9 how many Zeroes with which the
number 28 x 34 xg5 will end with 8
Answers
If we look at a number N, such that
TRAIL-1
Number of trailing zeroes is the Power of 10 in the expression or in other words, the number of times N is divisible by 10.
For a number to be divisible by 10, it should be divisible by 2 & 5.
Consider a number N, such that
TRAIL-2
For the number to have a zero at the end, both a & b should be at least 1.
If you want to figure out the exact number of zeroes, you would have to check how many times the number N is divisible by 10.
When I am dividing N by 10, it will be limited by the powers of 2 or 5, whichever is lesser.
Number of trailing zeroes is going to be the power of 2 or 5, whichever is lesser.
Let us look at an example to further illustrate this idea.
Question 1: What is number of trailing zeroes in 12000?
12000 = 25 x 3 x 53
When I divide it by 10, it would be divisible exactly thrice because I have only three 5s.
In this case, number of 5s has become the limiting factor and so, the power of 5, which is 3 is the answer.
Tip: The power of 5 will be the limiting factor in most cases of continuous distribution. It will happen because 5 is less likely to occur than 2.
Question 2: Find out the number of zeroes at the end of N
TRAIL-3
Looking at the expression, we can say that the power of 5 will be the limiting factor.
All we need to do is to figure out the number of 5s in the expression.
11, 22, 33, 1717, 8989,… will not give us any 5s.
55 will give us five 5s.
1010 will give us ten 5s.
1515 will give us fifteen 5s.
And so on.
So, the total number of 5s that I have is
TRAIL-4
But I have made a mistake in the above calculation.
I have assumed that 2525 will give me twenty-five 5s but that is incorrect.
It is incorrect because 2525 = 550 and will actually give me 50 5s.
Other errors are
5050 will actually give me 100 5s, whereas I have considered only 50 5s.
7575 will actually give me 150 5s, whereas I have considered only 75 5s.
100100 will actually give me 200 5s, whereas I have considered only 100 5s.
Considering the above, I have made an error of = 25 + 50 + 75 + 100 = 250 5s.
So the total number of 5s that I have are 1050 + 250 = 1300.
So the number of trailing zeroes at the end of the expression is 1300