グローバルビジネスで役立つ数学（66）与式が自然数となる整数解の数を求める（英語版）

Hi there!

I’ve seen many prodigious works of nature on my business trip abroad. To name just a few;

・ Onslaught of an isolated shower seen just scores of meters ahead of my driver’s seat in
　 San Francisco
・ A series of squeaky sounds of airplane wings by incredible turbulence when flying from
　 Singapore to Australia
・ Bunch of fallen trees in front of my hotel after the Great Earthquake in Guayaquil, Ecuador

How many times did I think we humans are nothing but small fry?

Even for COVID-19, we should figure out how not to eradicate viruses but to coexist with them, for we can’t beat them regarding their tremendous vitality. All we must do is stay healthy and get a lot of nutrition.

Today, awed about mother nature, let me challenge a word problem related to natural numbers and integers.

How many pieces of an integer \(m\) are there that make the given formula \(\frac{90}{45 – 3m}\) into a natural number?

Here is my solution to the problem.

First of all, let’s reduce the fraction.
\(\frac{90}{45 – 3m} = \frac{30}{15 – m}\)

Then, we factorize \(30\) into prime factors.
\(30 = 2・3・5\)
Therefore, \((1 + 1)(1 + 1)(1 + 1) = 8\).

Of course, we can get the same answer by finding \(15 – m\) that makes \(\frac{30}{15 – m}\) into a natural number:

\(15 – m\) must be either \(1, 2, 3, 5, 6, 10, 15,\) or \(30\),
In other words, there are \(8\) pieces of an integer \(m\) as \(14, 13, 12, 10, 9, 5, 0, -15\).

Caution is advised. You get into hot water if you belittle mother nature.

Stay tuned, and expect to see my next post.

Keep well.

Frank Yoshida

￣￣￣￣￣￣￣￣￣￣￣￣￣￣￣￣￣￣￣￣￣￣￣￣￣￣￣￣￣￣￣￣￣￣￣￣￣￣￣
【グローバルビジネスで役立つ数学】でもっと学習する b＾＾)
【参考図書】『もう一度高校数学』（著者：高橋一雄氏）株式会社日本実業出版社
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