Question

What is the total number of three-digit numbers with unit digit 7 and divisible by 11.
(a) 6
(b) 7
(c) 8
(d) 9

Answer 1

We start solving the by assuming the form of numbers as $xy7$, where x and y are digits between 0 and 9, also $x\ne 0$. We then use the property that the alternating sum of the digits in the given number must be divisible with 11 for a number to be divisible by 11. So, we assume the alternating sum of digits in number equal to the multiples of 11 and make the trial and error for the digits of ‘x’ and ‘y’ to get the total number of three digit numbers that were with unit digit 7 and divisible by 11.

Complete step-by-step answer:
According to the problem, we need to find the total number of three-digit numbers with unit digit 7 and divisible by 7.
Let us assume the form of these numbers be $xy7$, where the values of ‘x’ and ‘y’ lies in \left\\{ 0,1,2,3,4,5,6,7,8,9 \right\\} but $x\ne 0$.
We know that the alternating sum of the digits in the given number must be divisible with 11 for a number divisible by 11.
Let us first find the alternating sum of the number $xy7$.
So, the alternating sum is $x-y+7$ which should be divisible by 11.
Let us equate the alternating sum with the multiples of 11 and check the feasibility of digits for ‘x’ and ‘y’.
We know that the multiples of 11 are 0, 11, 22,…… .
Now, let us consider $x-y+7=0$.
$\Rightarrow x+7=y$.
Let us consider $x=1$, we get $y=8$. So, the number is 187.
Now, let us consider $x=2$, we get $y=9$. So, the number is 297.
Now, let us consider $x=3$, we get $y=10$ which is a contradiction. Similarly, we get the next terms greater than 9 which is not required for us.
Now, let us consider $x-y+7=11$.
$\Rightarrow x-4=y$.
Let us consider $x=1$, we get $y=-3$ which is a contradiction.
Now, let us consider $x=2$, we get $y=-2$ which is a contradiction.
Now, let us consider $x=3$, we get $y=-1$ which is a contradiction.
Now, let us consider $x=4$, we get $y=0$. So, the number is 407.
Now, let us consider $x=5$, we get $y=1$. So, the number is 517.
Now, let us consider $x=6$, we get $y=2$. So, the number is 627.
Now, let us consider $x=7$, we get $y=3$. So, the number is 737.
Now, let us consider $x=8$, we get $y=4$. So, the number is 847.
Now, let us consider $x=9$, we get $y=5$. So, the number is 957.
Now, let us consider $x-y+7=22$.
$\Rightarrow x-y=15$, which is not possible for digits between 0 and 9.
So, the numbers that were divisible by 11 with unit digit 7 is 187, 297, 407, 517, 627, 737, 847 and 957 which is a total of 8.
We have found that there are 8 three-digit numbers with unit digit 7 and divisible by 11.
The correct option for the given problem is (c).

So, the correct answer is “Option (c)”.

Note: We should keep in mind that we needed three-digit numbers that were divisible by 11 but if $x=0$, we get two-digit numbers which is not the desired result for us. We should know that the difference between two successive numbers satisfying the required result is 110. So, we can add 110 to the previous numbers to get the next numbers. Similarly, we can expect the problems to find the total number of three-digit numbers with last digit 7 that were divisible by 7.