
Write the number of zeroes in the end of a number whose prime factorization is \[{2^2}
\times {5^3} \times {3^2} \times 17\].
Answer
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Hint:We are asked to find the number the zeroes present in a number whose prime factorization is given. Find the value of each term of the prime factorization then put them together to find the value of the number. Observe the number obtained and check how many zeroes are there in the number.
Complete step by step solution:
Given, prime factorization of a number is \[P = {2^2} \times {5^3} \times {3^2} \times 17\].
(i)Let us find the values of each term separately for simplicity,
The first term \[{2^2}\],
\[{2^2} = 2 \times 2 = 4\]
The second term \[{5^3}\],
\[{5^3} = 5 \times 5 \times 5 = 125\]
The third term \[{3^2}\],
\[{3^2} = 3 \times 3 = 9\]
Now substituting these values of \[{2^2}\], \[{5^3}\] and \[{3^2}\] in equation (i), we get
\[P = 4 \times 125 \times 9 \times 17\]
\[ \Rightarrow P = 76500\]
Therefore, we can observe that there are two zeroes at the end of the number.
Alternative method:
We break all the terms in equation (i),
\[P = {2^2} \times {5^3} \times {3^2} \times 17\]
\[ \Rightarrow P = 2 \times 2 \times 5 \times 5 \times 5 \times 3 \times 3 \times 17\] (ii)
We know that \[5 \times 2 = 10\] and ten has one zero, we will make as many pairs of \[5 \times
2\]possible in equation (ii) to check how many zeroes will be there.
\[P = \left( {2 \times 5} \right) \times \left( {2 \times 5} \right) \times 5 \times 3 \times 3 \times 17\]
\[ \Rightarrow P = 10 \times 10 \times 5 \times 3 \times 3 \times 17\]
We observe that there are two tens, that is there are two zeroes. Therefore, the number will have two zeroes in the end.
Hence, the required answer is two.
Note: When questions are asked to find out the number of zeroes and the number \[2\] and \[5\] are present or multiple of \[2\] and multiple of \[5\] are present then always try to pair them up as their product will have zeroes in it. In this way without long calculations we can easily find out the number of zeroes.
Complete step by step solution:
Given, prime factorization of a number is \[P = {2^2} \times {5^3} \times {3^2} \times 17\].
(i)Let us find the values of each term separately for simplicity,
The first term \[{2^2}\],
\[{2^2} = 2 \times 2 = 4\]
The second term \[{5^3}\],
\[{5^3} = 5 \times 5 \times 5 = 125\]
The third term \[{3^2}\],
\[{3^2} = 3 \times 3 = 9\]
Now substituting these values of \[{2^2}\], \[{5^3}\] and \[{3^2}\] in equation (i), we get
\[P = 4 \times 125 \times 9 \times 17\]
\[ \Rightarrow P = 76500\]
Therefore, we can observe that there are two zeroes at the end of the number.
Alternative method:
We break all the terms in equation (i),
\[P = {2^2} \times {5^3} \times {3^2} \times 17\]
\[ \Rightarrow P = 2 \times 2 \times 5 \times 5 \times 5 \times 3 \times 3 \times 17\] (ii)
We know that \[5 \times 2 = 10\] and ten has one zero, we will make as many pairs of \[5 \times
2\]possible in equation (ii) to check how many zeroes will be there.
\[P = \left( {2 \times 5} \right) \times \left( {2 \times 5} \right) \times 5 \times 3 \times 3 \times 17\]
\[ \Rightarrow P = 10 \times 10 \times 5 \times 3 \times 3 \times 17\]
We observe that there are two tens, that is there are two zeroes. Therefore, the number will have two zeroes in the end.
Hence, the required answer is two.
Note: When questions are asked to find out the number of zeroes and the number \[2\] and \[5\] are present or multiple of \[2\] and multiple of \[5\] are present then always try to pair them up as their product will have zeroes in it. In this way without long calculations we can easily find out the number of zeroes.
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