Question

write the number of zeros in the end of a number whose prime factorization is 2^2×5^3×3^2×17

Answer 1

Given number,



$2^2 \times 5^3 \times 3^2 \times 17$

⇒as we know,.

2² = 4

5³ = 125

3² = 9

Now, substituting the values of 2², 5³ and 3² in the given number.

So, $2^2 \times 5^3 \times 3^2 \times 17$

⇒ 4 * 125 * 9 * 17

⇒ 76,500

So, 2² * 5³ * 3² * 17 = 76,500

Here, the number 76,500 is having two zeros at its end.

So, the number 2² * 5³ * 3² * 17 have two zeros at its ending.

Here, the prime factorisation of 76,500 is,

$\begin{array}{r | 1} 2 & 76500 \\ \cline{2-2} 2 & 38250 \\ \cline{2-2} 3 & 19125 \\ \cline{2-2} 3 & 6375 \\ \cline{2-2} 5 & 2125 \\ \cline{2-2} 5 & 425 \\ \cline{2-2} 5 & 85 \\ \cline{2-2} 17 & 17 \\ \cline{2-2} & 1 \end{array}$

Here, 76,500 is factorisation by using prime numbers to get the prime factorisation as -

2 * 2 * 3 * 3 * 5 * 5 * 5 * 17

⇒ 2² * 3² * 5³ * 17

⇒76,500

Thus, the number 2² * 3² * 5³ * 17 has 2 zeros at its ending.

Answer 2

Answer:

Two zeroes is the answer.

In the prime Factorization,we have to look the power of 2 and 5.Whoseever power is the smallest will the the number of zeroes in the end.

Step-by-step explanation: