANSAL
One of the divisors is 5 giving remainder of 3; means the unit digit of that two digit number is either 3 or 8
But 3 is not possible here as one of the divisor is 4 giving remainder of 2; because the unit digit then must not be odd.
So the possible numbers are 18, 28, 38, 48, 58, 68, 78, 88 & 98

As 3 is a factor of 6; so we need not check the numbers with 3.
Therefore we need to check them considering 4 and 6 here. Now it can easily be seen that 58 is such number.
Yeah Jayan bhaai has rightly said that if options were there to ye chutkiyon ka khel tha.

TRICK (see query 8)
When the difference between divisor and remainder is same like 3 - 1 = 4 - 2 = 5 - 3 = 6 - 4=2;
take LCM of divisors and subtract the common difference from it
Now the LCM of 3, 4, 5, 6 = 60

Therefore the required number here = 60 - 2 = 58


QUERY 19
The 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.

RONNIE BANSAL
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.


QUERY 20
64329 is divided by a certain number. While dividing, the numbers 175, 114 and 213 appear as three successive remainders. the divisor is:
  1. 184         
  2. 224         
  3. 234         
  4. 6250
RONNIE BANSAL
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 '3') is the divisor we need.


QUERY 21
7^8 - 5^8  is wholly divisible by which number?
  1. 25       
  2. 24       
  3. 23        
  4. 21
Vijay Bharath Reddy 
7^8 - 5^8 = (7^4)^2 - (5^4)^2

==> (7^4 + 5^4) (7^4 - 5^4)

==> (7^4 + 5^4) [(7^2)^2 - (5^2)^2]

==> (7^4 + 5^4) (7^2 + 5^2) (7^2 - 5^2) 

==> (7^4 + 5^4) (7^2 + 5^2) * 24

So the given expression is divisible by 24 (option '2')


QUERY 22
The greatest common divisor of (3^3^333) + 1 and (3^3^334) + 1
  1. 2          
  2. 1          
  3. (3^3^333) + 1          
  4. 20     
MAHA GUPTA
The first term of the second expression 3^3^334 can be re-written as 3^[(3^333)*(3^1)]

Now let 3^333 be y 
Therefore the given expressions are 3^y + 1 and 3^3y + 1

==> 3^y + 1 and (3^y)^3 + 1
If seen carefully they are in the form of x+1 and x^n +1

We know an expression in the form of x^n +1 is divisible by an expression of the form x +1 for all ODD values of 'n'
Here n = 3; which is odd.

So the second expression is divisible by the first; and by look we can see it's the greatest of all the given options

Hence (3^3^333) + 1 (option '3') is correct. 

NOTE: We don't need to check other options as it's the greatest of all.


QUERY 23
In (2^41 - 1)/9 what will be the remainder?

MAHA GUPTA
One should remember that one of the ways to find remainders of expressions with two or more terms is to find the remainder of each term and then add them.

The above can also be re-written as 2^41/9 - 1/9

==> [[[(2^3)^13]*2^2]]/9 - 1/9

==> [(8^13)*2^2]/9 - 1/9

By remainder theorem the remainder of 8^13 = -1; so the remainder of the first term = -4 (-1*4)
The remainder of the second term = -1 + 9 = 8

So the remainder that required = 8 + (- 4) = 4


QUERY 24
When a natural number divided by a certain divisor, we get 15 as remainder. But when 10 times of the same number is divided by the same divisor we get 6 as remainder  The maximum possible number of such divisors is?
  1. 6           
  2. 7           
  3. 15           
  4. can't be determined
MAHA GUPTA
In the first case the remainder is 15, means the divisor is greater than 15. And let the smallest dividend (the number to be divided) is 15 itself.

In the second case the dividend is 10 times the earlier one, means 10*15 = 150

But it gives the remainder as 6, means 144 (150 - 6) is fully divisible by the divisor. You know a number can only be divided fully by its factors only.

But the divisor is greater than 15, as it has already been shown above. So find all factors of 144 greater than 15.

They are: 144, 72, 48, 26, 24, 18, 16, which are 7 in number

So 7 is our answer.


QUERY 25
The greatest number by which the expression 7^2n - 3^2n is always exactly divisible?
  1. 4         
  2. 10         
  3. 20         
  4. 40
MAHA GUPTA
Here the given expression is in the form: x^2n - y^2n. You have to only remember if the powers are even (multiplication of n by 2 making them even), the expression of this form is always divisible by 
x^2 - y^2

So accordingly this should be divided by 7^2 - 3^2 = 49 - 9 = 40 
Though it can be divided by any options as it's divisible by 40, but we have to find the greatest number our answer is 40 (option '4').


QUERY 26
A common factor of (17^9 + 13^9) and (17^5 + 13^5) is?
  1. 30         
  2. 17^2 + 13^2         
  3. 17^5 + 17^5         
  4. 40
MAHA GUPTA
We see that both the given expressions are in the form of x^n + y^n; where n is odd. We know that if such is the case the expressions are divisible by an expression in the form x + y 

So both are divisible by 17 + 13 i.e. 30 (option '1')

QUERY 27
3^25 + 3^26 + 3^27 + 3^28) is divisible by
  1. 16        
  2. 11       
  3. 25        
  4. 30
MAHA GUPTA
3^25 + 3^26 + 3^27 + 3^28 = 3^24 (3 + 3^2 + 3^3 + 3^4) 
= 3^24 (3 + 9 + 27 + 81) = 3^24 (120)

It's clear from the above that 120 is a factor of
the given expression. We know that if a factor of an expression is divisible by any number that whole expression too is divisible by that number.

