Difference between revisions of "1970 Canadian MO Problems/Problem 4"
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Let the integer have n digits. Then the integer can be represented in the form of <math>6 *10^{n-1} + x</math> where x is a non-negative integer less than <math>10^{n-1}</math>. The problem is asking for integers such as <math>6 *10^{n-1} + x = 25x</math>. Solving for x results in <cmath>x = \frac{10^{n-1}}{4}</cmath> Since x has to be an integer, n has to be greater than or equal to 3. Thus, the answer is <cmath>625, 6250, 62500, \ldots</cmath> | Let the integer have n digits. Then the integer can be represented in the form of <math>6 *10^{n-1} + x</math> where x is a non-negative integer less than <math>10^{n-1}</math>. The problem is asking for integers such as <math>6 *10^{n-1} + x = 25x</math>. Solving for x results in <cmath>x = \frac{10^{n-1}}{4}</cmath> Since x has to be an integer, n has to be greater than or equal to 3. Thus, the answer is <cmath>625, 6250, 62500, \ldots</cmath> | ||
b) | b) | ||
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We use the same notation as part a. Then, the condition can be represented as <math>m*10^{n-1} + x = 35x</math> where <math>m</math> is an integer between 1 and 9 inclusive. Solving for x results in <cmath>x = \frac{m*10^{n-1}}{34}</cmath>For x to be an integer, the numerator of the expression has to be divisible by 34, and thus 17. <math>10^{n-1}</math> obviously does not have any factors of 17, meaning that <math>m</math> 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. | We use the same notation as part a. Then, the condition can be represented as <math>m*10^{n-1} + x = 35x</math> where <math>m</math> is an integer between 1 and 9 inclusive. Solving for x results in <cmath>x = \frac{m*10^{n-1}}{34}</cmath>For x to be an integer, the numerator of the expression has to be divisible by 34, and thus 17. <math>10^{n-1}</math> obviously does not have any factors of 17, meaning that <math>m</math> 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. | ||
Revision as of 00:15, 3 May 2020
Problem
a) Find all positive integers with initial digit 
such that the integer formed by deleting is 
of the original integer.
b) Show that there is no integer such that the deletion of the first digit produces a result that is 
of the original integer.
Solution
a)
Let the integer have n digits. Then the integer can be represented in the form of 
where x is a non-negative integer less than 
. The problem is asking for integers such as 
. Solving for x results in ![\[x = \frac{10^{n-1}}{4}\]](http://latex.artofproblemsolving.com/7/a/8/7a86ded68e96e8bba2b9c5acb94033f4f9823fa9.png)
Since x has to be an integer, n has to be greater than or equal to 3. Thus, the answer is ![\[625, 6250, 62500, \ldots\]](http://latex.artofproblemsolving.com/4/3/f/43fac99965dfdfa0b924c16c902b0d65033ba6a2.png)
b)
We use the same notation as part a. Then, the condition can be represented as 
where 
is an integer between 1 and 9 inclusive. Solving for x results in ![\[x = \frac{m*10^{n-1}}{34}\]](http://latex.artofproblemsolving.com/f/9/4/f94c788d14e8eedabf8978efe845ce26979d6573.png)
For x to be an integer, the numerator of the expression has to be divisible by 34, and thus 17. obviously does not have any factors of 17, meaning that 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.
