## Archive for May 10, 2016

### Problems at the 2016 MathCounts National Competition

Yesterday, Edwa rd Wan (WA) became the 2016 MathCounts National Champion. He defeated Luke Robitaille (TX) in the finals of the Countdown Round, 4-3. In the Countdown Round, questions are presented one at a time, and the first student to answer four correctly claims the title.

“This one is officially a nail-biter,” declared Lou DiGioia, MathCounts Executive Director and the moderator of the Countdown Round. Three times, Wan took a one-question lead; and three times, Robitaille tied the score on the following question. The tie was broken for good when Wan answered the following question:

What is the remainder when 999,999,999 is divided by 32?

This year’s winning question was relatively easy. What makes me say that? Well, for starters, when an odd number is divided by an even number, the remainder will be odd; and because 32 is the divisor, the remainder has to be less than 32. Consequently, the remainder is in the set {1, 3, 5, …, 31}, so there are only 16 possible answers.

But more importantly, most MathCounts competitors will be well trained for a problem of this type. It relies on divisibility rules that they should know, and it requires minimal insight to arrive at the correct answer. I suspect that the following explanation of the solution is the likely thought process that Wan used to solve this problem; of course, all of this occurred in his head in less than 7 seconds, which does make it rather impressive.

A fact that you probably know:

• A number is divisible by 2 if it’s even.

But said another way…

• A number is divisible by 2 if the last digit is divisible by 2.

There are then corollary rules for larger powers of 2:

• A number is divisible by 4 if the last two digits are divisible by 4.
• For example, we can conclude that 176,432,928 is divisible by 4 because the last two digits form 28, which is divisible by 4. The digits in the hundreds, thousands, and higher place values are somewhat irrelevant, because they represent some multiple of 100 — for instance, the 7 in the ten millions place represents 70,000,000, which is 700,000 × 100 — and every multiple of 100 is divisible by 4.
• A number is divisible by 8 if the last three digits are divisible by 8.
• For example, we can conclude that 176,432,376 is divisible by 8 because the last three digits form 376, which is divisible by 8 since 8 × 47 = 376.
• A number is divisible by 16 if the last four digits are divisible by 16.
• A number is divisible by 32 if the last five digits are divisible by 32.
• And so on.

These observations lead to a generalization…

• A number is divisible by 2n if the last n digits are divisible by 2n.

I won’t take the time to prove that statement here, but you can trust me. (Or maybe you’d like to prove it on your own.) I will, however, explain why it’s relevant.

A number will be divisible by 32 if the last five digits are divisible by 32. Consequently, any number that ends in five 0’s will be divisible by 32, which means that 1,000,000,000 is a multiple of 32. Since 999,999,999 is 1 less than 1,000,000,000, then it must be 1 less than a multiple of 32. Therefore, when 999,999,999 is divided by 32, the remainder will be 31.

The hardest part of solving that problem is recognizing that 999,999,999 is 1 less than a multiple of 32. But for most MathCounts students, that step is not very difficult, hence my contention that this was a relatively easy winning problem.

My favorite problem of the Countdown Round? Now, that’s another story, and it epitomizes what I generally love about MathCounts problems.

If a, b, c, and d are four distinct positive integers such that ab = cd, what is the least possible value of a + b + c + d?

This problem has several things going for it:

• It’s simply stated.
• It’s easily understood, even by students who don’t participate in MathCounts.
• It has an entry point for all students, since most kids can find at least one set of numbers that would work, even if they couldn’t find the set with the least possible sum.
• Finding the right answer requires convincing yourself that no lesser sum exists.

It’s that last point that I find so interesting. While I was able to find the correct answer, it took a while to convince myself that it was the least possible sum. But since I don’t want to deprive you of any fun, I’ll let you solve the problem on your own.

As a final point, I’ll show you a picture that I took at the event. Do you see the error? What can I say… it’s a math competition… you didn’t expect them to be good with numbers, did you? Full disclosure: The error was corrected halfway through the competition during a break.

## About MJ4MF

The Math Jokes 4 Mathy Folks blog is an online extension to the book Math Jokes 4 Mathy Folks. The blog contains jokes submitted by readers, new jokes discovered by the author, details about speaking appearances and workshops, and other random bits of information that might be interesting to the strange folks who like math jokes.

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