c
When none of the digits are repeated:
The hundred’s place can be filled by any of the digits: 2, 3, 5, 6, 7 or 8 except the one which has already been used at the thousand’s place, so it can be filled in 5 ways.
Similarly tens’ place can be filled in 4 ways: only those 4 numbers which have not been use either at hundred’s or thousand’s place.
Unit’s place can be filled in only 3 ways. So, total number of nos. Possible =4×5×4×3= 240
S=D/T
S=624KM/6.5HRS
S=96KM/HR
x+y+z
—– = 6800
3
x>=6400
6400 + y + z = 20400
y + z = 14000
to get the greatest of y and z, lets assume y = 6400
so, z = 7600
so ANS is 7600
Ans = 8
Use simple box method
[1][2][2][2]= 1x2x2x2 =8
Logic >
once place have 2 number (2,6)
Tens place have also 2 number (7,3)
So that number is divisible by 4
Now, 2 numbers fixed at once and tens place left 2 numbers which is choice at hundredth place
Now 3 number fix left 1 number which is placed at thousands place
Total number form is 4×4 = 16
But divisible by 4 is 1x2x2x2 =8
wake up guys and see modulo % can return only integer value.
So answer will be 1 and all other answers are wrong. You can
run your equation in C code and check the output.
c
C=150
G.c.d=num1 *num2/lcm
Num1=20
Gcd=5
Lcm=60
5=(20*num2)/60
Num2= 5*60/20
Num2=15
To determine how many consecutive zeros the product of S will end with, we need to find the highest power of 10 that divides the product. This is equivalent to finding the highest power of 5 that divides the product, since the number of factors of 2 will always be greater than the number of factors of 5.
The primes in S are {2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97}.
There are 24 primes in S, so the product of S is:
2 x 3 x 5 x 7 x 11 x 13 x 17 x 19 x 23 x 29 x 31 x 37 x 41 x 43 x 47 x 53 x 59 x 61 x 67 x 71 x 73 x 79 x 83 x 89 x 97
We need to find the highest power of 5 that divides this product. To do this, we count the number of factors of 5 in the prime factorization of each number in S.
5 appears once: 5
5 appears once: 25
5 appears once: 35
5 appears once: 55
5 appears once: 65
5 appears once: 85
So, there are six factors of 5 in the product of S. However, we also need to consider the powers of 5 that arise from the factors 25, 35, 55, and 65.
25 = 5 x 5 appears once: 25
35 = 5 x 7 appears once: 35
55 = 5 x 11 appears once: 55
65 = 5 x 13 appears once: 65
Each of these numbers contributes an additional factor of 5 to the product of S. Therefore, there are 6 + 4 = 10 factors of 5 in the product of S.
Since each factor of 5 corresponds to a factor of 10, we know that the product of S will end with 10 zeros. Therefore, the product of S will end with 10 consecutive zeros
₹1 today increases to (1+1/8) = 9/8 after 1year.
After 2 years it increases to 9/8 of 9/8 = (9/8)^2 =81/64 times.
Hence ₹64000 increase to 64000×81/64 =₹81000