Official Quant thread for CAT 2013

MBA CAT 2024
14263
@ravi.theja said:
How many three-digit natural numbers have their units digit neither less than the hundreds digit nor the tens digit?
330?

units digit = 1 => 1 * 2 = 2
units digit = 2 => 2 * 3 = 6
units digit = 3 => 3 * 4 = 12
.
.
sum = sum(n(n+1)) for n from1 to 9
= sum (n^2 + n)
= n(n+1)(2n+1)/6 + n(n+1)/2
= 285 + 45
= 330
14264
@getupsid said:
q1) what is the maximum value of HCF of [n^2+17] and [(n+1)^2+17]. a)69b)85c)170d)none of theseq2) how many integers N in the set of integers {1,2,3,......100} are there such that N^2+N^3 is a perfect square.a)5b)7c)9d)11q3) the number of natural numbers n such that [(n+1)^2/n+7] is an integer is,a)4b)6c)5d)none of theseq4) a natural number satisfies the following conditions A) NUMBER IS HAVING ALL THE 9's B) IT IS DIVISIBLE BY 13how many digits are there in N ?a)5b)6c)7d)8
pleazz someone answer these quwstions
14265
@getupsid said:
pleazz someone answer these quwstions
2)n^2(n+1)
n+1 should b a perfect square
n=3,8,15....99
hence total =0

3)(n+1)^2/n+7
(n+7)^2-48-12n/(n+7)

n+7-12(n+4)/(n+7)

n+7+[-12(n+7)+36]/(n+7)
n+7-12+36/n+7

factors of 36 of the form n+7 such that n is a natural no
9,12,18,36

hence 4

4)6
any number written 6n times is divisible by 7,11,13


14266
@getupsid said:
pleazz someone answer these quwstions
q1) d) none of these
(n+1)^2 + 17 = (n^2 + 17) + 2n + 1
Put 2n +1 = 1
n = 0
HCF[17,18] = 306

q2) c) 9
N^2 + N^3 = a^2
sqrt(N + 1) = a/N
So (N + 1) is a perfect square
N = 3,8,15,24,35,48,63,80,99
Total = 9

q3) a)4
(n+1)^2/(n+7) = a
(n^2 + 2n + 1)/(n+7) = a
(n-5) + 36/(n+7) = a
36/(n+7) = k
n + 7 = 36/k
k = 1,2,3,4
So 4 numbers.

q4) b) 6
13 divisibility rule
99999 = 9999 + 36 = 10035 = 1003 + 20 = 1023 = 102 + 12 = 114
999999 = 100035 = 10023 = 1002 + 12 = 1014 = 117
14267
@ravi.theja said:
Let 123abc231bca312cab be a 18 digit number in base 7. How many ordered pairs (a,b,c) exist such that the given 18 digit number is divisible by 4?
@anytomdickandhary saar ...
14268
@ravi.theja said:
Let 123abc231bca312cab be a 18 digit number in base 7. How many ordered pairs (a,b,c) exist such that the given 18 digit number is divisible by 4?
@Zedai
if we write this number in base of 10 then we would write it as

1*7^17 + 2*7^16 + 3*7^15 +.................+c*7^2 + a*7 + b
now if we represent 7^k as (8-1)^k then remainder when divided by 4 would be +1 if k is even and -1 if k is odd

using this we can write that remainder would be
1*-1 + 2*1 + 3*-1 ...........................c*1 + a*-1 +b

i.e sum of digits at odd places - sum of digits at even places.

=(1+3+b+2+1+c+3+1+a)-(2+a+c+3+b+a+1+c+b)
=(11+a+b+c) - (6+2a+2b+2c)
=5-a-b-c
=4 - (a+b+c-1)

so if (a+b+c) is of form 4k+1 then the number would be divisible by 4.
so possible vales of (a+b+c) = 5 or 9 or 13 or 17 or 21 or 25

hence no. of ordered solutions = C(7,2) + C(11,2) + C(15,2) + C(19,2) + C(23,2) + C(27,2)

21+ 55 + 105 + 171 + 253 +351 = 956.

