Official Quant thread for CAT 2013

MBA CAT 2024
17086
@CrookDinu said:
Find the sum of all two digit positive integers which excede the product of their digits by 12.. share the approach
10a+b-ab=12
ab-10a-b+12=0
a(b-10)-1(b-10)=-2
(a-1)(10-b)=2
Now 2=1 *2 =2*1
we get 39 and 28....
SUM=67
17087
@CrookDinu said:
Find the sum of all two digit positive integers which excede the product of their digits by 12.. share the approach
10a+b = 12+ab
10a+b-ab -12 = 0
a(10-b) - (10-b)*1 +10 - 12 = 0
(a-1)(10-b) = 2
so (a,b) = (2,8) and (3,9)
hence, sum = 28+39 = 67...
17088
@CrookDinu said:
Find the sum of all two digit positive integers which excede the product of their digits by 12.. share the approach
let the number be 10x+y
10x+y-xy = 12
10x-10+y-xy = 2
10(x-1)+y(1-x) = 2
10(1-x)-y(1-x) = -2
(1-x)(10-y) = -2
this is possible when one is -2 and the other is 1 or -1 and 2
when (1-x) = -2 =>10-y = 1 =>y=9 --1 number = 39
when (1-x) = -1 10-y = -2 number = 28
when (1-x) = 2 10-y = -1 not possible.
same is the case ith next
so sum = 28+39 = 67
17089

Find the least value of n such that n! has exactly 2394 zeroes.

17090

Find the smallest positive integer n such that n! is a multiple of 10^2009

17091

How many zeros are present at the end of 25! + 26! +27! +28! +30! ?

17092
@TootaHuaDil said:
How many zeros are present at the end of 25! + 26! +27! +28! +30! ?
[25/5] + [25/5^2] = 5+1 =6
17093
@TootaHuaDil said:
Find the least value of n such that n! has exactly 2394 zeroes.
9585??
17094
@ravi.theja said:
9585??
correct...approach please
17095
@TootaHuaDil said:
Find the smallest positive integer n such that n! is a multiple of 10^2009
8050??
17096
Suppose n is an integer such that the sum of the digits of n is 2, and
10^10

a) 11
b) 10
c) 9
d) 8
17097
@vijay_chandola said:
[25/5] + [25/5^2] = 5+1 =6
OA is 7
17098
@ravi.theja said:
8050??
correct....again approach :)
17099
Ques:
N persons stand on the circumference of a circle at distinct points. Each possible pair of persons, not standing next to each other, sings a two €“minute song after the other. If the total time taken for singing is 28 minutes, what is N?

a) 5
b) 7
c) 9
d) None of these

17100
@TootaHuaDil said:
Find the least value of n such that n! has exactly 2394 zeroes.
For base 10 to get n zeroes for n!, n is approx. aroung 4n. In general for base n it is approx. P*n where P is no of prime factors below that base n.
So here check for no of zeroes for 4*2394=>9576. We get 2392 zeroes. Need 2 extra zeroes so 9585
17101
@TootaHuaDil said:
correct...approach please
dono the correct approach ..but it will be close to four times the number of zeros...here its 2394*4 = 9576..

but wen u check for 9576 u get 2392 zero's... 2 zeros less...so make add 2 zeros i.e., 10.i.e.,10/5 = 2 we get...9576+9 is enough..if it is 9575 also u need to increase 10 so..that it becomes 9585..same for the next question wer u need 2009 zero's..
17102
@leonidas. said:
Ques:N persons stand on the circumference of a circle at distinct points. Each possible pair of persons, not standing next to each other, sings a two €“minute song after the other. If the total time taken for singing is 28 minutes, what is N?a) 5b) 7 c) 9d) None of these
nc2-n = 28 ==> n =9??
17103
@leonidas. said:
Ques:N persons stand on the circumference of a circle at distinct points. Each possible pair of persons, not standing next to each other, sings a two €“minute song after the other. If the total time taken for singing is 28 minutes, what is N?a) 5b) 7 c) 9d) None of these
n-2C2=28 =>N =9??
17104
@leonidas. said:
Suppose n is an integer such that the sum of the digits of n is 2, and10^10 a) 11b) 10c) 9d) 8
10??
17105

In how many ways can 5 different balls can be arranged in 3 different boxes such that each box has atleast one ball?