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
The formula to find number of diagonals (D) given total number of vertices or sides (N) is
N * (N – 3)
D = ———–
2
Using the formula, we get
1325 * 2 = N * (N – 3)
N2 – 3N – 2650 = 0
Solving the quadratic equation, we get N = 53 or -50
It is obvious that answer is 53 as number of vertices can not be negative.
Alternatively, you can derive the formula as triange has 0 diagonals, quadrangel has 2, pentagon has 5, hexagon has 9 and so on……
Hence the series is 0, 0, 0, 2, 5, 9, 14, …….. (as diagram with 1,2 or 3 vertices will have 0 diagonals).
Using the series one can arrive to the formula given above.
20/- per day
So, if he work one day
So,1*20=20
1/3*20=20/3
2/3*20=40/3
1/8*20=5/2
3/4*20=15
so,
20+20/3+40/3+5/2+15=57.5
So ans is 57.5 is the ans
For optimal size of a project team..
the % increase in staff size should be “Zero”
reason:
additional member directly proportional to increase in
staff size..
The probability that no one opens the door would be
0.5×0.5×0.5×0.5= 0.0625
So, the probability that atleast one opens the door is 1-0.0625= 0.9375
it is based on the right shift operation on the perticular digit
Statements :
Some men are educated. Educated persons prefer small families.
Conclusions :
I. All small families are educated.
II. Some men prefer small families.
MONKEY=YEKNOM then each letter is moved backward =XDJMNL so for TIGER=REGIT the each of the reversed letters are moved backward=QDFHS
-Total /n=80
– x/5=40
x=200
Total/n=y/n
y-x/n-5=90
80n-200=90(n-5)
80n-200=90n-450
450-200=10n
n =25 (students) write exam
C. 20
d
south east