Number System - Questions & Discussions

MBA CAT 2022
2436
unregisterjyoti Says
where from 77z came from...


Since 77 is the lcm of 7 & 11 therefore the number 77z will be a multiple of both such that 77z + 39 lie in the given interval. Now 39 will give the remainder required hence the number is of the form above.
2437
testdjjaydev Says
How many sets of two numbers will have the LCM as A^m * B^n where A and B are prime numbers and m and n are natural numbers??


guess it shud be (m+1)(n+1) because i can have as many different HCF's for the given number, and since lcm = constant, so those many different pairs. correct me if im wrong
2438

Approach whcih i used ...x^2-y^2=225
(x+y)*(x-y)=225
now since it is the product of two numbers (x+y)and (x-y)so we will find out the number of ways 225 can be represented as product of two numbers

it is 225*1,25*9,45*5,15*15 and 75*3
now 15*15 is not possible as x-y!=x+y
so we are left with four ways
equate x+y with one and x-y with the other ....4 is the right answer ...

2439

More 5 questions...

4> A= 1^101 + 2^102+ 3 ^101+........34^101, then find remainder is A is divided by 70......

Options --- 35,14,7,0
5>how many factors of 2^2x3^3x4^4x5^5x6^6 are even perfect squares that are also multiples of 72??

Options :- 96,84,72,120

6> A number when divided by N leaves a remainder of 4. When one -third of the number is divided by N , it leaves a remainder of 29, what is least such number greater than 1000.

Options:-1089,1152,1083,1079
7>The number of natural numbers less than 100 that can be expressed as the difference of two perfect squares in atleast one way is...

Options-750,792,810,749

8>Remainder when 32^32^33 is divided by 7......

Options:- 2 , 6 , 4 , 1

2440
More 5 questions...

4> A= 1^101 + 2^102+ 3 ^101+........34^101, then find remainder is A is divided by 70......

Options --- 35,14,7,0
5>how many factors of 2^2x3^3x4^4x5^5x6^6 are even perfect squares that are also multiples of 72??

Options :- 96,84,72,120

6> A number when divided by N leaves a remainder of 4. When one -third of the number is divided by N , it leaves a remainder of 29, what is least such number greater than 1000.

Options:-1089,1152,1083,1079
7>The number of natural numbers less than 100 that can be expressed as the difference of two perfect squares in atleast one way is...

Options-750,792,810,749

8>Remainder when 32^32^33 is divided by 7......

Options:- 2 , 6 , 4 , 1


7)750..all odd numbers and multiples of 4
using euler theorem ..
lambda(7)=6
solving 32^33 mod 7 =1

32 mod 7=4
2441
7)750..all odd numbers and multiples of 4
using euler theorem ..
lambda(7)=6
solving 32^33 mod 7 =1

32 mod 7=4


Can you elaborate how 750 come?

and i am unaware of this Euler's theorem
2442
Can you elaborate how 750 come?

and i am unaware of this Euler's theorem


first the question is how many numbers less than 1000 and not 100..
the number of odd numbers between 1 and 1000=500
and the multiples of 4=250
therefore 750...

learn euler,chinese reminder theorem to solve these type of problems ..TG IS the best source...;)

for the first question is the ans 35...?
2443
first the question is how many numbers less than 1000 and not 100..
the number of odd numbers between 1 and 1000=500
and the multiples of 4=250
therefore 750...

learn euler,chinese reminder theorem to solve these type of problems ..TG IS the best source...;)

for the first question is the ans 35...?



But yes it's 1000 not 100, but why odds and multiples of 4 only? how that thing clicked?

and can u provide link for remainder theorems?
Yes it is 35....!! gr88 now please tell method for it .......!!
2444

If the equation (x 2 - 2ax -4(a 2 + 1))(x 2 - 4x -2a(a 2 + 1)) = 0 has exactly 3 different roots, then how many values can "a" take?

2445

find the h.c.f of (2^100-1)and (2^120-1)?
a)2^10-1
b)2^20-1
c)1
d)none

2446

how many zeroes will be there at the end of 36!^36!
option 7*6!,8*6!,7*36!,8*36!

2447
how many zeroes will be there at the end of 36!^36!
option 7*6!,8*6!,7*36!,8*36!


is it 8^36!
2448
More 5 questions...

4> A= 1^101 + 2^102+ 3 ^101+........34^101, then find remainder is A is divided by 70......

Options --- 35,14,7,0
5>how many factors of 2^2x3^3x4^4x5^5x6^6 are even perfect squares that are also multiples of 72??

Options :- 96,84,72,120

6> A number when divided by N leaves a remainder of 4. When one -third of the number is divided by N , it leaves a remainder of 29, what is least such number greater than 1000.

Options:-1089,1152,1083,1079
7>The number of natural numbers less than 100 that can be expressed as the difference of two perfect squares in atleast one way is...

Options-750,792,810,749

8>Remainder when 32^32^33 is divided by 7......

Options:- 2 , 6 , 4 , 1




answer for question 7,
7)4-1=3,9-4=5,16-9=7,25-16=9,..,,

here,
1. all the odd numbers 2. all the multiples of 4 so ans is 750..
tell me if u need anything more..
2449

yes it is how you have solve it

2450
find the h.c.f of (2^100-1)and (2^120-1)?
a)2^10-1
b)2^20-1
c)1
d)none


since the base is same take the hcf of the powers ...
2^hcf(100,120)-1
2^20-1
2451
Is the remainder 1?

I'll explain if I'm right.


i got the answer.. thank u
2452

yes it is but how you have done it ?

2453

Originally Posted by unregisterjyoti
find the h.c.f of (2^100-1)and (2^120-1)?
a)2^10-1
b)2^20-1
c)1
d)none

since the base is same take the hcf of the powers ...
2^hcf(100,120)-1
2^20-1

unregisterjyoti Says
yes it is but how you have done it ?


100=10*10=2*5*2*5=20*5
120=10*12=2*5*2*2*3=20*6
HCF=(common terms in both)=2*5*2=20

as far as HCF of (2^100-1)and (2^120-1) is considered
we do the same thing we group common terms to get hcf
and

P.S: why does this place not have many doubts??
2454

How many divisors of 21600 are perfect squares?

Somehow I am getting answer as 15. Should it be 12? Can someone explain how?

2455
How many divisors of 21600 are perfect squares?

Somehow I am getting answer as 15. Should it be 12? Can someone explain how?


21600 = 2^5 * 3^3 * 5^2
so we can take 2^0, 2^2 nd 2^4 - 3 times, similarly for
3 its - 2 times
5 also 2 times
so total - 3*2*2 = 12