Number System - Questions & Discussions

MBA CAT 2022
2556
if there are 45 match played. A, B, C played 30, 23, and 18 respectively. Then, what is least no. of matches in which more than one of them played?
please suggest how to solve this and other min. - max. problems.


30+23+18 = 71

21 extra
for 3 together 2 cases will be redundant = 10
1 extra in 2

so i'm thinking 11 as the answer ??
oa plzz
2557
catoman Says
Find the maxium /minium value of f(x)= x^x



maximum infinity

minimum --1
2558
in a particular country all the members are expressed with the help of three alphabets a,b,c
15 is written as abc
6 is written as bc
60 is written as bcbc
how would one write 17 in that country?
a)abb
b)bab
c)baa
d)aba


base 3 representation hai bhaii

a = 1
b = 2
c = 0

so 17 = 122

hence my take is abb
2559
answer can be b) 177 or d)178 .. i dunno how to find which is the appropriate answer...

if options are close its better to leave the problem rather than attending it

Answer is 178
2560
in a particular country all the members are expressed with the help of three alphabets a,b,c
15 is written as abc
6 is written as bc
60 is written as bcbc
how would one write 17 in that country?
a)abb
b)bab
c)baa
d)aba


i think it is abb
plz confirm....
2561
in a particular country all the members are expressed with the help of three alphabets a,b,c
15 is written as abc
6 is written as bc
60 is written as bcbc
how would one write 17 in that country?
a)abb
b)bab
c)baa
d)aba


its in base 3
ax2+bx+c=15
bx+c=6
bx3+cx2+bx+c=60
we get x=3 is the base
and a=1, b=2, c=0

17=9+6+2=3^2*1+3*2+3^0*2=3^2*a+3*b+b=abb (in base 3)
2562
please help me with this problem.


1) LCM (1,2,3......m)=x
LCM(1,2,3....m+n)=x
m,m+n find max value of n.
2563

yar

i think 7

dont know vaise..

i got 113 and m+n =120

what is the answer ??/

1) LCM (1,2,3......m)=x
LCM(1,2,3....m+n)=x

2564

By the way We have to find the name difference between two prime number also considering the fact that power of existing prime number remain same

so by this way i am getting ...

89 and 97 so n is 8

2565

what is the remainder when 25! is divided by 10^7?

2566

what is the remainder when 128^500 is divided by 153?

2567

Can anybody explain the different remainder theorms we use like Elulers,chinese remainder and fermets. It would be great if u can post with examples also ..

i know that it would have been discussed in this thread, but i m not able to find it.

Thanks in advance

2568
ankurba Says
what is the remainder when 128^500 is divided by 153?


Euler number of 153 is 96 so the number reduces to 128^20

153 = 17 * 9
17 leaves a remainder of 17 and 9 leaves a remainder of 4
17A+16= 9B + 4

A = 3 and B = 7 is the first number that satisfies the above (chinese remainder theorem)

so remainder is 67.
2569
ankurba Says
what is the remainder when 25! is divided by 10^7?


25! has 6 zeros in the end. So the answer for above is nothing but finding the last digit of 25!

But i'm not sure if there is any method for that
2570

^^^^ this is a perfect way to do it but remeber when ur assuming to find last digit after division neglect the 2^6 also

2571
please help me with this problem.


1) LCM (1,2,3......m)=x
LCM(1,2,3....m+n)=x
m,m+n find max value of n.


Find out the prime numbers slightly less that 130,

2, 3, ..... 109, 113, 127...
Find 2 consecutive prime numbers with highest difference.. Here it should be 113 and 127.

So the LCM or 1, 2 , 3 ... 113 and LCM of 1, 2, 3, ... 126 will be same

113
113+ 13

So n is 13

Please correct me if i'm wrong.
2572

no yar actually wat happen is i think

when u find the LCM u actually add extra 11


becoz 121=11^2

so we need largest difference between two prime numbers

2573
ankurba Says
what is the remainder when 25! is divided by 10^7?


Answer to this will be 4*10^6.

25! has 6 zeros in the end ... (25=> 5+1 = 6)
So, remainder by 10^7 will be last non zero digit of 25! follwoed by six 0s.

Last non-zero digit of 25! = 2^5 * R(5) *R(0) = 2 * 2 = 4

Note: R(n) i.e. last non-zero digit of n! can be found by putting n in the form of 5a+b.
Then, R(n) = 2^a * R(a) * R(b)

ankurba Says
what is the remainder when 128^500 is divided by 153?


It will be 67.

128^500 = 2^(7*500) = 2^3500

Now, 153 = 9*17. So, first let's find out remainder by 9 and 17.

Remainder by 9:
2^3 will leave remainder by 9 as 8 i.e. (-1). And 2^6 will leave remainder by 9 as 1.
So, 2^3500 = 2^3498 * 2^4 = 2^6k * 2^2 will leave remainder as 1*4 = 4

Remainder by 17:
2^4 will leave remainder by 17 as 16 or (-1).
So, 2^3500 = 2^(4*25*35) = (2^4)^(some odd number) = (-1)^(some odd number) = -1

So, we need to find a number of form, 9x+4 = 17y-1 => 67 at x=7 and y=4

So, remainder by 153 for 128^500 = 67
2574

Hi Puys,

Can anyone pls explain how to solve below problem by Chinese theorem:

Whats the remainder when 123412341234......(upto 400 digits) is divided by 909.


thanks,
Gaurav K

2575

last non zero digit of 25! is 2 so remainder will be 2 because 25! has 6 zero's