Number System - Questions & Discussions

MBA CAT 2022
1004
(22^33 +10^35)/45= find the rmndr

quite easy --but mah ans is not matching (courtesy nishit sinha )


22^33/45

22 and 45 are co-primes

E(45)=24

22^24k = 1mod45

22^9 mod 45

22*484^4 mod 45

22* (-11)^4 mod 45

22*(-14)^2 mod 45

22*196 mod 45

22*16 mod 45 = 37

10^35 mod 45

5*(2*10^34 mod 9) = 10

so we have (37 + 10)/45

47 mod 45 = 2
1005

ya its ok-- i got the same ans too but ans in nishit sinha was 8 . a mistake probably

1007
Hi Puys
Ques> There are 1000 lockers(all closed) and thousand children. The first child unlocks all the doors. The second child locks open doors and opens locked doors that are multiple of 2. Third child locks open doors and opens locked doors that are multiples of 3 and so on till 1000 students have taken their turns.


All Perfect Squares upto 1000, i.e. 961 would be open....
So Doors Open = 31
Doors Closed = 1000-31 = 969
1008
Hi Puys
I'm posting a question from one of the mocks that I gave. Try solving it. I'l post the answer tonight.
Ques> There are 1000 lockers(all closed) and thousand children. The first child unlocks all the doors. The second child locks open doors and opens locked doors that are multiple of 2. Third child locks open doors and opens locked doors that are multiples of 3 and so on till 1000 students have taken their turns.
Find the number of locked and open doors at the end.

Please post your approach. I'l share mine tonight.


All Perfect Squares upto 1000, i.e. 961 would be open....
So Doors Open = 31
Doors Closed = 1000-31 = 969


@divishth, kindly xplain ur soln..
1009

Quick question from my side ....

what is the greatest power of 6 that divides number 34! exactly.
Please elaborate the approach.

Thanks

1010

This is what I did ....
If you list down the all doors from 1- 10 , AND mark doors as Open and close for first 9 child .... u'll see the pattern such that only door numbers, whihc are PERFECT SQUARES will remain open, others will get closed.

As we have total 31 perfect squares less than 1000, so 31 doors will be open and rest will be closed.

1011
Hi Puys
I'm posting a question from one of the mocks that I gave. Try solving it. I'l post the answer tonight.
Ques> There are 1000 lockers(all closed) and thousand children. The first child unlocks all the doors. The second child locks open doors and opens locked doors that are multiple of 2. Third child locks open doors and opens locked doors that are multiples of 3 and so on till 1000 students have taken their turns.
Find the number of locked and open doors at the end.

Please post your approach. I'l share mine tonight.





This is what I did ....
If you list down the all doors from 1- 10 , AND mark doors as Open and close for first 9 child .... u'll see the pattern such that only door numbers, whihc are PERFECT SQUARES will remain open, others will get closed.

As we have total 31 perfect squares less than 1000, so 31 doors will be open and rest will be closed.
1012
Quick question from my side ....

what is the greatest power of 6 that divides number 34! exactly.
Please elaborate the approach.

Thanks



is the ans 5?
1013
abhgud Says
is the ans 5?


Nopes, Even i got 5 .... but its 15 ....
1014
Quick question from my side ....

what is the greatest power of 6 that divides number 34! exactly.
Please elaborate the approach.

Thanks

abhgud Says
is the ans 5?


Power of 2 in 34! => 32
Power of 3 in 34! => 15
So Power of 6 in 34! => 15
1016

what is the remainder of 7^99 when divided by 2400.

1017
abhgud Says
is the ans 5?

Power of 2 in 34! => 32
Power of 3 in 34! => 15
So Power of 6 in 34! => 15



Do we hav any shortcut way to find that Power of 2 in 34! => 32 .... OR we have to manually check ?
1019
harsh_hbti Says
what is the remainder of 7^99 when divided by 2400.

2400 = 3*25*32
7^99 mod 3 = 1
7^99 mod 2^5 = 7^3 mod 32 = 23
7^99 mod 25 = 7.49^49 mod 25 = -7

25k -7 = 32p + 23,==>p = 10
800q + 343 = 3r + 1,==>q = 0

remainder = 343
1020
harisin_47 Says
Do we hav any shortcut way to find that Power of 2 in 34! => 32 .... OR we have to manually check ?


harisin we have the shortcut method
it goes like this,
+ + ....
1021
2400 = 3*25*32
7^99 mod 3 = 1
7^99 mod 2^5 = 7^3 mod 32 = 23
7^99 mod 25 = 7.49^49 mod 25 = -7

25k -7 = 32p + 23,==>p = 10
800q + 343 = 3r + 1,==>q = 0

remainder = 343


thanks the answer is correct..
1022
harsh_hbti Says
what is the remainder of 7^99 when divided by 2400.

7^4 = 2401 = 1 mod 2400

Thus, 7^96 = 1 mod 2400
7^99 = 343 mod 2400

Thus, remainder is 343
1024

well u shud have posted @ quant official thread.
Anyway my take is 93C3. I hope you have written all the conditions given.
in present form: its just a+b+c+d = 90

1025
The AMS CAT MOCK has 4 sections with 45 marks each. Find the number of ways in which a student can qualify if 90 is the qualifying score.
a>34566
b>79869
c>64906
d>98777


Solved Many Times In Quant Thread.....Have A Look....
1026

"Number of Ways He Cannot Score = 4*47c3 "

plz explain this part

1027
harsh_hbti Says
what is the remainder of 7^99 when divided by 2400.


7^4=2401
7^99/2400=R(7^3/2400)=343