Number System - Questions & Discussions

MBA CAT 2022
2836
How many numbers from 7, 13, 14, 21, 26, 39 and 91 divide (69^12 38^12)?
OPTIONS

1)7
2)5
3)3
4)1
5)None

My take is 3.

It is divisible by 7,13 and 91.
Used Euler's.

kannabal Says
Can you please explain.


See. The power is 12.
E(7) = 6.
12 mod 6 = 0.
=> 69^0 - 38^0 mod 7 = 1 - 1 mod 7 = 0.
Similarly, E(13) = 12.
And since 91 = 7*13. So it is divisible by all 3.

Hope it helps.
2837

hi...i have a doubt...
can we use euler's method to find the reminder of a number with some power of when divided by 9.
reason for my confusion -

eulers number of 9 = 6.
so eg 1 : number (2^6)/9 --- reminder of this would be 1...since it becomes 2^0 = 1(0 - reminder when power 6 is divided by euler number) so 1/9(reminder) = 1.
eg : number (3^6)/9 --- reminder as per the above step using euler's method,would be 1,but in actually the reminder is 0..
Can you please clarify my doubt..

2838
hi...i have a doubt...
can we use euler's method to find the reminder of a number with some power of when divided by 9.
reason for my confusion -

eulers number of 9 = 6.
so eg 1 : number (2^6)/9 --- reminder of this would be 1...since it becomes 2^0 = 1(0 - reminder when power 6 is divided by euler number) so 1/9(reminder) = 1.
eg : number (3^6)/9 --- reminder as per the above step using euler's method,would be 1,but in actually the reminder is 0..
Can you please clarify 2 and 9 are co-prime to each other bt 3 and 9 are nt co-prime and dats the reason
2839
hi...i have a doubt...
can we use euler's method to find the reminder of a number with some power of when divided by 9.
reason for my confusion -

eulers number of 9 = 6.
so eg 1 : number (2^6)/9 --- reminder of this would be 1...since it becomes 2^0 = 1(0 - reminder when power 6 is divided by euler number) so 1/9(reminder) = 1.
eg : number (3^6)/9 --- reminder as per the above step using euler's method,would be 1,but in actually the reminder is 0..
Can you please clarify my doubt..


No, you cant always use euler.
What you need to check is that the numerator and denominator should be co-prime.

Eg - 3^6 mod 9.
Here 3 and 9 are not prime so we wont be able to apply euler.

Hope it is clear now. 😃
2840

A two digit no when added to another number which is its reverse
gives a perfect square .
How many such numbers are there .. ??

2841
A two digit no when added to another number which is its reverse
gives a perfect square .
How many such numbers are there .. ??


only one number 56 and 65.i.e121...logic is 10a+b+10b+a=11
(a+b),so 11square only posibility
2842
A two digit no when added to another number which is its reverse
gives a perfect square .
How many such numbers are there .. ??


10a+b+10b+a = x^2.
=> 11a+11b = x^2.
=> a+b = x^2/11.
So the perfect square should be a multiple of 11.
=> 11(a+b) = 121.
or (a+b) = 11.
=> (a,b) = (9,2);(8,3);(7,4);(6,5).
Hence, total 4 unordered or 8 ordered pairs.
2843

Thanks enceladus

2844

What is the remainder of 32^32^32 / 7
ie 32 raised to power 32 raised to power 32.
Thanks

2845
What is the remainder of 32^32^32 / 7
ie 32 raised to power 32 raised to power 32.
Thanks


4^32^32/7

4^3k+1/7

Remainder is 4
2846
4^32^32/7

4^3k+1/7

Remainder is 4


How did u arrive at 3k+7
2847
only one number 56 and 65.i.e121...logic is 10a+b+10b+a=11
(a+b),so 11square only posibility


ohh...yes 121 is the only perfect square bt we have to luk for two digit number...shht..question half done..
2848
4^32^32/7

4^3k+1/7

Remainder is 4


yes 4^32^32/7 nw 4/7=4,4^2/7=9,4^3=64/7=1 cyclicity is 3 ,32^32/3=(33-1)^32/3=1 remainder,,thats y 4/7=4 which is remainder
2849
AlbusDumbeldore Says
How did u arrive at 3k+7


Powers of 4 i.e. 4^x when divided by 7 has a cyclicity of 3
x in this case is 32^32 and hence, 32^32 is represented as 3k+1
so as to get remainder as rem(4^3k/7)*rem(4^1/7)= (1)(4)=4
2850
What is the remainder of 32^32^32 / 7
ie 32 raised to power 32 raised to power 32.
Thanks

Soln)

R(32^32^32/7)=R(4/7)=4
2851
What is the remainder of 32^32^32 / 7
ie 32 raised to power 32 raised to power 32.
Thanks


It should be 4.

E(7) = 6.
32^32 mod 6 = 4
=> 32^4 mod 7 = 4^4 mod 7 = 4.
2852

Q1:
100 students of a class appeared for a test that had 30ques.Each Ques carried 1 mark for correct answer and there were no -ve markings for wrong answers.
The no. of students in the class who attempted at least n ques wrong is given by An = 60-2n(n>0).
The average marks of the class can be:

1- 21.3
2- 22.0
3- 24.5
4- 25.33

Q2:
Unequal side of isosceles triangle is 2cm long.The medians drawn to the equal sides are perpendicular to each other.
Area of triangle is:
1- 3 cm^2
2- sqrt(10) cm^2
3- 2*sqrt(3) cm^2
4- 2*sqrt(2) cm^2

2853

My answers in bold

Q1:
100 students of a class appeared for a test that had 30ques.Each Ques carried 1 mark for correct answer and there were no -ve markings for wrong answers.
The no. of students in the class who attempted at least n ques wrong is given by An = 60-2n(n>0).
The average marks of the class can be:

1- 21.3
2- 22.0
3- 24.5
4- 25.33

Q2:
Unequal side of isosceles triangle is 2cm long.The medians drawn to the equal sides are perpendicular to each other.
Area of triangle is:
1- 3 cm^2
2- sqrt(10) cm^2
3- 2*sqrt(3) cm^2
4- 2*sqrt(2) cm^2
2854
silzeal Says
My answers in bold

Answer is 21.3

Students who has at least 29 questions wrong = 60-2n= 60-2*29= 2
Students who has at least 28 questions wrong = 60-2n= 60-2*28= 4
.
.
.
Students who has at least 1 questions wrong = 60-2n= 60-2*1= 58

therefore there are 2 people of each of 1,2 ,3.........29 marks

till now we know marks of 58 people and remaining 42 people have 30 marks each...

average= (2*1+2*2+..........+2*29+ 42*30)/100= 21.3

ans for the other question is 3 cm2
2855
Q1:
100 students of a class appeared for a test that had 30ques.Each Ques carried 1 mark for correct answer and there were no -ve markings for wrong answers.
The no. of students in the class who attempted at least n ques wrong is given by An = 60-2n(n>0).
The average marks of the class can be:

1- 21.3
2- 22.0
3- 24.5
4- 25.33



Soln)

Av.=42*30+2*(29+28+27+.............+1)/100=21.3