Question

write the number of zeros in the end of a number whose prime factorization is 2^2x5^3x3^2x17​

Answer 1

Answer:

There will be 2 zeroes in the given expression.

Step-by-step explanation:

The given expression,  = 2² x 5³ x 3² x 17

we know that zeroes in an expression are a result of number of 10's in it. The only way to make a 10 is the product of 2 and 5. Hence the number of zeroes in an expression will be the number of 2 or 5 whichever is minimum.

2² x 5³ x 3² x 17

= 2x2 x 5x5x5 x 3² x 17

= (2x5) x (2x5) x 5 x 3² x 17

= 10 x 10 x 5 x 3² x 17

Hence from the expression we can see that there will be 2 zeroes in the given expression.

Answer 2

concept : if we multiply 10 with any number, we get a zero in unit digit of resultant number.similarly when we multiply 100 with any number, we get two zeros in the end of resultant number and so on. for example ; 10 × 45 = 450.

100 × 45 = 4500.

means, for getting number of zeros in the end of a number we have to check how much multiple of 10 are present in that number.

here, given prime factorisation of number is 2² × 5³ × 3² × 17

= 2² × 5² × 5 × 3² × 17

= (2 × 5)² × 5 × 3² × 17

= (10)² × 5 × 3² × 17

= 10 × 10 × 5 × 3² × 17

here , there are two 10 are present in the prime factorisation of number so, $\textbf{two zeros}$ are present in the end of the given number.