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# Have you figured it out? Are you someone who thinks outside the box?

Earlier today, I presented these questions from the book “Lateral Solutions to Mathematical Problems” by Des MacHale. Here they are once more, along with their corresponding solutions.

1. Three cloves on an orange

What is the likelihood that three points on a sphere lie on the same hemisphere?

Avoid performing any mathematical operations. Imagine this scenario: place two arbitrary points on the surface of a sphere (similar to inserting two cloves randomly into an orange). Then, cut the sphere (or orange) in half, with the cut passing through these two points (cloves). This results in two hemispheres, each containing the two points. The third random point (or clove) will be located on one of these hemispheres.

2. A big number

What is the last digit of the product when all prime numbers below one million are multiplied together?

A prime number is a number that can only be divided by itself and 1, such as 2, 3, 5, 7, 11, 13, and so on.

The final digit will be 0.

Do not attempt to multiply all of them together. The number is the product of 2, 3, 5, 7, and so on.

Rewording:
As 2 multiplied by 5 equals 10, we can express this as 10 multiplied by 3 multiplied by 7 multiplied by …

All integers that are divisible by 10 will have a 0 as the last digit.

3. The three threes

Can you create the number 20 using three instances of the number three and any mathematical operations of your choice?

[i.e you need to find an expression that includes 3, 3 and 3, and no other digits, but may include any other mathematical symbol, such as +, -, x, ÷, (, ), √, ., etc. An example might be 3√3/3, although this would be wrong since it does not equal 20.]

One way to approach this is by utilizing a decimal point.

20 = (3 + 3)/.3

4. Square are you?

What is the method for dividing this shape into four sections that can then be put back together to create a square?

Solution

There are 20 smaller squares within the shape. If we were to combine them to create a larger square, the side length of this square would be equal to √20. Let’s attempt to locate a line within the diagram that measures √20.

is added to 3, the sum of 2 squared and 3 squared is equal to the square of 5.

The Pythagorean Theorem states that in a right triangle, the square of the longest side (hypotenuse) is equal to the sum of the squares of the other two sides. In this case, when 2 and 3 are squared and added together, their sum is equal to the square of 5.2 + 42 = 20 = (√20)2

In the given image, the hypotenuse of a right triangle with sides measuring 2 and 4 is √20, which is the desired length. As we are creating a square, it is likely that the cut at P and Q will form a 90-degree angle. This yields the desired result.

5. Roamin’ numerals

Rearrange two matchsticks to make the equation correct.

Here is a possible solution.

Here is another example:

If you have discovered any additional ones, kindly inform me in the section below!

Thanks to Des MacHale for today’s puzzles, which are all taken from his excellent new book Lateral Solutions to Mathematical Problems.

Since 2015, I have been posting a puzzle here every other Monday. I am constantly searching for interesting puzzles. If you have any suggestions, please email me.

Source: theguardian.com 