Earlier today I set you the following puzzles:

1. *In this word-sum, each letter stands for one of the digits 0–9, and stands for the same digit each time it appears. Different letters stand for different digits. No number starts with 0.*

*Find all the possible solutions of the word-sum shown above.*

**Solution**

To find *O*, we add three *O*s to find a number that has an *O *in the units place. The only digits which satisfy this condition are 0 and 5.

If *O *= 0, there are no tens to “carry” to the middle column and so we are seeking a digit for *M*, for which three *M*s have an *M *in the units place. Therefore, *M *is 0 or 5. However, *O *= 0 and so *M *= 5 (as different letters stand for different digits).

If *O *= 5, the total of three *O*s is 15 and so we have a ‘1’ to carry to the middle column. However, there are no digit values of *M*, where 3*M *+ 1 gives *M *in the units column.

Hence, *O *= 0 and *M *= 5. If *J *= 1, *I *= 4. If *J *= 2, *I *= 7. If *J *> 2, 3 × ‘*JMO*’ does not give a 3-digit number and so there are only 2 possible values for *J*.

The only solutions are ‘*JMO*’ = 150 with ‘*IMO*’ = 450 and ‘*JMO*’ = 250 with ‘*IMO*’ = 750.

2*. In the diagram below, a quarter circle with radius 3cm is positioned next to a quarter circle with radius 4cm.*

*What is the total shaded area bounded by the blue lines, in cm2.*

**Solution**

Extend the bottom line to make a rectangle with height 1cm and width 3cm:

The area of a circle is *πr*2, where *r *is the radius, so the area bounded by the perimeter of the shape in the diagram above is the area of the 3x1 rectangle plus the two quarter circles:

3 + (1/4)9π + (1/4)16π = 3 + (25/4)π

The shaded area bound by the blue lines is this value minus the area of the long triangle of height 1cm and width (3 + 4)cm.

Which is 3 + (25/4)π – 7/2. So the answer is:

(25/4)π – (1/2).

3. *An equilateral triangle is divided into smaller equilateral triangles.*

*The figure on the left shows that it is possible to divide it into 4 equilateral triangles. The figure on the right shows that it is possible to divide it into 13 equilateral triangles.*

*What are the integer values of n, where n > 1, for which it is possible to divide the triangle into n smaller equilateral triangles?*

**Solution**

Let’s call the equilateral triangle which is to be divided into smaller equilateral triangles the original triangle.

Let’s start with *n *= 2. Is there a way of dividing the original triangle into two smaller ones? We can’t draw a line from a vertex, since this would produce angles less than 60°. But if we draw a single line that draws an equilateral triangle, as in (a) below, we create another quadrilateral in the original triangle, not a triangle. So *n *cannot be 2.

If we were going to be able to divide the quadrilateral in (a) into equilateral triangles, we could only divide it into an *odd *number of equilateral triangles, as in (b), which divides into 5 smaller equilateral triangles. Since there is no way to divide the quadrilateral into an even number of triangles, we can’t divide it into either 2 or 4 triangles, so we can also rule out *n *= 3 and *n *= 5.

The case for *n *= 4 is given in the question and so *n *= 4 is possible.

Diagram (b) shows the case for *n *= 6. Any of the smaller equilateral triangles in this diagram can be spilt into 4 equilateral triangles, which has the effect of adding 3 to the total number of triangles. (Diagram (c) is (b) with the top triangle divided into 4.) We can continue in this fashion in order to create cases when *n *= 9, 12, 15, . . . . Therefore, it is possible to split the original triangle into *n *smaller equilateral triangles where *n *is a multiple of 3, (excluding *n *= 3).

Returning to the diagram for *n *= 4, we can see that any equilateral triangle can be split into 4 equilateral triangles. Hence, *n *can be 7, 10, 13, . . . . Therefore, it it is possible to divide the original triangle into *n *smaller equilateral triangles where *n *is one more than a multiple of 3.

Diagram (d) shows the construction for splitting the original triangle into 8 smaller equilateral triangles. From this diagram, we can show that *n *= 11, 14, 17, . . . are possible by splitting any of the existing equilateral triangles into 4 equilateral triangles. Therefore, is is possible to split the original triangle into *n *smaller equilateral triangles where *n *is two more than a multiple of 3 (excluding *n=* 5.)

Therefore, is is possible to divide the triangle into *n *smaller equilateral triangles for all positive integer values of *n*, excluding 2, 3 and 5.

*Thanks again to the United Kingdom Mathematics Trust for letting me use today’s problems, which come from this year’s Junior Mathematical Olympiad. *

*I’ll be back in two weeks.*

*I set a puzzle here every two weeks on a Monday. I’m always on the look-out for great puzzles. If you would like to suggest one, email me.*