ATDH.
14269
@Logrhythm thanks the ans is correct could u explain more about it
14270
@ravi.theja said:
Let 123abc231bca312cab be a 18 digit number in base 7. How many ordered pairs (a,b,c) exist such that the given 18 digit number is divisible by 4?
7^0 mod 4 = 1
7^1 mod 4 = 3
7^2 mod 4 = 1
7^3 mod 4 = 3
Group the digits at odd places
(1 + 3 + b + 2 + 1 + c + 3 + 2 + a) * 3
Group the digits at even places
(2 + a + c + 3 + b + a + 1 + c + b) * 1

Total = (12 + a + b + c)*3 + (6 + 2a + 2b + 2c) * 1 mod 4
= 5a + 5b + 5c + 2 mod 4
So
5a + 5b + 5c = 4k - 2
a + b + c = (4k - 2)/5

k = 3
a + b + c = 2
c = 0 => 3 ways
c = 1 => 2 ways
c = 2 => 1 way
total = 6 ways

k = 8
a + b + c = 6
c = 0 => 7 ways
c = 1 => 6 ways
..
c = 6 => 1 way
total = 28 ways

k = 13
a + b + c = 10
c = 0 => 3 ways
c = 1 => 4 ways
c = 2 => 5 ways
c = 3 => 6 ways
c = 4 => 7 ways
c = 5 => 6 ways
c = 6 => 5 ways
total = 36

k = 18
a + b + c = 14
c = 2 => 1
c = 3 => 2
.
c = 6 => 5
total = 15

k = 23
a + b + c = 18
total = 1

Grand Total = 6 + 28 + 36 + 15 + 1 = 86
14271
@ravi.theja said:
How many three-digit natural numbers have their units digit neither less than the hundreds digit nor the tens digit?
330?

H......T.......U
1-9......0-9....9 = 9*10
1-8......0-8....8 = 9*8
.....
...........
so on

9*10+8*9+7*8+6*7+5*6+4*5+3*4+2*3+1*2 = 330
14272
@grkkrg can u plz explain d 1st solution... hcf( 17, 18) = 306 ??
14273
@getupsid said:
pleazz someone answer these quwstions
1. a. 69. use euclidean algorithm to find hcf --> hcf (a,b) = hcf ( b, a % b)
14274
@pirateiim478 said:
For case you need to divide by 2!. Rest is correct!! 14 is correctIn how many ways 200 can be represented as1. sum of consecutive numbers2. Sum of consecutive even numbers3.Sum of consecutive odd numbers
200 - consecutive numbers - no. of odd factors - 1
=> 3-1 = 2 ways..
b) sum of consecutive even no. - 2 ways (36,38,40,42,44) & (18,20,22,24,26,28,30,32)
C) sum of consecutive odd no - 3 ways (99,101), (47,49,51,53), (11,13,15,17,19,21,23,25,27,29)
14275
if 15x67y is divisible by 56, find x+y
14276
@bs0409 said:
if 15x67y is divisible by 56, find x+y
154672 is the number. 56 =8x7
last three digits must be divisible by 8 for number to be divisble by 8. so, y=2
and for x=4, number is divisble by 7.
so OA= 6?
14277
@bs0409 said:
if 15x67y is divisible by 56, find x+y
y = 4
x = 2

=> 6 ....?
14278
@bs0409 said:
if 15x67y is divisible by 56, find x+y
is it 6
14279
@ravi.theja Yes... 7a + b divisible by 4.

When a=0 , b =0,4,8
a= 1 , b = 1,5,9
a= 2, b = 2,6
a= 3 , b=3,7
a= 4, b = 0,4,8
a= 5, b= 1,5,9
6 , b= 2,6
7, b= 3,7
8, b= 0,4,8
9, b = 1,5,9
10, b = 2,6
.....
Till a=14, b = 2

I think aise Honda cahaheyez.. Baki I m not sure..
14280

If m is a natural no. and N = 2^m, what is the remainder when N! is divided by 2^N

14281

15x67y

56=8*7
Must apply divisiblity rules for both 8 and 7.
for 8..last three digits should be divisible by 2..
672 divisible by 8.
y=2.
15x672 -> 15x67-4 -> 15x63 == 15x6-6 -> 15x0==15x-0=15x
15-2x=7
solving x=4..
x+y=6
14282
@pavimai said:
15x67y
56=8*7
Must apply divisiblity rules for both 8 and 7.
for 8..last three digits should be divisible by 2..
672 divisible by 8.
y=2.
15x672 -> 15x67-4 -> 15x63 == 15x6-6 -> 15x0==15x-0=15x
15-2x=7
solving x=4..
x+y=6
Perfect method..........