Number System - Questions & Discussions

MBA CAT 2022
1693
Some Specifications of a password
a) 7 digit odd no.
b) 1st 2 digits = 16 or 18 ;
c) 6 occurs twice ; 8 occurs once .
Max trials to crack such password =

a) 7680
b) 1250
c) 3840
d) 5760


case 1: 18_ _ _ _ _
case 2: 16 _ _ _ _ _
C(4,2)*8*8*5 + C(4,2)*2*8*8*5
=5760
d)5760
1694

3^4k divided by 5 always yeilds remainder 1(eulers theorem.)




1695

yes the answer is a.

38 and 17 are coprime , any number of the form 38^16K will give remainder 1....

1697
Some Specifications of a password
a) 7 digit odd no.
b) 1st 2 digits = 16 or 18 ;
c) 6 occurs twice ; 8 occurs once .
Max trials to crack such password =

a) 7680
b) 1250
c) 3840
d) 5760


Okie,

16 _ _ _ _ _

now for odd we have 1,3,5,7,9
so,unit digit has 5 choices, 5 * 12 * 8*8 = 3840

18 _ _ _ _ _
5*6 * 8*8 = 1920

adding them i am getting 5760 is it rite ??

Suja
1698

Here is a problem I'm facing.

( 3^5^7 ) / 4 remainder?

I'm trying to figure it out by getting cycle or 3/4 and then whether 5^7 is of the form 3n / 3n+1 / 3n+2 etc
I'm getting 1 as answer.
But in a note it says the answer is 3 and I am not being able to understand where I'm making the mistake.

Please help!

1699

write 3 as (4-1)

using binomial theorem .... the Rem will be (-1)^5^7

-1 to the power of an odd no. = -1
hence rem = -1 or 3

1700
write 3 as (4-1)

using binomial theorem .... the Rem will be (-1)^5^7

-1 to the power of an odd no. = -1
hence rem = -1 or 3


Thanks but I am looking for the reason - whats wrong in the way I was trying.

Its def easier ur way in this case but how to do this for cases like Remainder of ( 4^32^32 )/7 ?
1701

4^3 = 64 ... its near to 7*9 i.e. 63....

1702
charchit040216 Says
4^3 = 64 ... its near to 7*9 i.e. 63....


totally lost u !!
1703

i have confusion in the question


what is the remainder when 2222^5555+5555^2222 is divided by 7.

I got it as 1 but the answer in the book is 0.

Could anybody explain if I am right or wrong.

1704
i have confusion in the question


what is the remainder when 2222^5555+5555^2222 is divided by 7.

I got it as 1 but the answer in the book is 0.

Could anybody explain if I am right or wrong.


2222^5555 mod 7

3^5555 mod 7

Euler number of 7 is 6.

3^5 mod 7

R1=5

5555^2222 mod 7

4^2222 mod 7

4^2 mod 7

R2=2

R1+R2=7

7 mod 7 is 0

So, remainder will be 0.

Hope that helps.
1705
Q. A certain country has currencies with denominations 6, 9, 20, 50 and 100. What is the largest amount that a person can not make using these currencies?

1. 37
2. 16
3. 23
4. 43
5. None of these

PaGaLGuY.com - The Everything of MBA in India and Abroad, CAT 2009, GMAT, XAT, MAT - Cracking CAT 2010

none of these. largest no. will be 103
1706

Q. What is the remainder of ((6!)^(7!))^13333 when divided by 13 .

1)5
2)1
3)8
4)9

Can someone help me out to solve this ..

1707
Q. What is the remainder of ((6!)^(7!))^13333 when divided by 13 .

1)5
2)1
3)8
4)9

Can someone help me out to solve this ..

remainder is1 . apply euler, for 13 it is 12 so check 7!*13333 is divisible with 12 or not, it is divisible in this case , so remainder is 1
1708
red.wall Says
remainder is1 . apply euler, for 13 it is 12 so check 7!*13333 is divisible with 12 or not, it is divisible in this case , so remainder is 1

method - 7! % 4 = 0
6! ^ 0 = 1
Ans shud be option 2
1709

Ansser for the second ques posted in this thread

answer a)1

[ (38^(16*(15!*1777)) ] % 17 = 1

since 16 is the Euler number of 17

1710

can you guys give me the resource from which i can learn "base" related sums????//
i m not finding myself comfortable with those sums.......

1711

answer is 9

1712

hey puys, plz solve dis 4 me...!
Q.Find the remainder when 1212..1212(300 digits) is divided by 999 ?

1713
hey puys, plz solve dis 4 me...!
Q.Find the remainder when 1212..1212(300 digits) is divided by 999 ?


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Sorry wrong post