Number System - Questions & Discussions

MBA CAT 2022
2596
Swatysachdeva Says
N=323232......................32323...if N has exactly 101 digits find the remainder when N is divied by 440


what is the answer to this question.... is it 163
2597

If 1/(13!0!)+1/(12!1!)+1/(11!2!)............1/(2!11!)+1/(1!12!)+1/(0!13!)=2^a/b!

then what is (a+b)??

2598
saksham123 Says
what is the answer to this question.... is it 163


yes its 163
2599
If 1/(13!0!)+1/(12!1!)+1/(11!2!)............1/(2!11!)+1/(1!12!)+1/(0!13!)=2^a/b!

then what is (a+b)??



1/(13!0!)+1/(12!1!)+1/(11!2!)............1/(2!11!)+1/(1!12!)+1/(0!13!)

or ( 13! / 13 !) x
( 1/(13!0!)+1/(12!1!)+1/(11!2!)............1/(2!11!)+1/(1!12!)+1/(0!13!))
(13c0 +13c1 +13c2...............13c13) /13 !

2^13 /13 ! So, a=b=13 and a+b=26
2600

Find the number of divisors of 1080 excluding the throughout divisors, which are perfect squares.

a) 28
b) 29
c) 30
d) 31

2601

The equation 4x-Ay=B has a number of integral solution .If HCF(A,4)=1 and the number of solutions of (x.y) for 0??

2602
Find the number of divisors of 1080 excluding the throughout divisors, which are perfect squares.

a) 28
b) 29
c) 30
d) 31


1080= 2^3 * 3^3 * 5

no. of factors = 4*4*2 = 32
no. of perfect square factors = 1,4,9,36

so ... remaining factors= 32-4 = 28 ...
2603

how many no.s below 100 can be exxpressed as a difference of two perfect squares in only one way.??
a. 25
b. 26
c. 34
d. 35

2604

do u think that these are the options??


i think we can express every odd number in terms of difference of two squares


and every multiple of 4 also as difference of two??

2605

Each alphabet stands for a digit. Try and determine which letter stands for which digit.

2606

Please explain the method behind the following question

Find the remainder when (51)^203 is divided by 7

a) 4
b) 2
c) 1
d) 6

2607
Please explain the method behind the following question

Find the remainder when (51)^203 is divided by 7

a) 4
b) 2
c) 1
d) 6

my take 4

2*2.....(203 times) %7

(2*2*2) % 7 is 1

so 2*2*2........up to 201 digit remainder is 1

2*2% 7 = 4
2608

9456
1087
------
10543

2609

It can have multiple answers
it is quite clear that
S = 9
m = 1
R = 8
O=0
e+1 = n
now e can be 3/4/5;n ==> 4/5/6

2610
my take 4

2*2.....(203 times) %7

(2*2*2) % 7 is 1

so 2*2*2........up to 201 digit remainder is 1

2*2% 7 = 4


correct
thankyou brother
2611
9456
1087
------
10543


check your answer , answer almost there

and please post the approach also
2612
It can have multiple answers
it is quite clear that
S = 9
m = 1
R = 8
O=0
e+1 = n
now e can be 3/4/5;n ==> 4/5/6


thankyou brother:D
2613
please explain the method behind the following question

Find the remainder when (51)^203 is divided by 7

a) 4
b) 2
c) 1
d) 6


e(7)=6

203%6=5
51%7=2

2^5%7=4
2614
e(7)=6

203%6=5
51%7=2

2^5%7=4


(51)^203/7 = (2)^203/7 ...
Now according to Chinese Remainder Theoram:-
(2)^1%7=2
(2)^2%7=4
(2)^3%7=1

Total Cycles=3
=> 203%3=2
=>In cycle option 2 i.e;4 is the answer.
2615
(51)^203/7 = (2)^203/7 ...
Now according to Chinese Remainder Theoram:-
(2)^1%7=2
(2)^2%7=4
(2)^3%7=1

Total Cycles=3
=> 203%3=2
=>In cycle option 2 i.e;4 is the answer.



yeah the cyclicity method works fine with smaller numbers but what if the question is something like 497^34526 mod 504. Then euler is the best way for it. 😃