Difference between revisions of "1970 Canadian MO Problems/Problem 4"

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(Solution)
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==Solution==
 
==Solution==
a) Find all positive integers with initial digit <math>6</math> such that the integer formed by deleting <math>6</math> is <math>1/25</math> of the original integer.
 
b) Show that there is no integer such that the deletion of the first digit produces a result that is <math>1/35</math> of the original integer.
 
 
 
a)
 
a)
  

Revision as of 23:15, 2 May 2020

Problem

Solution

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.