Question

In the decimal system of numeration the number of six digit numbers in which the sum of the digits is divisible by 5 is
(a) 180000
(b) 540000
(c) $5\times {{10}^{5}}$
(d) none of these

Answer 1

Hint: Here, we will find the total number of ways in which we can form a 6 digit number that is divisible by 5. Any number which is divisible by 5 always ends with 0 or 5. The total number of ways can be found using the Fundamental principle of counting.

Complete step-by-step answer:
Fundamental principle of counting states that if there are n ways of doing something and m ways of doing another thing after that, then there are $n\times m$ ways to perform both of these actions.
Here, we are given that a 6 digit number have to be formed.
Now, total number of digits which form up any number are 10, i.e., 0, 1, 2, 3, 4, 5, 6, 7, 8 and 9.
A six digit number will always start from any of the digits from 1 to 9. So, we have 9 choices to fill up the place for first digit.
The next 4 digits of the number can be any of the numbers from 0 to 9. So, we have 10 ways each for filling the places for the second, third, fourth and fifth digits.
Now, the sum of the 5 digit number obtained after filling the places for these 5 digits is of the form $5k,5k+1,5k+2,5k+3\text{ or }5k+4$ because any integer can be represented in these four forms only where ‘k’ is an integer.
When the sum is of the form $5k$, the last digit can be 0 or 5 because any number divisible by 5 always end with 0 or 5.
When the sum is of the form $5k+1$, then the last digit can be 4 or 9, as $5k+1+4=5k+5$ which is divisible by 5 and $5k+1+9=5k+10=5\left( k+2 \right)$ which is also divisible by 5.
When the sum is of the form $5k+2$ , the last digit can be 3 or 8 because $5k+2+3=5k+5=5\left( k+1 \right)$ which is divisible by 5 and $5k+2+8=5k+10=5\left( k+2 \right)$ which is also divisible by 5.
When the sum is of the form $5k+3$ , the last digit can be 2 or 7 because $5k+3+2=5k+5=5\left( k+1 \right)$ which is divisible by 5 and $5k+3+7=5k+10=5\left( k+2 \right)$ which is also divisible by 5.
When the sum is of the form $5k+4$ , the last digit can be 1 or 6 because $5k+4+1=5k+5=5\left( k+1 \right)$ which is divisible by 5 and $5k+4+6=5k+10=5\left( k+2 \right)$ which is also divisible by 5.
So, we can see that in all cases, there are two ways of finding the last digit.
Therefore, by fundamental principle of counting, total number of six digit numbers for which the sum of digits is divisible by 5 is = $9\times 10\times 10\times 10\times 10\times 2=18\times {{10}^{4}}=180000$.

So, the correct answer is “Option A”.

Note: Students should note here that we can represent all integers in the form $5k,5k+1,5k+2,5k+3,5k+4$ by just varying k. A chance of mistake here is that the student considers the first digit to be filled in 10 ways that is 0 to 9 but if the digit starts from 0 it would become a 5 digit number. So, the first digit will be selected from 1 to 9 only. All the calculations must be done properly to avoid mistakes.