Quant by Arun Sharma

MBA CAT 2022
1074

Here is another one,
Amitabh has a certain number of toffes, such that if he distributes them amongest ten childern he has nine left, if he distribtes among nine childern he has eight left, if he distributes among 7 children he wold have left 6... and so on until if he distributes amongest 5 children he should have 4 left. What is the second highest number of toffes he could have with him?
a) 2519 b) 7559 c) 8249 d) None of these.

Any answer + explaination. I am fed up with this, once I thought it is (a) as the second highest after 7559 that number is satisfying the conditions. But answer is d) None of these. What is the number anyway, any help to find out in proper way?

1075
Here is another one,
Amitabh has a certain number of toffes, such that if he distributes them amongest ten childern he has nine left, if he distribtes among nine childern he has eight left, if he distributes among 7 children he wold have left 6... and so on until if he distributes amongest 5 children he should have 4 left. What is the second highest number of toffes he could have with him?
a) 2519 b) 7559 c) 8249 d) None of these.

Any answer + explaination. I am fed up with this, once I thought it is (a) as the second highest after 7559 that number is satisfying the conditions. But answer is d) None of these. What is the number anyway, any help to find out in proper way?


Amitabh has a certain number of toffes, such that if he distributes them amongest ten childern he has nine left, if he distribtes among nine childern he has eight left


friend this bold part is correct ?

first distribute in 10 children then 9 then directly 7 u forgot to write down 8 or its correct?
1076
Here is another one,
Amitabh has a certain number of toffes, such that if he distributes them amongest ten childern he has nine left, if he distribtes among nine childern he has eight left, if he distributes among 7 children he wold have left 6... and so on until if he distributes amongest 5 children he should have 4 left. What is the second highest number of toffes he could have with him?
a) 2519 b) 7559 c) 8249 d) None of these.

Any answer + explaination. I am fed up with this, once I thought it is (a) as the second highest after 7559 that number is satisfying the conditions. But answer is d) None of these. What is the number anyway, any help to find out in proper way?


Is it the second least no of toffees when there is equal distribution ??

Coz least no is LCM (5,6,7,8,9,10) -1
Hence Second least no is 2*LCM (5,6,7,8,9,10) -1
= 2*2520 -1
=5039

Not sure abt d ans ...pls correct d approach ...
1077
Here is another one,
Amitabh has a certain number of toffes, such that if he distributes them amongest ten childern he has nine left, if he distribtes among nine childern he has eight left, if he distributes among 7 children he wold have left 6... and so on until if he distributes amongest 5 children he should have 4 left. What is the second highest number of toffes he could have with him?
a) 2519 b) 7559 c) 8249 d) None of these.

Any answer + explaination. I am fed up with this, once I thought it is (a) as the second highest after 7559 that number is satisfying the conditions. But answer is d) None of these. What is the number anyway, any help to find out in proper way?



If i write down in my previous post its correct then

10x+9
9y+8
8z+7
----

--- so on upto 5

now find lcm of (10*9*8*7*6*5)-1

so ans is 2519

but here u need second least (not highest) so 2520*2-1=5039:nono:
1078
If i write down in my previous post its correct then

10x+9
9y+8
8z+7
----

--- so on upto 5

now find lcm of (10*9*8*7*6*5)-1

so ans is 2519

but here u need second least (not highest) so 2520*2-1=5039:nono:


Can you tell me why u subtract 1 from the LCM and for the second least number u have multiplied by 2?
If you can give an explainatory example, that would be helpful
1079

Hi,
Question: When 2222^5555 + 5555^2222 is divided by 7 the remainder is a) 0 b)2 c) 4 d) 5

There are many questions related to this kind. If there is any general way to approach, pls tell me.:|

1080
Hi,
Question: When 2222^5555 + 5555^2222 is divided by 7 the remainder is a) 0 b)2 c) 4 d) 5

There are many questions related to this kind. If there is any general way to approach, pls tell me.:|


2222^5555 + 5555^2222

2222 & 7 prime to each other also 5555 & 7 prime to each other

so 5555=6k+5 and 2222=6p+2

so 2222^5 and 5555^2

now 2222/7 gives reminder =3

5555/7 gives reminder = 4

so remain 3^5 + 4 ^ 2

3^5 gives reminder 5 when divided by 7

4^2 gives remidner 2 when divided by 7

so final ans 7/7 so reminder =0
1081
Can you tell me why u subtract 1 from the LCM and for the second least number u have multiplied by 2?
If you can give an explainatory example, that would be helpful


here multiplied by 2 because u need to find second least if in question if they ask for least then ans is 2519 not 5039

and subtract 1 because there is difference of 1 between reminder (means in data reminder 9,8,7....5)
1082
here multiplied by 2 because u need to find second least if in question if they ask for least then ans is 2519 not 5039

and subtract 1 because there is difference of 1 between reminder (means in data reminder 9,8,7....5)