We can see 120 is divisible only by 30, so the given expression too is divisible by 30

Hence 30 (option '4') is the answer.


QUERY 28
If x^5 - 9x^2 - 12x - 14 is divided by (x -3), what is the remainder?
  1. 0        
  2. -184        
  3. 184        
  4. 56
MAHA GUPTA
By the remainder theorem x = 3 will give you the remainder.

So the remainder = 3^5 - 9*(3^2) - 12*3 - 14 = 243 - 81 - 36 - 14

= 112 (not given in the options)


QUERY 29
(4^97)/25. Find remainder

   
SUCHI SHARMA
4^97
= (4^3^32)*4 
= (64^32)*4
Remainder obtained after dividing 64 by 25 is 14 (so when (14^32)*4 is divided by 25 will give us the same remainder when (64^32)*4 is divided by 25)

Now (14^32)*4
= (14^2^16)*4
= (196^16)*4
Remainder obtained after dividing 196 by 25 is 21 (so when (21^16)*4 is divided by 25 will give us the same remainder when (196^16)*4 is divided by 25)

Now (21^16)*4
= (21^2^8)*4
= (441^8)*4
Remainder obtained after dividing 441 by 25 is 16 (so when (16^8)*4 is divided by 25 will give us the same remainder when (441^8)*4 is divided by 25)

Now (16^8)*4
= (16^2^4)*4
= (256^4)*4
Remainder obtained after dividing 256 by 25 is 6 (so when (6^4)*4 is divided by 25 will give us the same remainder when (256^4)*4 is divided by 25)

Now (6^4)*4
= (6^2^2)*4
= (36^2)*4
Remainder obtained after dividing 36 by 25 is 11 (so when (11^2)*4 is divided by 25 will give us the same remainder when (36^2)*4 is divided by 25)

Now (11^2)*4 
=121*4
= 484

You see when 484 is divided by 25 the remainder is 9 (answer)


QUERY 30
19^23^25 divided by 16. Find remainder.

   
SUCHI SHARMA
Remainder obtained after dividing 19 by 16 is 3 (so when 3^23^25 is divided by 16 will give us the same remainder when 19^23^25 is divided by 16)

Now 3^23^25 
= [(3^4^5)*3^3]^25
= [(81^5)*3^3]^25
Remainder obtained after dividing 81 by 16 is 1 (so when [(1^5)*3^3]^25 is divided by 16 will give us the same remainder when [(81^5)*3^3]^25 is divided by 16

Now [(1^5)*3^3]^25
= 3^3^25
= 3^75
= (3^4^18)*3^3
= (81^18)*3^3

So according to the step-II said above when [(1^18)*3^3]^25 is divided by 16 will give us the same remainder when [(81^18)*3^3 is divided by 16

Now (1^18)*3^3
= 27
You see when 27 is divided by 16 the remainder is 11 (answer)


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?

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


QUERY 32
When a number is divided by 24, the reminder is 16. The reminder when the same number is divided by 12 is?

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 (answer)


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?

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


QUERY 34
Find remainder if 32^32^32 is divided by 7
  1. 1
MAHA GUPTA
Remainder obtained after dividing 32 by 7 is 4 (so when 4^32^32 is divided by 7 will give us the same remainder when 32^32^32 is divided by 7)

Now 4^32^32 
= [(4^3^10)*4^2]^32
= [(64^10)*4^2]^32

Remainder obtained after dividing 64 by 7 is 1 (so when [(1^10)*4^2]^32 is divided by 7 will give us the same remainder when [(64^10)*4^2]^32 is divided by 7

Now [(1^10)*4^2]^32
= 4^2^32
= 4^64
= (4^3^20)*4
= (64^20)*4

Remainder obtained after dividing 64 by 7 is 1 (so when [(1^20)*4 is divided by 7 will give us the same remainder when (64^20)*4 is divided by 7

Now (1^20)*4
= 4
You see when 4 is divided by 7 the remainder is 4 (option '3')


QUERY 35
Find the least possible 5 digit number which when divided by 2, 4, 6, 8 it leaves the remainder 1, 3, 5, 7 respectively.

MAHA GUPTA
Here you see that the difference between the divisors and their remainders is same i.e. 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


QUERY 36
Find the number when it is divided by 9, 11, 13 leaves remainders 1, 2, 3 respectively.

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 

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. 

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


QUERY 37
If the number 23583ab is completely divisible by 80, then the value of a - b is
  1. 2
  2. 3
  3. 4
  4. 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 ‘1’)


QUERY 38
Find the smallest number when is divide by 8, 5 and 11 leaves remainders 5, 2 and 6 respectively.

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

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.


QUERY 39

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

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 (answer)


QUERY 40
16^15 + 2^15 is divisible by?
  1. 31
  2. 13
  3. 27
  4. 33
MAHA GUPTA
16^5 = (2^4)^5 = 2^20 = (2^15)(2^5) = 32(2^15) 

Thus, 16^5 + 2^15 = 32(2^15) + (2^15) = (2^15)(32 + 1) = (2^15)*33

Hence, 33 is a factor of the given expression; means divides it fully (option '4')


QUERY 41
   

MAHA GUPTA

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 = 2963 - 1312 = 1651
Factors of 1651 are 1, 13, 127 and 1651
But we have to find the three digit number here, so 127 is the required number and their sum is 1+2+7 = 10 (option 'D')
Visit this user's website Find all posts by this user
Quote this message in a reply
Post Reply 


Forum Jump:


User(s) browsing this thread: 1 Guest(s)
This forum uses Lukasz Tkacz MyBB addons.