Answer
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Hint: Focus on the point that any number ending with 5 can be written as 10m+5, where 5 represents its unit place and m is the number formed by the digits occupying the tens and the higher places. Similarly, numbers ending with zero can be written as 10k+0. Now use the fact that 10 is a multiple of 5 to reach the required answer.
Complete step-by-step answer:
We know that any number ending with 5 can be represented as 10m+5, where 5 represents its unit place and m is the number formed by the digits occupying the tens and the higher places. So, if we simplify and take 5 common, we get
$10m+5=5\left( 2m+1 \right)$ , and as m is a constant integer, so 2m +1 is also a constant integer and we take this constant integer to be p. So, the numbers ending with 5 can be written as 5p, where p is integer and from this form we can clearly say that such numbers are divisible by 5.
Similarly, numbers ending with zero can be written as 10k+0, i.e., 10k and k is an integer. Now we can write 10 as $5\times 2$ and let 2k to be q, where q will be a constant integer too. So, the numbers ending with zero are $5\times 2k=5q$ , which clearly shows that such numbers are also divisible by five.
So, looking at the above results, we can easily conclude that a number is divisible by 5 if the last digit is 0 or 5.
Hence the answer to the above question is option (d).
Note: We have taken 2m+1 and 2k to be integers as m and k were integers according to our assumptions and we know that 2 and 1 are also integers and we know that if we multiply, add or subtract two integers the result is always an integer. However, division of two integers may or may not yield an integer as the answer.
Complete step-by-step answer:
We know that any number ending with 5 can be represented as 10m+5, where 5 represents its unit place and m is the number formed by the digits occupying the tens and the higher places. So, if we simplify and take 5 common, we get
$10m+5=5\left( 2m+1 \right)$ , and as m is a constant integer, so 2m +1 is also a constant integer and we take this constant integer to be p. So, the numbers ending with 5 can be written as 5p, where p is integer and from this form we can clearly say that such numbers are divisible by 5.
Similarly, numbers ending with zero can be written as 10k+0, i.e., 10k and k is an integer. Now we can write 10 as $5\times 2$ and let 2k to be q, where q will be a constant integer too. So, the numbers ending with zero are $5\times 2k=5q$ , which clearly shows that such numbers are also divisible by five.
So, looking at the above results, we can easily conclude that a number is divisible by 5 if the last digit is 0 or 5.
Hence the answer to the above question is option (d).
Note: We have taken 2m+1 and 2k to be integers as m and k were integers according to our assumptions and we know that 2 and 1 are also integers and we know that if we multiply, add or subtract two integers the result is always an integer. However, division of two integers may or may not yield an integer as the answer.
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