Solving Mathematical Problems

a personal perspective by Terence Tao. This is a new edition of a book which was written by Tao more than 15 years ago, which means when he was only 15! It’s a thin little book that takes a leisurely look at solving some competition type problems. The coverage is not huge, but the author take pains to go through in great detail various strategies one can adopt in solving problems. Quite a nice book but very pricey for 102 pages.

I found exercise 2.1 quite fun.

In a parlour game, the ‘magician’ asks one of the participants to think of a three-digit number [tex]abc_{10}[/tex]. Then the magician asks the participant to add the five numbers [tex]acb_{10}, bac_{10}, bca_{10}, cab_{10}[/tex] and [tex]cba_{10}[/tex], and reveal their sum. Suppose the sum was 3194. What was [tex]abc_{10}[/tex]?

My solution is this. If we add all the six permutations, we know that the sum equals
[tex] (2a+2b+2c) \times 100 + (2a+2b+2c) \times 10 + (2a+2b+2c) [/tex]
[tex]= (a+b+c) \times 222[/tex].
So we just need to know the multiples, [tex]1 \times 222, \ldots, 27 \times 222[/tex]. Take the smallest multiple larger than the given number, and check by subtracting the difference and summing the digits. You do not have to do it with more than 5 different multiples.

[tex]15 \times 222 = 3330; 3330 – 3194 = 136[/tex].
But [tex]1+3+6 = 10[/tex], so incorrect.
now [tex]16 \times 222 -3194 = 136 + 222 = 358[/tex].
And [tex]3+5+8 = 16 [/tex]and we found our number.

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3 Responses to Solving Mathematical Problems

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  2. Carlos says:

    Wrong, the correct first equation is
    100(a+2b+2c)+10(c+2b+2a)+b+2a+2c=3194

    Even though that’s not very useful…

  3. Carlos says:

    Forget it, I confused abc_10 with acb_10

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