Official Quant thread for CAT 2013

MBA CAT 2024
25028
@aditi88 said:
ans s nt given but n is gvn as 2^5*3^3
yeah because it is given 2n has 28 factors
assume n=2^a*3^b
number of factors will be (a+1)*(b+1)
for 2n you will have 2^(a+1)*3^b
number of factors will be (a+2)*(b+1)=28...........(1)
for 3n you will have 2^a*3^(b+1)
number of factors will be (a+1)*(b+2)=30...........(2)
solve the two equations you will get the answer
25029
@viewpt said:
Q: 37^47^57^101 div by 103 R?
E(103) = 102
E(102) = 32

57^101 mod 32 = 25

47^(32k +25) mod 102 = 47^25 mod 102 = 47

37^47^57^101 mod 103 = 37^(102k+47) mod 103 = 37^47 mod 103 = 42

So, is the answer = 42 ? ..
25030
how many 3 digit numbers have their unit digit neither less than the hundred digit nor less than the tens digit ??
25031
@ChirpiBird said:
total codes = 9*10= 90digits that can cause confusion 0,1,6,8,95C2 = 10ways.and they can be mirrored so .... 10*2 = 20but here 69 when mirrored is 69 only.90-20 +1 = 71 numbers?EDIT : 11, 88 should also be considered just like 69.. total 73?
the "0" you have considered for the mirror effect that will cause confusion will effectively create confusion only if you have considered 01,02,.... to be valid numbers. but according to the question they arent valid so you shouldnt consider "0" for the confusion thing as it wouldnt lead to any confusion. i am not able to put into appropriate words hope this is clear enough
25032
@swapnil4ever2u said:
how many 3 digit numbers have their unit digit neither less than the hundred digit nor less than the tens digit ??
720?
25033
@iLoveTorres nah... OA is 330.. !!
25034
@iLoveTorres ya answer is correct.......plz post the method
25035
@swapnil4ever2u said:
how many 3 digit numbers have their unit digit neither less than the hundred digit nor less than the tens digit ??
When last digit is 9, total ways = 9*10
When last digit is 8, total ways = 8*9 and so on...

So, Total = 1*2 + 2*3 ....+9*10
= n(n+1) = n^2 + n = n(n+1)(2n+1)/6 + n(n+1)/2 for n = 9
= 9*10*19/6 + 9*5 = 285 + 45 = 330 ??
25036
@Abir1103 said:
Q. a boy n a girl r riding bicycles on the edge of a circular lake .the boy rides at a speed of 5mps n the girl wid a speed of 4mps . if they travel in same direction the boy wld overtake the girl for the second time aftr 14 mins n 40 sec . hw far is the girl frm the starting pt , wen they meet for the third time in case they r travelling in oppo dir ??soln needed .
look the speed is in miles/sec and the time given is in minutes.. so first convert it to seconds you will get 880.
now when they travel in the same direction the relative speed is 5-4=1 mile per second.
Now for the person to meet or over take he must cover a distance of the entire circular track i.e "d".
so 1mps = d/t
the second time the boy will over take when the distance is 2d
hence 1mps = 2d/880 ==> d=440miles
Now when they are travelling in opposite direction relative speed = 4+5=9mps
each meeting point will be when they together travel the distance d. So third meeting will be when the total distance traveled is 3d=1320
the girl will travel 4/9 part of the distance i.e 4/9*1320 ~587. To find the distance from the starting point subtract d from it = 587-440=147
25037
What is the remainder when 3^1001 is divided by 1001?
25038
@Abir1103 said:
Q. a boy n a girl r riding bicycles on the edge of a circular lake .the boy rides at a speed of 5mps n the girl wid a speed of 4mps . if they travel in same direction the boy wld overtake the girl for the second time aftr 14 mins n 40 sec . hw far is the girl frm the starting pt , wen they meet for the third time in case they r travelling in oppo dir ??soln needed .
let the girl and boy move in same direction
let d be circumference
d/(5-4) * 2=880
d=440
now let them move in same direction
440/9 -- time when they meet once
440*3*4/9--- distance moved by girl in third meeting
so 48440/3 - 440 = 440/3 = 147 -- distance from the start point
25039
@The_Loser said:
What is the remainder when 3^1001 is divided by 1001?
1001 = 7*11*13

3^1001 mod 7 = 5
3^1001 mod 11 = 3
3^1001 mod 13 = 9

No of the form = 7k + 5 = 11l + 3 = 13m + 9

No 971 satisfies! Hence, remainder is 971 ( Edited )
25040
@The_Loser said:
What is the remainder when 3^1001 is divided by 1001?
E(1001)=720
3^281 mod 1001
1001=7*11*13

3^281 mod 7=3^5 mod 7 = -1*9 mod 7=5
3^281 mod 11 =3
3^281 mod 13=3^5 mod 13=9

7a+5=11b+3

77c+47=13d+9
1001n+971

971??
25041
@ScareCrow28 said:
1001 = 7*11*13 3^1001 mod 7 = 53^1001 mod 11 = 33^1001 mod 13 = 33^1001 mod 1001 = 45Hope kahi mishtake na hui ho
shouldn't mod 13 be 9??
25042
@Subhashdec2 said:
shouldn't mod 13 be 9??
Yeah...
25043
@ScareCrow28 said:
1001 = 7*11*13 3^1001 mod 7 = 53^1001 mod 11 = 33^1001 mod 13 = 33^1001 mod 1001 = 45Hope kahi mishtake na hui ho
mistake ho gayi sir.
13 se remainder 9 hai
25044
@Subhashdec2 said:
E(1001)=7203^281 mod 10011001=7*11*133^281 mod 7=3^5 mod 7 = -1*9 mod 7=53^281 mod 11 =33^281 mod 13=3^5 mod 13=97a+5=11b+377c+47=13d+91001n+971971??
77c+47=13d+9

after that, how you straight way clicked 971. ?
25045
@The_Loser said:
mistake ho gayi sir.13 se remainder 9 hai
Haan bhai.. I was feeling that some mistake has happened! May be I should just stick to pen and paper and not emulate @scrabbler :P
25046

remainder when 25! is div by 10^7 is?

25047
@viewpt said:
Q: 37^47^57^101 div by 103 R?
Phi (Phi(Phi(103)))=16; Phi(Phi(103))=32; Phi(103)=102, where Phi(n) represents the Euler Totient function of n.

57^101= 25(mod32);
47^25=(47^8)^3 * 47(mod102)=1*47(mod102)=47(mod102);
37^47=(37^8)^5 * 27000*37(mod103)=14*27000*37(mod103)=42(mod103)

Hence the remainder is 42

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