Question

In the decimal system of numeration of sixth-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: What we will do here is, we will make different cases for each digit place and then we will take the sum of the first 5 digit number and then we will choose the values of the last digit according to different cases. Also, we will use the concept of what are the endings of a number, which is divisible by 5 which is every number divisible by 5 that ends with 0 or 5.

Complete step-by-step solution:
We know that there are ten digits which are 0, 1, 2, 3, 4, 5, 6, 7, 8, 9.
Now, we have to find those types of six-digit numbers whose sum of digits are divisible by 5.
So, for a six-digit number, the first digit cannot be equal to 0 as it will lead to a five-digit number.
So, the number of which we can put in the first place will be equal to 9 as zero has been excluded.
For, second, third, fourth, and fifth digit there will be no restriction.
So, numbers which we can put one, second, third, fourth, and fifth digits will be equal to 10.
Lastly, for the last digit that is the sixth digit, we can put only 0 or 5 as it is given that the six-digit number is divisible by 5.
So, the number which we can put on the sixth digit will be predicted on the basis of the sum of the first five-digit numbers.
Cases are as follows,
If the sum of the first five-digit numbers is of form 5k, then the sixth digit will take value 0f 0 or 5.
If the sum of the first five-digit numbers is of the form 5k + 1, then the sixth digit will take a value of 4 or 9.
If the sum of the first five-digit numbers is of the form 5k + 2, then the sixth digit will take value of 3 or 8.
If the sum of the first five-digit numbers is of the form 5k + 3, then the sixth digit will take value of 2 or 7.
If the sum of the first five-digit numbers is of the form 5k + 4, then the sixth digit will take value of 1 or 6.
Now, the number of ways of getting 5 digit numbers according to the above statements, will be equals to 9 $\times {{10}^{4}}$
The first digit cannot be equal to 0 as it will lead to a five-digit number, so the number of which we can put in the first place will be equal to 9 as zero has been excluded.
Now, for each different case, we get at most 2 values for the sixth digit.
So, : In the decimal system of numeration of sixth-digit numbers in which the sum of the digits is divisible by 5 will be equals to = 9 $\times {{10}^{4}}$$\times 2$
On simplifying,
9 $\times {{10}^{4}}$$\times 2$=180000
Hence, option ( a ) is correct.

Note: Always remember, there may exist different cases in different conditions which lead to more possible ways to select objects. Numbers, which are divisible by 5 always end with digit 0 or 5. Also, if we need an n - digit number then its first digit cannot be equal to 0 as it will make the number of ( n – 1) digit.