Hi there!
In my heyday of overseas business consultancy, the Lanchester Strategy caught on among small and medium-sized companies; I had participated in several early-morning study meetings of several enterprises with their C.E.O.s.
I also capitalized on its strategy when I helped a small raw hides tanning factory with its globalization. Under the Lanchester Strategy, the more effective approach for a company is to target one aspect or location of its rival to destabilize a potential monopoly. I took the helm as a trailblazer to start marketing and sales promotions to a firm of standing, a prestigious Japanese sports shoe company, stealthily and aggressively.
To my consternation, a half year later, my tactics, including sales pitch and fieldwork, bore fruitful, and in a year or so, the sales turnover reached over 100 million yen. I talked to one of the big clients with whom our competitors used to have a business into changing their supply source to us, and finally, I managed to convince them to purchase a massive amount of leather from us.
Truth be told, I believe our man-to-man talk led to the crystallization of big business.
The Lanchester Strategy helped me develop my career path. To be agile, nimble, and resourceful towards business is a prerequisite for success. We shouldn’t be afraid of adopting a new strategy if we want to differentiate ourselves from others.
Strike while the iron is hot. Let’s give it a shot to maneuver a “mod” or “hot” strategy to make it happen.
Today, let’s challenge an integer word problem with the modulo or mod in mathematics.
We set the multiplication of all positive integers whose digit in the ones place is three or seven, \(x\). Find the integer in the tens place of \(x\). (Source: Math Olypiad, qualifying session)
Here is my solution to the problem.
First of all, let’s make the following multiplication given the condition of the problem.
\((10k + 3)(10k + 7)\)
\(= \displaystyle\left\{10k + (5 – 2)\right\}\displaystyle\left\{10k +(5 + 2)\right\}\)
\(= (10k + 5)^2 – 4\)
Now, \((10k + 5)^2 = 100k^2 + 100k + 25 ≡ 25\)(mod\(100\)),
then \((10k + 5)^2 – 4 ≡ 21\)(mod\(100\)).
Whenever we plug in the positive integer for \(k\) of the formula \((10k + 5)^2 – 4\) to meet the given condition, we find the integer “2” in the tens place. Therefore, the answer is “2.”
I have yet to become a master, for I have considerable number of mathematical questions to solve.
Stay tuned, and expect to see my next post.
Keep well.
Frank Yoshida
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【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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