Thank alot Jigar, that's really nice explaination, I got it.
1083

32^32^32 when divided by 9 the remainder is a) 4 b) 7 c) 1 d) 2

have these finding remainder type questions some clue to answer, I don't know how to solve it can u help me pls.:

1084
32^32^32 when divided by 9 the remainder is a) 4 b) 7 c) 1 d) 2

have these finding remainder type questions some clue to answer, I don't know how to solve it can u help me pls.:


32^32^32/9

32 and 9 coprime to each other so 32^6/9 gives reminder=1(as per euler's theorem)

so 32^32=32*32.....32 times

so u can write 6k+2*6k+2.....32 times

so remain 2^32 now 2^32=1024*1024*1024*4

32/9= gives reminder=5

so 5^1024*1024*1024*4

1024/6=4

so 5^4*4*4*4

so final reminder =4

:nono:
1085
32^32^32 when divided by 9 the remainder is a) 4 b) 7 c) 1 d) 2

have these finding remainder type questions some clue to answer, I don't know how to solve it can u help me pls.:


what Jigar has done is correct,but,another approach.for starters:

consider 32^32

its (6k+2)^32

now,6k for the Euler number of 9 is 6.

it becomes

(6m+2^32)

so,

(6m+16^ mod 9

16^8 mod 9

7^8 mod 9

49^4 mod 9

5^4 mod 9

625 mod 9

answer is 4.

try to reduce the number as much as possible and then go for the kill in these type of problems.once you try out some Euler problems you will be able to do it proficiently.
1086

Hi, i have a question... from lod1 number systems...

1. which among the following is not a perfect square..

1)1,00,856 2)3,25,137
3)9,45,729 4)All of these(AOT)
5)None of these(NOT)

Pls someone ans...

1087
Hi, i have a question... from lod1 number systems...

1. which among the following is not a perfect square..

1)1,00,856 2)3,25,137
3)9,45,729 4)All of these(AOT)
5)None of these(NOT)

Pls someone ans...


now the square of a number ends in 1,4,9,6,5,0

so,clearly,option 2 is out.and subsequently option 5.

100856 is 2*2*25219=so,we have to find out whether 25219 is perfect square or not.

150^2=22500 160^2=25600

so,possible options are 153 and 157

153 is not possible as it has to be divisible by 3.

157^2=>22500+49+2100=24649

945729=9*3*35027 as,there is no further 3 in the number,it is not a perfect square.

so,option (4)
1088

thanks for the ans...

150^2=22500 160^2=25600

so,possible options are 153 and 157

153 is not possible as it has to be divisible by 3.

157^2=>22500+49+2100=24649


din get this part, can you pls explain this part?

1089
thanks for the ans...

150^2=22500 160^2=25600

so,possible options are 153 and 157

153 is not possible as it has to be divisible by 3.

157^2=>22500+49+2100=24649


din get this part, can you pls explain this part?


see,the number is 25219 so,for it to be a square,it has to be a square of either 153 or 157,as these are the numbers the squares of which end in 9 and are between 22500 and 25600.so,i considered just 2 numbers which reduces calculations.

hope that helps.
1090
see,the number is 25219 so,for it to be a square,it has to be a square of either 153 or 157,as these are the numbers the squares of which end in 9 and are between 22500 and 25600.so,i considered just 2 numbers which reduces calculations.

hope that helps.



thanks..got it
1091

Hi,
can someone let me know the easy way of solving this kind of problems..
1)Find the last two digits of 3 to the power of 15678.

I tried to solve this one by dividing it by 100,but it is a tedious approach :lookround:and i got stuck ...Pls ans this question.

1092
Hi,
can someone let me know the easy way of solving this kind of problems..
1)Find the last two digits of 3 to the power of 15678.

I tried to solve this one by dividing it by 100,but it is a tedious approach :lookround:and i got stuck ...Pls ans this question.


3^15678=(10-1)^7839=.....all with term atleast 100..+ 7839c1*10-1
=78390-1=..........89

So the anser is 89..
1093
hi,
can someone let me know the easy way of solving this kind of problems..
1)find the last two digits of 3 to the power of 15678.

I tried to solve this one by dividing it by 100,but it is a tedious approach :lookround:and i got stuck ...pls ans this question.



3^15676*3^2

(81)^3919*3^2

now multiply 3(tens digit) with power's last digit (6)

so got 21

so final and=21*9=89