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).

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!!

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

3. Ans=246

4. 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

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

6. 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