Official Quant thread for CAT 2013

MBA CAT 2024
13743
@bs0409 said:
If intelligent, they shld be satisfied with 1 diamond. Otherwise they are gonna die........!!!!!!!
Find solutions N^4+4=p where N is natural number and p is prime number.
Is it 1??
N^4 = p - 4
p is prime, hence odd > 3 as N^4 cannot be odd
p - 4 = odd - even = odd
n^4 can end in either 1,6,5 and 0
so we can get only 1 and 5 as per our equation, also we can exclude gettin 1 as all other numbers ending in 5 would be multiples of 5 (not prime)
(N,P) = (1,5)
13744
@bs0409 said:
Counting the vote of the oldest thief, I think we will get 97..........
Ah this one didn't consider the vote of the eldest thief, didn't notice. But then I don't think A will ever get the coins. B will never agree to vote for A :/ or does the question mean after A dies, B becomes the oldest thief and hence his vote doesn't count :splat:
13745
@Estallar12 said:
Method I -264 = 2^3*3*11(1,other) = 15 (total factors -1)(2,other) = 2(4,other) = 2(8,other) = 2{(2*3,11),(4*3,11),(8*3,11),(2*11,3),(4*11,3),(8*11,3),(2,3*11),(4,3*11),(8,3*11)} = 9(3,11) = 1Total = 15+6+9+1=31Method II -Condition for two divisors of any number n to be co-prime to each other(i) Let N=a^m*b^n has (m+1)(n+1)-1+mn numbers of factors which are co-prime to each other.If N=12, then as 12=2^2*3,so it has (2+1)*(1+1)-1+2=7 sets which are co-prime to eachother. They are: (1,2),(1,3),(1,4),(1,6),(1,12),(2,3),(3,4).(ii) Similarly we can deduce the formula for higher orders.If N=a^m*b^n*c^p then it will have (m+1)(n+1)(p+1)-1+mn+np+mp+3mnp co prime factors.
Awesome :clap:
13746
@ankita14 said:
Ah this one didn't consider the vote of the eldest thief, didn't notice. But then I don't think A will ever get the coins. B will never agree to vote for A :/ or does the question mean after A dies, B becomes the oldest thief and hence his vote doesn't count
I think I just got this question, after some possibly permanent over-heating of the brain :splat:
Correct me, my thinking may have gone wrong somewhere :embarrassed:

by the backward process thing: if D+E are left, D will definitely lose and die, and E gets all 100.
if C+D+E are left, D would vote for C to survive, E would not.
if B+C+D+E are left- C has a chance to win so will vote against, D and E's votes could be bought by a diamond each - because if B died, C would win and D, E both get 0.
if A,B,C,D,E are there, since B has a chance will vote against, for C - if A dies and B proposes, C would get nothing so C could be bought by 1 diamond, and D,E could also be bought by 1 diamond each (same reason as in before vala case)
 is this the reasoning @gyrodceite ?
13747

although it makes little sense when i remember that only 50% votes are required for A to win :splat: :banghead: :banghead: :banghead:

13748
For N values of t, where t=12
13749

N! is completely divisible by 13^52.what is sum of the digits of the smallest such number N

13750
@pavimai said:
N! is completely divisible by 13^52.what is sum of the digits of the smallest such number N
16 is the answer
13751


A teacher wanted to adminster a multiple choice(each question having six choices) based quiz of high difficulty level to a class of sixty students.The quiz had sixty questions.The probability of selecting the correct answer for a good and brilliant student was 0.2 and 0.25 respectively.All the students were seated serially in 10 rows and 6 columns.

q) three good students were seated next to each other.What is the probability of them having the same incorrect choice for four consecutive questions ?

i did it like

to get one incorrect answer, probabilty is 1/5

so all INCORRECT for one student is 1/5*1/5*1/5*1/5

so overall for 3 students (1/5*1/5*1/5*1/5)^3 but it is not in options.where did i go wrong ?

answer is 4/3125.
13752
@pavimai said:
N! is completely divisible by 13^52.what is sum of the digits of the smallest such number N
Is it 16??
637! is the smallest such N!
13753

What are the materials/books that would be good..i have done arun sharma and time which fetched me a 93% in QA..want to increase my speed and exposure to varied type of questions..pls suggest some nice books/materials..

13754
@Logrhythm said:
Is it 12?? 390! is the smallest such N!
It wud be 637! I guess.
13755
@Calvin4ever said:
It wud be 637! I guess.
yea. corrected it....i misread 52 as 32 at first
13756
@pavimai said:
N! is completely divisible by 13^52.what is sum of the digits of the smallest such number N
N is 637
Hence, sum of digits = 16
13757
@nirzone said:
What are the materials/books that would be good..i have done arun sharma and time which fetched me a 93% in QA..want to increase my speed and exposure to varied type of questions..pls suggest some nice books/materials..
This thread is for quant discussion , ask your question here http://www.pagalguy.com/posts/4406166

13758
@doctoriim logic pls
13759

Find a number such that after dividing it by 13 once we get 49 as quotient... smallest such number is 49X13=637... and 649 is the greatest such number!!

13760
@nirzone said:
@doctoriim logic pls
Minimum N! which is divisible by 13^52 should have 52 number of 13's in it.
Now, 13! has one 13
26! has two 13's
169! has fourteen 13's . {(13*13) * (168) * (167) ......(13*12)......}
hence going by this logic (13*49)! has 52 number of 13's.
13*49 = 637
Sum of digits = 16
P.S. Hope u understood!
13761
@nirzone said:
@doctoriim logic pls
52 = 50+2, 49+3, 48+4....
49/13 = 3.x
hence 49*13 would be the smallest...

13762

Approach for N! is completely divisible by 13^52.what is sum of the digits of the smallest such number N?