B
Pearl.. Head cut then earl.. tail cut then pear.. Head and tail cut then ear
18
8+1=9
9+2=11
11+3=14
14+4=18
18+5=23
and finally
23+6=29
16.8 sec
Spade 13/52 = 1/4
King 4/52 = 1/13
Spade OR King 13/52 + 4/52 – 1/52 (because the King of Spades is in both groups) = 16/52 = 4/13
8400 ? 120 x 15 + 150 = ?
ans:
8400 % 120 x 15 + 150 = ?
1200
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
10 x 27=270
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
Diagonal of square=1.414* a, where a is side of square.
ans=1.414*20
ans=28.28