Saturday, August 12, 2006

Radek SMASH! Topic: Number Theory. Level: AMC.

Problem: Find a three-digit number such that the sum of the digits is $ 9 $, the product of the digits is $ 24 $, and the number read in reverse is $ \frac{27}{38} $ of the original.

Solution: Now, we could go about mathizing all this nonsense, but instead we're just going to guess-and-check because in this case it is probably much simpler. $ 24 $ only has a few factorizations into three numbers; furthermore, it's not hard to find one that sums to $ 9 $. We see this to be $ 2 \cdot 3 \cdot 4 = 24 $.

Now, to complete the problem, we use the last piece of information. One useful thing this tells us is that the number read in reverse is divisible by $ 27 $. Knowing your powers of $ 3 $, we remember that $ 243 $ is divisible by $ 27 $. Luckily, $ 342 $ is also divisible by $ 38 $ and $ \frac{243}{342} = \frac{27}{38} $. So our number is $ 342 $. QED.

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Comment: This is the type of problem that will show up on an AMC or easier tests at smaller competitions. Speed is the main thing you're probably worried about for a problem like this; being able to take in the problem statement and analyze it quickly often is the most important skill for winning local competitions.

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Practice Problem: Use your number theory skills to devise another problem similar to this one - make sure the numerator and denominator of the fraction are both less than $ 50 $ or so (or it just gets pointless; the lower the better).

7 comments:

  1. Find a three-digit number such that the sum of the digits is 12, the product of the digits is 48, and the number read in reverse is of the original 41/107.

    yay!! this is the first problem you posted that i'm actually able to answer!!

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  2. Haha nice. Can you find one so that the fraction has numerator and denominator less than 50?

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  3. Find a three-digit number such that the sum of the digits is 3, the product of the digits is 1, and the number read in reverse is 111/111 of the original.

    =P

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  4. Gee Qiaochu, that was a tough one. How did you manage to come up with it?

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  5. wut numerator/denominator?? you can still solve my problem with out that right? 246 yay moo!

    ahh... i wish all problems were as tough as qc's

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