Quant by Arun Sharma

MBA CAT 2024
4201
Q)What will be the remainder when 111111111111111....................729digits is divided by 728?
a)1
b)11
c)111
d)None of these


My take is 111

Any digit repeated 6k times is divisible by 7, 11 and 13.
728 = 7*8*13
So, 1 repeated 726 times will be divisible by 7 and 11.
And if we place three 0s ahead of 726 times 1; number will remain divisible by 7 and 13 and it will also become divisible by 8 as last 3 digits are 0s.

So, remainder will be 111 as number is divisible 728
4202

it should be C

728 is nothing but 91*8

ohk...

when 11111111... / 91= remainder is 10
and when 111111..../8 remainder is = 7 ( Last three digits should be divisible to be completely divisible)

so u need to find the 91k+10 and 8b+7

By options and by value substitution 111 is the answer

4203
Q)What will be the remainder when 111111111111111....................729digits is divided by 728?
a)1
b)11
c)111
d)None of these



Puys is there any general method ..or some specific techniques which will help to tackle such digit repitition problems ?? I am having a tough time solving these kind of problems ..It would be very kind of you to list these methods or provide a link where these methods are explained

Thanks,
4204

Any digit repeated 6k times is divisible by 7, 11 and 13

best approach
it is possible for any multiple of 111111

and u can find for other...

4205
here's my explanation:

look there is a theorem called as Chinese remainder theorem

suppose u have to find the remainder of say 435 when divided by 26

u can do it by this way

435/2=1
435/13=6

so u start finder the remainder
13k+6 and 2B+1 form

so the remainder is 19

...


now what i did in that question for 2 is

2222/2 =0
7777/2 = 1

there is no such number possible that will be in two different forms....

moreover y i left there it is because

ODD+even = Always ODD

it could never be Even so it could never be divisible by 2


Hey Thanks...Yea i knw abt Chinese Remainder Theoram...Just forgot dt funda of two diffrnt forms n finding the remainder....got it...thnks again...
4206

Pls give full explanation :
A number when successively divided by 3, 5 and 8 leaves a remainder 1,4, and 7 respectively. Find the respective remainders if the orders of divisors be reversed

4207

approach would be u have to find the number first then divide it in reverse order...

otherwise find in general terms :like this

3(5(8k+7)+4)+1

now take k = 0 or 1 ..

i take 0 first
number would be 118

divide it in reverse order u will get 6, 4,2

check it with k=1

Thanks
Ashish Patial

4208
approach would be u have to find the number first then divide it in reverse order...

otherwise find in general terms :like this

3(5(8k+7)+4)+1

now take k = 0 or 1 ..

i take 0 first
number would be 118

divide it in reverse order u will get 6, 4,2


check it with k=1

Thanks
Ashish Patial


Thanks for the quick reply..but why did u take K=1 and 0 only..can we take other numbers as well???
4209

yup u can take any number ....it depends upon question

but the smallest number would be at k=0

but yes for k=1

the answer would be same too

k=1

238

remainder 238/8=6
remainder 29/5=4
5/3=2

4210
it should be C

728 is nothing but 91*8

ohk...

when 11111111... / 91= remainder is 10
and when 111111..../8 remainder is = 7 ( Last three digits should be divisible to be completely divisible)

so u need to find the 91k+10 and 8b+7

By options and by value substitution 111 is the answer


Hey ashish, it shld be 111111.....%91=Remainder shld be 20...

Therefore, 91a+20 & 8b+7 will result in 111 as answer...
By substituting a=1 & b=13...
P.S. Correct If m wrong somewhere....
4211

Pls help with explanation

1. What will be the remainder when (67^67 + 67) is divided by 68?

2. Which of the following will completely divide (49^15 - 1)?
a) 8 b) 14 c) 48 d) 50

3. What will be the remainder when 17^200 is divided by 18?

4) It is being given that 2^32 + 1 is completely divisible by a whole number. Which is the number?
a) (2^16+1) b) (2^16-1) c)7x2^33 d)(2^96+1)

5) Which one of the following will completely divide 3^25+3^26+3^27+3^28 ?
a) 11 b) 16 c) 25 c) 30

4212
Hey ashish, it shld be 111111.....%91=Remainder shld be 20...

Therefore, 91a+20 & 8b+7 will result in 111 as answer...
By substituting a=1 & b=13...
P.S. Correct If m wrong somewhere....

Awesum

thanxx
4213
Pls help with explanation

1. What will be the remainder when (67^67 + 67) is divided by 68?

i think it should be 66

67^67 / 68= -1^67 + 67 =66

2. Which of the following will completely divide (49^15 1)?
a) 8 b) 14 c) 48 d) 50

u can write it as 49^15-1^15 therefore a^n-b^n is always divisible by a-n

hence 48

but it should be divisible by 8 also

:D

3. What will be the remainder when 17^200 is divided by 18?

remainder is -1^200=1

4) It is being given that 2^32 + 1 is completely divisible by a whole number. Which is the number?
a) (2^16+1) b) (2^16-1) c)7x2^33 d)(2^96+1)

i think it should be first that is my guess


i still dont know binomial expression...


