Official Quant thread for CAT 2013

MBA CAT 2024
5041
@audiq7 said:
what is the highest power of x to yield an integral ans for 167!/(24!)^x?
7?
5042

i have n biscuits which i want to distribute among 7 children. but the condition is, at the most only 4 children can get equal no.of biscuits and no two of the remaining children can get same no.of biscuits. Each child get at least one biscuit. minimum value of n??

9
11
13
22

5043
@audiq7 said:
i have n biscuits which i want to distribute among 7 children. but the condition is, at the most only 4 children can get equal no.of biscuits and no two of the remaining children can get same no.of biscuits. Each child get at least one biscuit. minimum value of n??9111322
13 - 1+1+1+1+2+3+4=13

edit - took only 6 children at first
5044
@audiq7 said:
i have n biscuits which i want to distribute among 7 children. but the condition is, at the most only 4 children can get equal no.of biscuits and no two of the remaining children can get same no.of biscuits. Each child get at least one biscuit. minimum value of n??
9
11
13
22

13..
5045
@audiq7 said:
i have n biscuits which i want to distribute among 7 children. but the condition is, at the most only 4 children can get equal no.of biscuits and no two of the remaining children can get same no.of biscuits. Each child get at least one biscuit. minimum value of n??9111322
4*1 + 2 + 3 + 4 = 13.

ATDH.
5046
@krum said:
9
OA:13
5047
@krum said:
9
13 ayega !
5048

last two digits of (1^3+2^3......49^3)?

5049
@audiq7 said:
last two digits of (1^3+2^3......49^3)?
25

Direct formula...
{n(n+1)/2}^2
5050
@audiq7 said:
last two digits of (1^3+2^3......49^3)?
25?

49^2 * 50^2 / 4
= 2401 * 2500 / 4
= 6002500/4

last two digits = 25
5051
@audiq7 said:
i have n biscuits which i want to distribute among 7 children. but the condition is, at the most only 4 children can get equal no.of biscuits and no two of the remaining children can get same no.of biscuits. Each child get at least one biscuit. minimum value of n??
9
11
13
22

1+1+1+1+ 2+ 3+ 4 =13
5052
@audiq7 said:
last two digits of (1^3+2^3......49^3)?
25
5053
@audiq7 said:
last two digits of (1^3+2^3......49^3)?
25?
5054
@audiq7 said:
i have n biscuits which i want to distribute among 7 children. but the condition is, at the most only 4 children can get equal no.of biscuits and no two of the remaining children can get same no.of biscuits. Each child get at least one biscuit. minimum value of n??9111322
13 ?
1,1,1,1,2,3,4 ?
5055
@audiq7 said:
last two digits of (1^3+2^3......49^3)?
n.(n+1)/2 ^2
49*25 ^2
25 ?
5056
@audiq7 said:
last two digits of (1^3+2^3......49^3)?

n2(n+1)2 /4
sum contains 25 in last


5057
@krum said:
24!=2^22*3^10*5^2*7^2*11^2*13*17*19*23here 23 will be the limiting value so 7
yes 23 will limit the value of x,but we need to find out the max value of x,...so we need to divide it by sm other no. which will maximize the value of x is dat right or am i misinterpretin it..
5058
@Ashmukh said:
yes 23 will limit the value of x,but we need to find out the max value of x,...so we need to divide it by sm other no. which will maximize the value of x is dat right or am i misinterpretin it..
tell me what u think will be the answer, than i can confirm it
5059
@krum said:
tell me what u think will be the answer, than i can confirm it
it shud be 12 if i divide 167! by 13 will maximize the value of x,and clearly by dividing it by other prime no.s they are
5060
@Ashmukh said:
it shud be 12 if i divide 167! by 13 will maximize the value of x,and clearly by dividing it by other prime no.s they are
the question says "integral ans for 167!/(24!)^x? " ..we need an integral quotient

if we take x=12, we will get some fraction, 167! has only 7 factors of 23 so it will be 7 only