Question

If the prime factorization of a natural number $N$ is ${2^4} \times {3^4} \times {5^3} \times 7$, write the number of consecutive zeros in $N$.

Answer 1

Hint: To find the number of zeros, we need to look out for the factors which on multiplying gives a zero. We can check one by one for all the factors. Then count the number of powers of the factors which on multiplication gives zero. The minimum power is the number of consecutive zeros.

Complete step by step solution:
We are given the natural numbers ${2^4} \times {3^4} \times {5^3} \times 7$, in which the factors are $2,3,5,7$.
To count the number of zeros we are checking by multiplying two factors at once, which gives at least one zero and we get:
$2 \times 3 = 5$, which does not give zero.
$2 \times 5 = 10$, which gives zero.
$2 \times 7 = 14$, which does not give zero.
$3 \times 5 = 15$, which does not give zero.
$3 \times 7 = 21$, which does not give zero.
$5 \times 7 = 35$, which does not give zero.
Since, we are getting only two numbers or factors whose product gives one zero and rest factors would not give a single zero on multiplication.
And, we have ${2^4}$ which have four $2's$, which can be written as \[{2^4} = 2 \times 2 \times 2 \times 2\]. Similarly, we have three $5's$, which can be written as \[{5^3} = 5 \times 5 \times 5\].
Multiplying, four $2's$ and three $5's$ to get the total number of zeros in the equation, and we get:
\[{2^4} \times {5^3} = 2 \times 5 \times 2 \times 5 \times 2 \times 5 \times 2 = 10 \times 10 \times 10 \times 2 = 2 \times 1000\].

Therefore, there are three consecutive zeros in ${2^4} \times {3^4} \times {5^3} \times 7$.

Note:
We can also directly find the number of zeros by comparing the powers of the factors which on multiplication are giving zeros. The minimum value of the power would be the count of the number of zeros. For example: - the power of $2$ is $4$ and power of $5$ is $3$. Comparing both for minimum value we get: $\min \left( {4,3} \right) = 3$, which is the consecutive number of zeros.