5) Which one of the following will completely divide 3^25+3^26+3^27+3^28 ?
a) 11 b) 16 c) 25 c) 30
it should be 30

3^25 ( 1+3 +9+27 )

=so 3^25(40)

hence divisible by 30


that my take buddy
4214
Pls help with explanation

1. What will be the remainder when (67^67 + 67) is divided by 68?

2. Which of the following will completely divide (49^15 - 1)?
a) 8 b) 14 c) 48 d) 50

3. What will be the remainder when 17^200 is divided by 18?

4) It is being given that 2^32 + 1 is completely divisible by a whole number. Which is the number?
a) (2^16+1) b) (2^16-1) c)7x2^33 d)(2^96+1)

5) Which one of the following will completely divide 3^25+3^26+3^27+3^28 ?
a) 11 b) 16 c) 25 c) 30


1. (67^67)/68 is equal to ((-1)^67)/68. Therefore=> -1+67=66 is remainder.
2.48....StraightForward using Formula as the power is ODD
3.1 ...S//ly as question 1...((-1)^200)%18=1
4. 😞 Finding a better way....
5.3^25(40). So its divisible by 30
4215
1. (67^67)/68 is equal to ((-1)^67)/68. Therefore=> -1+67=66 is remainder.
2.48....StraightForward using Formula as the power is ODD
3.1 ...S//ly as question 1...((-1)^200)%18=1
4. 😞 Finding a better way....
5.3^25(40). So its divisible by 30


i dont understand the funda of ((-1)^67)/68
4216
nitya2903 Says
i dont understand the funda of ((-1)^67)/68

See, the number was 67...wich is 1 less than 68(the number with which ur doing operation on 67)..so u can write (67^anything)/68 as (-1)^anything)/68 ....This step is done w.r.t. the number ur operating with.

For E.g.:- Its 67^300/50 =((+17)^300)/50 (s//ly for positive case as 67 is 17 greater than 50)
OR
67^56/70 = (-3)^56)/70
P.S. Hope ur doubt is clear upto some extent..
4217
See, the number was 67...wich is 1 less than 68(the number with which ur doing operation on 67)..so u can write (67^anything)/68 as (-1)^anything)/68 ....This step is done w.r.t. the number ur operating with.

For E.g.:- Its 67^300/50 =((+17)^300)/50 (s//ly for positive case as 67 is 17 greater than 50)
OR
67^56/70 = (-3)^56)/70
P.S. Hope ur doubt is clear upto some extent..


yes it is!! thanks 😃
4218
Ques:- (2^1990)/1990 . Remainder??
4219
sid_roy09 Says
Ques:- (2^1990)/1990 . Remainder??


It should be 1024

First, we can take out 2 common from numerator and denominator and multiply by it in the end.

So, we have to first find out remainder of 2^1989 by (5*199)
Remainder by 5:
Euler's number for 5 is 4.
2^4k leaves remainder by 5 as 1. 1989 = 4k+1 => Remainder by 5 = 2^1 = 2

Remainder by 199:
Euler's number for 199 is 198.
2^198k leaves remainder by 199 as 1.
So, 1989 = 198*10 + 9 => Remainder by 199 = 2^9

So, we need a number of form 5a + 2 = 199b + 2^9 => 2^9 at b=0

So, remainder of 2^1989 by 5*199 is 2^9
=> Remainder of 2^1990 by 2*5*199 is 2*2^9 = 2^10 = 1024
4220
It should be 1024

First, we can take out 2 common from numerator and denominator and multiply by it in the end.

So, we have to first find out remainder of 2^1989 by (5*199)
Remainder by 5:
Euler's number for 5 is 1.
2^4k leaves remainder by 5 as 1. 1989 = 4k+1 => Remainder by 5 = 2^1 = 2

Remainder by 199:
Euler's number for 199 is 198.
2^198k leaves remainder by 199 as 1.
So, 1989 = 198*10 + 9 => Remainder by 199 = 2^9

So, we need a number of form 5a + 2 = 199b + 2^9 => 2^9 at b=0

So, remainder of 2^1989 by 5*199 is 2^9
=> Remainder of 2^1990 by 2*5*199 is 2*2^9 = 2^10 = 1024


Points:- 1. Euler number of 5 is 4 nt 1.
2. 5a+2=199b+2^9 .....if b=0,then a=? I mean what is the common term(number) for both the equations???
3. You have taken 2 common from both denominator n numerator initially but in the last step u have multiplied the term wid 2(numerator wala) but wat abt the 2(denominator wala) wich u also taken as common in the initial step...????

P.S. Clear the Points plz....