1970 Canadian MO Problems/Problem 4


a) Find all positive integers with initial digit $6$ such that the integer formed by deleting $6$ is $1/25$ of the original integer.

b) Show that there is no integer such that the deletion of the first digit produces a result that is $1/35$ of the original integer.


a) Let the integer have n digits. Then the integer can be represented in the form of $6 *10^{n-1} + x$ where x is a non-negative integer less than $10^{n-1}$. The problem is asking for integers such as $6 *10^{n-1} + x = 25x$. Solving for x results in \[x = \frac{10^{n-1}}{4}\] Since x has to be an integer, n has to be greater than or equal to 3. Thus, the answer is \[625, 6250, 62500, \ldots\]

b) We use the same notation as part a. Then, the condition can be represented as $m*10^{n-1} + x = 35x$ where $m$ is an integer between 1 and 9 inclusive. Solving for x results in \[x = \frac{m*10^{n-1}}{34}\]For x to be an integer, the numerator of the expression has to be divisible by 34, and thus 17. $10^{n-1}$ obviously does not have any factors of 17, meaning that $m$ must have a factor of 17, which is also impossible since m is an integer between 1 and 9. Therefore, an integer required by the problem is not possible.

Sol by Vwall