Question

If there is a natural number n relative prime with 10. Then show that there exists another natural number m such that all digits are 1's and m is div. by 'n'.

Answer 1

Hint: First, start with the smallest value of m whose all digits are 1 and check for the condition. If it satisfies all the conditions, then that will be the number. If not, check for another number whose all digits are 1. Again, check for the condition, and so on. Check until you didn’t find the number which satisfies all the conditions.

Complete step-by-step answer:
Given:- The natural number n is relatively prime with 10 and the digits of m consist 1’s only.
The relative prime is those numbers that cannot be divided by any number greater than 1 or there are no common factors other than 1.
Consider $m = 1$,
Then, the only number dividing m is 1.
So, $n = 1$.
Since, n cannot be 1, as 1 is not a relative prime with 10.
Thus, m can’t be 1.
Now, consider $m = 11$,
The factors of 11 are 1 and 11.
Since, n cannot be 1, as 1 is not relative prime with 10.
Also, n cannot be 11, as $m = 11$ and m cannot be equal to n.
Thus, m can’t be 11.
Now, consider $m = 111$,
The factors of 111 are 1, 3, 37, and 111.
Since, n cannot be 1 or 111, as 1 and 111 is not a relative prime with 10.
Also, n cannot be 3 and 37, as all digits are not 1.
Thus, m can’t be 111.
Now, consider $m = 1111$,
The factors of 1111 are 1, 11, 101, and 111.
Since, n cannot be 1, 101, or 1111, as 1, 101, and 111 is not relative prime with 10.
But m is divisible by 11 which is relatively prime with 10.
Thus, m will be 1111.

Hence, the value of m is 1111 and the value of n is 11.

Note: The students might make mistakes by choosing random numbers instead of starting from the smallest number which fulfills all the conditions. Choosing a random variable will take time to find the numbers. The students may take the same value of m and n which is not correct. As it is mentioned in the question that there exists another number m.