The Puzzle Archives, II

This week, I’ll continue with some more problems from the contests for the 2014 conference of the International Group for Mathematical Creativity and Giftedness.  We’ll look at problems from the Intermediate Contest today.  Recall that the first three problems on all contests were the same; you can find them here.

The first problem I’ll share is a “ball and urn” problem.  These are a staple of mathematical contests everywhere.

You have 20 identical red balls and 14 identical green balls. You wish to put them into two baskets — one brown basket, and one yellow basket. In how many different ways can you do this if the number of green balls in either basket is less than the number of red balls?

Another popular puzzle idea is to write a problem or two which involve the year of the contest — in this case, 2014.

A positive integer is said to be fortunate if it is either divisible by 14, or contains the two adjacent digits “14” (in that order). How many fortunate integers n are there between 1 and 2014, inclusive?

The other two problems from the contest I’ll share with you today are from other contests shared with me by my colleagues.

In the figure below, the perimeters of three rectangles are given. You also know that the shaded rectangle is in fact a square. What is the perimeter of the rectangle in the lower left-hand corner?


I very much like this last problem.  It’s one of those problems that when you first look at it, it seems totally impossible — how could you consider all multiples of 23?  Nonetheless, there is a way to look at it and find the correct solution.  Can you find it?

Multiples of 23 have various digit sums. For example, 46 has digit sum 10, while 8 x 23 = 184 has digit sum 13. What is the smallest possible digit sum among all multiples of 23?

You can read more to see the solutions to these puzzles.  Enjoy!

First we’ll look at the solution to the ball and urn problem.  Since the number of green balls is less than the red in any one basket, we know that every red ball has a green partner, so we consider the 14 pairs of red/green balls along with the six remaining red balls.

There are 15 ways to put the 14 pairs into the two baskets; the brown basket may contain 0, 1,…,14 pairs. There are 5 ways to put the remaining six red balls in the brown basket — 1, 2, 3, 4, or 5 red balls. We cannot put either 0 or 6, or else that would leave one basket with the same number of red as green balls.

Thus, there are 15 x 5 = 75 ways to put the balls in the baskets to fulfill the conditions of the problem.

Now let’s see how many fortunate numbers there are between 1 and 2014.  We first count the number of integers which contain the digits “14” in that order. There are 100 fortunate numbers of the form 14AB, 20 of the form A14B (A could be 1 or 0; in the latter case, the number has just three digits), and 21 of the form AB14 (A or B may be 1 or 0, and we have the lone case with A=2 being 2014). However, the number (and the only number) counted twice in this scheme is 1414. This gives 100 + 20 + 21 – 1 = 140 numbers so far.

There are 143 multiples of 14 between 0 and 2014. Now 140 + 143 = 283, but some numbers have been counted twice — those which both contain the digits “14” and are multiplies of 14. There are just 12 of these: 14, 140, 714, 1148, 1400, 1414, 1428, 1442, 1456, 1470, 1484, and 1498. Thus, the final count of fortunate numbers is 283 – 12 = 271.

The perimeter of the shaded rectangle is 68.  To see this, we label various segments in the figure as follows:


The information about the perimeters of the smaller rectangles results in the following system of equations: be = 24, abc = 68, and acd = 78.  We are looking for the quantity 2(+ e). But adding the first and third equations results in abcde = 102, and then subtracting the second equation from this gives de = 34.  So the perimeter of the rectangle in the lower left corner is 2(34) = 68.

As for the smallest possible digit sum — it’s just 2!  It should be clear that no multiple of 10 can be a multiple of 23. However, it is possible to have a digit sum of 2: 100,000,000,001 is a multiple of 23.  But how would you go about finding this multiple?

Just start a long division problem where a number beginning 10000…. is divided by 23. 23 goes into 100 four times, with a remainder of 8, bring down a 0. 23 goes into 80 three times with a remainder of 11, bring down a 0, and so forth.

Now bringing down a 0 will never result in an even division (because no power of 10 is a multiple of 23), but perhaps bringing down a 1 will help. Of the numbers 11, 21, 31,…, 211, 221, only 161 is a multiple of 23. So if at any time we have a remainder of 16, we may add a 1 (instead of a 0) to the divisor, with the result an even division. This in fact happens after 10 zeros, giving a minimum digit sum of 2.  Amazing!

I hope you enjoyed this set of puzzles.  Stay tuned for the next installment of The Puzzle Archives for more problem-solving fun!

Published by

Vince Matsko

Mathematician, educator, consultant, artist, puzzle designer, programmer, blogger, etc., etc. @cre8math

One thought on “The Puzzle Archives, II”

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