Official Quant thread for CAT 2013

MBA CAT 2024
24908
@MANJULNEOGI said:
41!^2 find remainder by 83 ?
Is it 1?
Applied Wilson's theorem
If p is a prime number, (p-1)! will give a remainder of -1 when divided by p
So 82! divided by 83 gives remainder -1
41! should also give a remainder -1
So answer -1^2 = 1
24909

@nramachandran thnx
24910
@MANJULNEOGI said:
41!^2 find remainder by 83 ?
82!/83 =82

now (1.2.3. ....41)(42.43.....82) /83=82

-(1.2.3.. 41)^2 /83=82

- 41!^2/83=82

so reminder =1


24911
@MANJULNEOGI said:
41!^2 find remainder by 83 ?
Extending the wilson's theorem we get (p-2)!=1modp for any prime p
(81)!=1mod83=> 41!*(-41*-40*....-3*-2)=1mod83 =>(41!) ^2 = 1mod 83

Team BV - Vineet
24912
If n is an integer from 1 to 96, what is the probability for n*(n+1)*(n+2) being divisible by 8?
24913
@MANJULNEOGI said:
If n is an integer from 1 to 96, what is the probability for n*(n+1)*(n+2) being divisible by 8?
Either n is even i.e. 48 values Or n+1 is multiple of 8 i.e. 12 values....so total favorable cases for values of n are 48 + 12 = 60.
Thus the required probability is = 60/96 = 5/8

Team BV - Kamal Lohia
24914
@jain4444 said:
What are the last three digits of the greatest common divisor of 2^(2^100−1) − 1 and 2^(2^105−1) − 1 ?
For non negative numbers m and n, gcd (x^m - 1, x^n - 1)= x^gcd(m,n) - 1. We need to use this recursively in the problem at hand.

So, the required gcd = 2^{2^gcd(100,105) - 1} - 1=2^{2^5 -1}-1 =2^31 - 1

Last three digits= (2^31 - 1)(mod 1000)= (2*24^3-1)(mod1000)=647(mod1000)

Team BV - Vineet
24915
@techgeek2050 said:
What is the remainder when 1^2 * 2^3 + 2^2 * 3^3 ..... + 99^2 * 100^3 divided by 36?
Clearly all the terms are multiples of 4 and all the terms that are of the form 3k are divisible with 36

For the remaining terms if we take remainder by 9, we have 15*11(mod9) = 3mod9

9a+3=4b=> min(a)=1 => Required remainder 12

Team BV - Vineet
24916
@MANJULNEOGI said:
If n is an integer from 1 to 96, what is the probability for n*(n+1)*(n+2) being divisible by 8?
n has to be a multiple of 2 so that n+2 becomes a multiple of 4 -> 96/2 = 48
or n+1 can be a multiple of 8 -> 96/8 = 12
total = 60

so, prob = 60/96 = 5/8
24917

can anybody explain why the no. of diagonals formed in a decagon is--> 10c2-10 ?


p.s.- explanation for the "-10" needed

24918
@MANJULNEOGI said:
If n is an integer from 1 to 96, what is the probability for n*(n+1)*(n+2) being divisible by 8?
we can choose 5 forms of n from every 8 forms

8k , 8k + 2 , 8k + 4 , 8k + 6 , 8k + 7

and 96 is multiple of 8

so , prob. = 5/8
24919
@MANJULNEOGI said:
If n is an integer from 1 to 96, what is the probability for n*(n+1)*(n+2) being divisible by 8?
The expression will be divisible for n even and n=8k-1. So the required probability = (48+12)/96 = 60/96 = 5/8

Team BV -Vineet
24920
@vishcat said:
can anybody explain why the no. of diagonals formed in a decagon is--> 10c2-10 ?p.s.- explanation for the "-10" needed
coz in the selection 10c2 you are also selecting a set of 2 points which are colinear to each other, and those are 10 (adjacent sides)
24921
@vishcat said:
can anybody explain why the no. of diagonals formed in a decagon is--> 10c2-10 ?p.s.- explanation for the "-10" needed
Diagonal is a line segment formed by joining two non-adjacent vertices of a polygon.
In a decagon, there are 10 vertices and total number of line segments formed by joining any two of the vertices is = C(10, 2).
And in these lines, included are the 10 sides of the decagon which are not diagonal.
So total number of diagonals is = C(10, 2) - 10.

Alternatively, you can count as every vertex is used in 10 - 3 = 7 diagonals as it is not to be joined with itself and its two adjacent vertices. So total number of diagonals is = 10(10 - 3)/2. :)

Team BV - Kamal Lohia
24922

one from my side: Find the largest number which cannot be written as 3x + 5y + 7z where x, y, z are non-negative integers.

one more on similar lines: Find the largest number which cannot be written as 13x + 5y + 17z where x, y, z are non-negative integers.

Team BV - Kamal Lohia

24923
"And in these lines, included are the 10 sides of the decagon which are not diagonal."
i am not able to understand this line.
@bodhi_vriksha u may take another example to explain this
24924
@bodhi_vriksha said:
one from my side: Find the largest number which cannot be written as 3x + 5y + 7z where x, y, z are non-negative integers.one more on similar lines: Find the largest number which cannot be written as 13x + 5y + 17z where x, y, z are non-negative integers.Team BV - Kamal Lohia
1) 4
2) 29

Not sure though...
24925
@Logrhythm i am having trouble to visualize the explanation.
24926
@vishcat said:
"And in these lines, included are the 10 sides of the decagon which are not diagonal."i am not able to understand this line.@bodhi_vriksha u may take another example to explain this
C(10, 2) means all lines formed by joining any two vertices at a time. Isn't it?

Now the sides of the decagon are also such lines only which are formed by joining two vertices at a time. But they are not "diagonals". Right.

That's why subtracting 10 from C(10, 2)

Team BV - Kamal Lohia
24927
@bodhi_vriksha got it... thanks.