# How many zeroes at the end of the first 100 multiples of 10.

Last updated date: 19th Mar 2023

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Answer

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Hint- Here, we will proceed by firstly finding out all the first 100 multiples of 10 and then evaluating the number of zeroes by observing the pattern which will exist and then using the formula i.e., Total number of zeros at the end of first 100 multiples of 10$\left( {1 \times {\text{Numbers of multiples with one zero at the end}}} \right) + \left( {2 \times {\text{Numbers of multiples with two zeros at the end}}} \right) + \left( {3 \times {\text{Numbers of multiples with three zeros at the end}}} \right)$

Complete step-by-step answer:

As, we know that the first 100 multiples of 10 are given by

$

\left( {10 \times 1} \right),\left( {10 \times 2} \right),\left( {10 \times 3} \right),\left( {10 \times 4} \right),......,\left( {10 \times 97} \right),\left( {10 \times 98} \right),\left( {10 \times 99} \right),\left( {10 \times 100} \right) \\

\Rightarrow 10,20,30,40,........970,980,990,1000 \\

$

Clearly from the above multiples, we can observe the pattern as under

For the first nine multiples i.e., 10,20,30,40,50,60,70,80,90 there is only one zero occurring at the end of each multiple.

For tenth multiple i.e., 100 there are two zeros occurring at the end.

Similarly, for the next nine multiples i.e., 110,120,130,140,150,160,170,180,190 there is only one zero occurring at the end of each multiple.

For twentieth multiple i.e., 200 there are two zeros occurring at the end.

So, finally we can say that for these nine multiples i.e., 100,200,300,400,500,600,700,800 and 900 there are two zeroes at the end, for multiple 1000 there are 3 zeroes at the end and rest of the multiples (90 multiples) i.e., 10,20,30,….,90,110,120,….,180,190,210,220,…,280,290,…..,910,920,…,980,990 there are only one zero occurring at the end.

Numbers of multiples with one zero at the end = 90

Numbers of multiples with two zeros at the end = 9

Numbers of multiples with three zeros at the end = 1

Therefore, Total number of zeros at the end of first 100 multiples of 10$\left( {1 \times {\text{Numbers of multiples with one zero at the end}}} \right) + \left( {2 \times {\text{Numbers of multiples with two zeros at the end}}} \right) + \left( {3 \times {\text{Numbers of multiples with three zeros at the end}}} \right)$

$ \Rightarrow $Total number of zeros at the end of first 100 multiples of 10$ = \left( {1 \times {\text{90}}} \right) + \left( {2 \times {\text{9}}} \right) + \left( {3 \times {\text{1}}} \right) = 90 + 18 + 3 = 111$

Hence, there are a total 111 zeroes at the end of the first 100 multiples of 10.

Note- In this particular problem, the multiples of 10 are easily determined by multiplying 10 with 1,2,3,4,5, etc which depends on how many multiples we need. Multiples of any number like that of 10 represents the numbers which have that number (i.e., 10) as one of factors i.e., a number is called multiple of 10 only if that number is exactly divided by 10 without leaving any remainder.

Complete step-by-step answer:

As, we know that the first 100 multiples of 10 are given by

$

\left( {10 \times 1} \right),\left( {10 \times 2} \right),\left( {10 \times 3} \right),\left( {10 \times 4} \right),......,\left( {10 \times 97} \right),\left( {10 \times 98} \right),\left( {10 \times 99} \right),\left( {10 \times 100} \right) \\

\Rightarrow 10,20,30,40,........970,980,990,1000 \\

$

Clearly from the above multiples, we can observe the pattern as under

For the first nine multiples i.e., 10,20,30,40,50,60,70,80,90 there is only one zero occurring at the end of each multiple.

For tenth multiple i.e., 100 there are two zeros occurring at the end.

Similarly, for the next nine multiples i.e., 110,120,130,140,150,160,170,180,190 there is only one zero occurring at the end of each multiple.

For twentieth multiple i.e., 200 there are two zeros occurring at the end.

So, finally we can say that for these nine multiples i.e., 100,200,300,400,500,600,700,800 and 900 there are two zeroes at the end, for multiple 1000 there are 3 zeroes at the end and rest of the multiples (90 multiples) i.e., 10,20,30,….,90,110,120,….,180,190,210,220,…,280,290,…..,910,920,…,980,990 there are only one zero occurring at the end.

Numbers of multiples with one zero at the end = 90

Numbers of multiples with two zeros at the end = 9

Numbers of multiples with three zeros at the end = 1

Therefore, Total number of zeros at the end of first 100 multiples of 10$\left( {1 \times {\text{Numbers of multiples with one zero at the end}}} \right) + \left( {2 \times {\text{Numbers of multiples with two zeros at the end}}} \right) + \left( {3 \times {\text{Numbers of multiples with three zeros at the end}}} \right)$

$ \Rightarrow $Total number of zeros at the end of first 100 multiples of 10$ = \left( {1 \times {\text{90}}} \right) + \left( {2 \times {\text{9}}} \right) + \left( {3 \times {\text{1}}} \right) = 90 + 18 + 3 = 111$

Hence, there are a total 111 zeroes at the end of the first 100 multiples of 10.

Note- In this particular problem, the multiples of 10 are easily determined by multiplying 10 with 1,2,3,4,5, etc which depends on how many multiples we need. Multiples of any number like that of 10 represents the numbers which have that number (i.e., 10) as one of factors i.e., a number is called multiple of 10 only if that number is exactly divided by 10 without leaving any remainder.

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