Question

Find the number of zeroes in the end in product ${{5}^{6}}{{.6}^{7}}{{.7}^{8}}{{.8}^{9}}{{.......30}^{31}}$.

Answer 1

Hint: We solve this question by realising that the number of zeros at the end is equal to the power of all 10’s in it, there by the power of fives in the product. Then we consider the multiples of 5 in the product 5, 10, 15, 20, 25 and 30 raised to their respective powers. Then we find the number of 5’s present in each multiple and raise it to the power of 5 respectively. Then we add all the numbers of 5’s to find the total power to which 5 is raised. Then as the number of zeros at end is equal to the power of 5, we get our required answer.

Complete step by step answer:
Let us start solving the question by starting with when the product of numbers results in zero at the end.
If the end of a product or the unit digit of a number is zero, it means it is divisible by 10, that is it is a multiple of 10.
So, the number of zeros at the end of any number is equal to the number of times that number can be factored into the power of 10.
For example, we can write 200 as \[200=2\times 10\times 10=2\times {{\left( 10 \right)}^{2}}\]. In it we can see that the number of zeroes is equal to the number of times 10 is raised to power.
As $10=2\times 5$, we can say that the number of zeroes in the end is equal to the power of 5 in the product.
Now, let us consider the product ${{5}^{6}}{{.6}^{7}}{{.7}^{8}}{{.8}^{9}}{{.......30}^{31}}$.
As we need to find the number of zeros at the end of the product, we can find the number to which 5 is raised in the product as they are equal.
So, to find that we need to consider all the multiples of 5 present in the given product. They are 5, 10, 15, 20, 25, 30.
In ${{5}^{6}}$, as it is 5 itself it has 6 fives.
In ${{10}^{11}}$, as $10=2\times 5$, it has 11 fives.
In ${{15}^{16}}$, as $15=3\times 5$, it has 16 fives.
In ${{20}^{21}}$, as $20=2\times 2\times 5$, it has 21 fives.
In ${{25}^{26}}$, as $25=5\times 5$, it has 26 fives two times, that is 52 fives.
In ${{30}^{31}}$, as $30=2\times 3\times 5$, it has 31 fives.
As we need the total number of fives in the product, let us add all the above possible fives. Then we get,
$6+11+16+21+52+31=137$
So, number of zeros at the end of the product of ${{5}^{6}}{{.6}^{7}}{{.7}^{8}}{{.8}^{9}}{{.......30}^{31}}$ are 137.

So, the correct answer is 137.

Note: While solving this problem one might make mistake by considering only the multiples of 10 and finding their power and adding them to get the answer as,
In ${{10}^{11}}$, as it is 10 itself, it has 11 10’s.
In ${{20}^{21}}$, as $20=2\times 10$, it has 21 10’s.
In ${{30}^{31}}$, as $30=3\times 10$, it has 31 10’s.
Adding them we get 11+21+31=63. One might solve it as above. But it is wrong as there are multiples of 5 that are neglected like 5, 15, 25, which when multiplied by an even number gives zero at the end. So, it is better to consider the power to which 5 is raised to find the answer.