@Budokai001 right....
one person is wearing black and white shoes as his pair.If he is having 20 white and 20 black shoes. what will be the least number of time he has try to take his correct pair of shoes??
@nagpal9 said:one person is wearing black and white shoes as his pair.If he is having 20 white and 20 black shoes. what will be the least number of time he has try to take his correct pair of shoes??
minimum =2
maximum= 20+1
@nagpal9 said:@Budokai001 can u explain??
for minimum
First he can take any one shoe out of 40 ->assume its white
Next he can take black in next time itself as we want to minimize it..
so 1+1=2
For maximum
The first 20 he takes out of 40 turns out to be of a single colour only .
The next one he takes will be of different colour for sure as the 20 of the original colour he drew is already taken ,so 20+1
If ans is wrong,i mustve misread the question
dinesh is seated 7th from left and satish is seated 12 from right. when they interchange their positions, dinesh is seated at 22nd from left. how many people are there?
@nagpal9 said:dinesh is seated 7th from left and satish is seated 12 from right. when they interchange their positions, dinesh is seated at 22nd from left. how many people are there?
33 ?
Q1.N=7!^3. How many factors of N are multiples 10?
a) 736
b) 1008
c) 1352
d) 894
Q2.how many zeroes are at the end of36!^36!?
a)7*6!
b)8*6!
c)7*36!
d)8*36!
@getupsid said:Q1.N=7!^3. How many factors of N are multiples 10?a) 736b) 1008c) 1352d) 894Q2.how many zeroes are at the end of36!^36!?a)7*6!b)8*6!c)7*36!d)8*36!
1)
N=(2*3*4*5*6*7)^3=(2^12*3^6*5^3*7^3)=10*(2^11*3^6*5^2*7^3)
so 12*7*3*4 = 1008
2)
36/5=7
7/5=1
so 8*36!
N=(2*3*4*5*6*7)^3=(2^12*3^6*5^3*7^3)=10*(2^11*3^6*5^2*7^3)
so 12*7*3*4 = 1008
2)
36/5=7
7/5=1
so 8*36!
any tricks to make additions fast in DI??????
@getupsid there is a short cut for no of zeros of any factorial of a number say( n!) all you need to do is
divide the no with 5 , 5^2,5^3..untill the value after division is less than 1(considering the integer part only)...
36!...>36/5+36/5^2 (considering only integer part) = 7 + 1 =8 ..since the next value is
therefore answer is 8^36!
divide the no with 5 , 5^2,5^3..untill the value after division is less than 1(considering the integer part only)...
36!...>36/5+36/5^2 (considering only integer part) = 7 + 1 =8 ..since the next value is
therefore answer is 8^36!
Q. Each computer system has a password which is 6 to 8 digits long, and contains of alphabets and digits. Each password must have atleast one digit. How many passwords are possible ?
@krum said:1)N=(2*3*4*5*6*7)^3=(2^12*3^6*5^3*7^3)=10*(2^11*3^6*5^2*7^3)so 12*7*3*4 = 10082)36/5=77/5=1so 8*36!
1. B
2. D
2. D
N is the largest 3-digit number which when divided by 11 and 9 leaves 9 and 7 as the respective remainders. The sum of the digits of N is
@aimiift2012 said:N is the largest 3-digit number which when divided by 11 and 9 leaves 9 and 7 as the respective remainders. The sum of the digits of N is
988?
@aimiift2012 said:N is the largest 3-digit number which when divided by 11 and 9 leaves 9 and 7 as the respective remainders. The sum of the digits of N is
990-2=988
@getupsid said:@krumpleaz explian q.2 a bit......Q2.how many zeroes are at the end of36!^36!?a)7*6!b)8*6!c)7*36!d)8*36!
36/5=7
7/5=1
no of zeros in 36!/5=8 and (8no. of zeros) ^36!= total no. of zeros
so 8*36!
eg:
no. of 0s means div by 5
and suppose Q asks find the no of zeros in 10^2 then ans will be 2 .
7/5=1
no of zeros in 36!/5=8 and (8no. of zeros) ^36!= total no. of zeros
so 8*36!
eg:
no. of 0s means div by 5
and suppose Q asks find the no of zeros in 10^2 then ans will be 2 .
@aimiift2012 said:N is the largest 3-digit number which when divided by 11 and 9 leaves 9 and 7 as the respective remainders. The sum of the digits of N is
this means that if we add 2 to the number then it should be divisible by both 11 and 9 .....so we do the reverse process i.e first find the largest 3 digit number that is divisible by 11 and 9 and then we subtract 1 out of that.
i.e 990 - 2 = 988
ATDH.