16, 33, 65, 131, 261, (…..)
441=21^2
Sunday
Monday
Tuesday
3600
Robin =72
Dravid = 98
Azhar = 22
Sachine = 98
Total score = 290
4155
avg of 10nos.=23==>23*10=230
if each no increased by 4 ==> 4*10=40
then new avg is giveen by : 230+40=270
270/10=27
hence the new avg =27
150
Smart Question. What was the total gain by both of them, not Just Krishan
Perimeter of semi circular = πr+2r
Where r is radius , π = 22/7
πr+2r= 144 (given)
(22/7)r +2r = 144
22r+14r = 144*7 (multiply both sides by 7)
36r = 144*7
r= 144*7/36 = 28
radius =28 cm
Area =( 1/2)πr^2
Area=( 1/2 )*(22/7)*28*28
= 1232cm^2
Ans : area is 1232 cm^2
SI = PTR / 100
7/5 * 5/q = 7/10
35/5q = 7/10
q=10
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
( b ) 216.90
1/5
523