Official Quant thread for CAT 2013

MBA CAT 2024
16043
@TootaHuaDil said:
(1) All the divisors of 360, including 1 and the number itself, are summed up. The sum is 1170. What is the sum of the reciprocals of all the divisors of 360?
thats easy , 1170 is a taboo.. add up 1 , 1/2 , 1/4 , 1/8 , and 1/3 , 1/9 , and 1/5 , 1/360.
16044
@dragster said:
let a = xN+4 a/3= yN=29 ;solve u get N(3y-x)= 83 since x>y , N*any number - N*any other no. , must be a mutliple of N ( x n y can be 2 , 1 or 5 , 2 n so on...), and 83 is a prime no , u get N= 83..1st such nearby no = 1000 , next 1083.
thnks
16045
@TootaHuaDil said:
(5) N is a (n + 1) digit positive integer which is in the form anan - 1an - 2 . . . a2a1a0, where ai (i = 0, 1, 2, . . ., n) are digits and an0, thus N = an 10n + an-1 10n - 1+ . . . + a1 10 + a0, where 0 ai 9 and an0. We define F (N) = (an + 1) (an - 1 + 1) . . . (a1 + 1) (a0 + 1) For Example F(3407) = (3 + 1) (4 + 1) (0 + 1) (7 + 1) = 160. Identify the number of two digit numbers such that F(N) = N + 1.a. 9b. 1c. 6
b..
16046
16047
@swapnil4ever2u said:
how to solve above question using mass point geometry ??
16048
@TootaHuaDil said:
(5) N is a (n + 1) digit positive integer which is in the form anan - 1an - 2 . . . a2a1a0, where ai (i = 0, 1, 2, . . ., n) are digits and an0, thus N = an 10n + an-1 10n - 1+ . . . + a1 10 + a0, where 0 ai 9 and an0. We define F (N) = (an + 1) (an - 1 + 1) . . . (a1 + 1) (a0 + 1) For Example F(3407) = (3 + 1) (4 + 1) (0 + 1) (7 + 1) = 160. Identify the number of two digit numbers such that F(N) = N + 1.a. 9b. 1c. 6
ab is a 2 digit number

10a + b + 1 = (a+1)*(b+1) = ab + a + b + 1
ab = 9a
b = 9

a can take any digit from 1 to 9

so 9 numbers (19, 29, 39, 49, 59, 69, 79, 89, 99)

16049
Which of the following can be written as the sum of SQUARES of three odd natural numbers?
a) 5021 b)4445 c)3339 d)1233
16050
@ishu1991 all odd squares are of the form 8k+1...so the sum will be of the form 8k+3...only option c satisfies this...so answer is c)3339
16051
@sbharadwaj said:
1002.??
no answer is 1083
16052
@ishu1991 said:
Which of the following can be written as the sum of SQUARES of three odd natural numbers?
a) 5021 b)4445 c)3339 d)1233
Is it 3339??
square of odd natural numbers is of the 8k+1
so, 3(8k+1) = x
hence, the number should be div by 3
option options c and d hold...and after division by 3 the number minus 1 should be div by 8.
only option c holds...
16053
@The_Loser said:
is R = 19?If yes than i wud explain.
yes R=19 :)
16054
@swapnil4ever2u said:
how to solve above question using mass point geometry ??
22?
16055
@ishu1991 said:
Which of the following can be written as the sum of SQUARES of three odd natural numbers?a) 5021 b)4445 c)3339 d)1233
3339?
square of odd no. is of the form = 8k + 1
3(8k + 1) = 3339
k = 139
only 3339 holds true for the above eqn
16056
@swapnil4ever2u said:
how to solve above question using mass point geometry ??
are u sure about the options? m getting 46 sq units.
16057
If K = 231 — 319, then how many positive divisors of K^2 are less then K but do not divide K?
some one please elaborate the solution.
16058
@negiSannu said:
22?
@techgeek2050 said:
are u sure about the options? m getting 46 sq units.
actually i dont have the OA .. but can u explain ur approaches ..
16059
@TootaHuaDil said:
If K = 231 — 319, then how many positive divisors of K^2 are less then K but do not divide K?some one please elaborate the solution.



K=3*11^2*7*29 SO NO OF DIVISORS ARE 24
NOW K^2=3^2*11^4*7^2*29^2 SO NO OF DIVISORS ARE 135
NOw for k^2 leaving aside k halg of factors are lyingbelow k and half of the factors are lying above k so number of factors of k^2 lying below k =134/2 =67 there numbers which are multiples k but not multiples og k^2 =67-23(not considering k here)=44
16060
@swapnil4ever2u said:
actually i dont have the OA .. but can u explain ur approaches ..
I used ladder theorem
16061
@swapnil4ever2u said:
actually i dont have the OA .. but can u explain ur approaches ..
answer is 22.
see the attached file
ED = d
DB = c
DC = a
DF = b
from the given areas
since the heights are same, area will be in ratio of bases
a/b = 2
c/d = 5/4
area of AED = y and ADF = x
using the same concept as above
(5 + x) / y = 5/4

(8+y)/ x = 2

solving them.. x = 10 and y = 12
area remaining part = x + y = 10 + 12 = 22
16062

1000!/10^n .What can be the largest possible integral value of n??