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Calendar Puzzle

```
Date: 12/02/1999 at 19:31:12
From: amanda fredricks
Subject: Math puzzle

A student cuts a three-by-three square from a calendar page. If the
sum of the nine dates is divisible by 19, what is the date in the
lower left-hand corner of the square?
```

```
Date: 12/02/1999 at 21:57:55
From: Doctor Wilkinson
Subject: Re: Math puzzle

This is a cute problem, Amanda! If you don't know any algebra, you can
do it by trial and error. The number in the bottom right cannot be
greater than 31, so the number in the upper left cannot be greater
than 31 minus 16 or 15, so you have only 15 possibilities to check,
which is not too awfully many.

A little algebra makes it easier, though, so I'll assume you can at
least follow an argument that starts "let x be ..."

So here goes:

Let x be the sum of the numbers in the top row of the square.

The numbers in the second row of the square are all 7 greater than the
corresponding numbers in the top row. For example, it might look like

3    4    5
10   11   12
17   18   19

For example, the 10 is 7 more than 3 because it's in the next week,
and each week has 7 days.

So the sum of the numbers in the second row is 21 (3 times 7) more
than the sum of the numbers in the first row, and the sum of the
numbers in the third row is 21 more again, or 42 more. So if the sum
of the numbers in the first row is x, the sum of the numbers in the
second row is x + 21, and the sum of the numbers in the third row is
x + 42, so the sum of all the numbers is

x + (x + 21) + (x + 42)

or

3x + 63

and this is supposed to be divisible by 19.

Now 63 is divisible by 3, so we can write

3x + 63 = 3*(x + 21)

and you can see that x + 21 has to be divisible by 19 also.

Now let's take a closer look at x.  If you have three consecutive
numbers, like

6 7 8

for example, their sum is always three times the middle number,
because the first number is one less, and the last number is one
greater. So we can write

x = 3m

where m is the middle number.

Now we have 3m + 21 is divisible by 19, and we can do the same trick
again, writing this as

3*(m + 7)

so m + 7 has to be divisible by 19, and in fact it has to be equal to
19, because the next multiple of 19 is 38, and that's too big
(remember we're talking about days of the month here, and no month has
more than 31 days).

So m must be 12, and now you can fill out the square yourself and
figure out what's in the lower left.

- Doctor Wilkinson, The Math Forum
http://mathforum.org/dr.math/
```

```
Date: 8/30/2000 at 17:40:50
From: Doctor TWE
Subject: Re: Math puzzle

A similar, but somewhat simpler approach, is to let x be the number in
the middle of the 3x3 square. Then the numbers in the 3x3 square can
be written as:

x-8   x-7   x-6
x-1    x    x+1
x+6   x+7   x+8

The number in the upper left-hand corner is x-8 because it is a week
and a day, or 8 days, before day x. Similarly, the number in the
middle of the upper row is x-7 because it is a week, or 7 days, before
day x. And so on.

The sum of the 9 squares is then

(x-8) + (x-7) + (x-6) + (x-1) + x + (x+1) + (x+6) + (x+7) + (x+8)
= 9x

(It's convenient that the plusses and minuses cancel out.) Now we are
told that the sum, 9x, is divisible by 19. Since 9 and 19 relatively
prime - i.e. they share no common factors other than 1 - we know
that x is divisible by 19. Since x is the date in the middle of the
3x3 square, it can only be 19. (As Dr. Wilkinson pointed out, the next
multiple of 19 is too large to be a date.) Now we can see that the
number in the lower left-hand corner is

x+6 = 19+6 = 25.

- Doctor TWE, The Math Forum
http://mathforum.org/dr.math/
```
Associated Topics:
Middle School Algebra
Middle School Puzzles

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