Kiss those Math Headaches GOODBYE!

Archive for November 8, 2008


The problem:

It’s your friend’s 73rd birthday. You’ve put together a surprise party and baked a special coconut meringue cake. But at the last minute you realize that — golly gee! — you forgot to get candles.

Rummaging through your drawers with just five minutes before your friend’s scheduled arrival, you find that you do have 14 candles. And being a brilliant mathematician, you realize that you can represent the number 73 with these 14 candles, using every candle. How do you do it?

As a hint, here’s a model showing how to do a problem like this, if you are celebrating someone’s 44th birthday, when you have just 13 candles. Notice that each dot on the top row is one candle.


Note that you may use icing to create the symbols: +, –, x, ÷, and you may also put in exponents, using candles to show the value of the exponent.

Have fun!

Send answers to:

Make the Subject line: POTM

Please include your full name, where you live, and if you don’t mind, describe your connection to math and math education (for example: teacher, tutor, math enthusiast, etc.).

The first person to send in a correct answer receives a $20 gift certificate toward the purchase of any Singing Turtle Press products. I’ll fill the winner in on the details by email.

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ANSWER: October Problem of the Month

The problem, once again, was this:

Here’s the OCTOBER POTM —


A) Each of the variables x, y, and z, stands for a unique natural number from 1 thru 9 inclusive.

B) The following, therefore, represents a number in the millions place: 3,x2y,1z3

C) This number: 3,x2y,1z3 is divisible by 9.

Question: What are three other 7-digit numbers containing all of these same digits: 3, x, 2, y, 1, z, and 3 that are also divisible by 9?

The correct answer was provided by Kevin Pickard, 48, a computer programmer and homeschooling parent of “two very smart girls,” ages 7 and 10. Kevin lives in Markham, ON, Canada.
Here is Kevin’s answer, in his own words:
Given that 3,x2y,1z3 is divisible by 9, then all of its digits
will add up to 9 eventually by repeated sums (eg. if x=4, y=2, z=3
then 3 + 4 + 2 + 2 + 1 + 3 + 3 = 18 and 1 + 8 = 9). So any arrangement
of these digits will still add up to 9 by repeated sums. These new
arrangements will therefore also be divisible by 9. So 3 other 7-digit
numbers that are also divisible by 9 would be as follows.
Plugging in the example values of x,y & z from above confirms
the result.
Josh’s note: In this problem, ANY new combination of the 7 given digits will produce a number divisible by 9. So in a sense this is a “trick problem.” The way the problem was written made some people think that only certain combinations of the digits would result in a number divisible by 9. The problem shows that the rule for divisibility by 9 is very flexible. When applying this divisibility rule, you need not take into account the order of the digits whatsoever. All that matters is that the sum of the digits is divisible by 9.

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