amarillo of kerria

ライティング練習。ブラウザがChromeなら画面を右クリックからも翻訳できるよ。

数学の問題に出会った

I was reading a blog I've frequnted, and in its entry yesterday, there was a sort of quiz to readers. It was a mathematical one. The author of the blog was prompting readers to try to find by themselves a well-formulated solution to a mathematical question he came upon with, as the author himself couldn't formulate well the mathematical proof of it.


So for a while I pondered on that quiz, tinkering equations. I seem to have arrived, though somewhat dubious, at an answer ... so I decided to write it down as this post of my English diary (though the topic is not about English).


The question which was presented in the entry could be retold as follows:


You have a number which is a multiple of 3 (for instance, 525 which is 3 times 175). By exchanging the digits in the ones place and the tens place, you get another number (552). This new number is also a multiple of 3 (because 552 = 3*184). But how come this relation should seemingly always be true of any multiple of 3?


---


My answer was as follows (though it might be wrong, of course...):


Let n = 100x + 10y + z, where n, x, y, z are all non-negative integers and y and z are not greater than 9.


If n is divisible by 3,


(100x + 10y + z)/3 = p, where p is a non-negative integer. (A)


The equation (A) can be rewritten as:


(33 + 1/3)x + (3 + 1/3)y + (1/3)z = p


(33x + 3y) + (x + y + z)(1/3) = p


(x + y + z)(1/3) = p - (33x + 3y)


Here (33x + 3y) and p are both non-negative integers. Therefore, (x + y + z)(1/3) must also be an integer, because the right side of the equation is an integer. This means that (x + y + z) is divisible by 3.


Now, by exchanging the ones and tens digits of (A), you have the following equation:


(100x + 10z + y)/3 = q ... (B)


What you want to prove is that q is an integer.


The equation (B) can be rewritten as:


(33 + 1/3)x + (3 + 1/3)z + (1/3)y = q


(33x + 3z) + (x + y + z)(1/3) = q


Here (33x + 3z) is an integer, because x and z are integers; and we've known that (x + y + z) is divisible by 3, making (x + y + z)(1/3) amounting to an integer. Therefore q is an integer. Q.E.D.


Looking at the structure of the equation, it appears that it's also likely that all the digits of the multiple of 3 can be randomly rearranged to be a number divisible by 3.