Slips and Sums

Today's question is a pure thinking question. Its answer is very short, but it's a great question nonetheless. Read it carefully, then try to answer. Don't look at the answer just yet, it's easy when you think.

A jar is filled with a thousand paper slips which have the numbers 1 - 1000. Each time two slips are taken out of the jar, and the difference between the numbers on them is written on a new slip of paper. The difference is put back in the jar and the two original numbers are taken out. Will the number on the last slip of paper be odd or even?

Like many other questions, this one looks impossible at first. There are so many papers, and it's hard to keep track on them. Trying out all possibilities will take way too long, but luckily, using a very powerful tool, your brain, will solve this question easily.

Try looking at the sum of all numbers in the jar. At first, it's 500,500 (using the sum formula of natural numbers from 1 to n: n(n + 1)/2). Then you take out two numbers and put back their difference. However, the parity does not change.

If the sum of two numbers is odd, their difference is also odd. The same goes for even sums. When you take out two numbers and put back their difference, the sum goes down, but the parity does not change. You either take out an even sum and put back an even number, or take out an odd sum and put back an odd number.

Since the parity of the sum of the numbers in the jar is always kept even (because the initial sum is even), the last two slips will have an even sum. Because their sum is even, their difference is also even, and the number on the last slip will be even.

Hope you liked it,
Nadav

nadavs

Magic T

Remember "Magic C"? The one where you had to put the digits 1 - 5 to create sums? Well, today I have Magic T. The question is nearly identical, but with a little twist (besides the other letter).

The digits 1, 2, 3, 4, 5, 6, and 7 are placed in squares that form a T shape. The top row has 3 squares and the middle column has 5 squares. The intersection square, the one that appears on both the row and the column, is shaded. The digits are placed in such way that the sum of the digits in the row and the sum of the digits of the column are equal.

Show that the shaded square must have an even number to create equal sums, and find an example for each possible digit on the shaded square.

One way to do that is find every possible combination for these squares and see that only even numbers can be in the shaded square. However, there is a mental condition that describes doing such thing: insanity. There are many possible combinations (5040 to be exact), and finding the right ones will take forever. It seems like we need a better method: creative thinking.

First, try making a correct combination. It shouldn't take you too long. A good method you can use is number replacement. Check the difference between the two sums and switch digits accordingly. Notice what happens when you change digits: the difference changes by an even number. Always. That happens because one sum goes up by a number, and the other down goes down by the same number, so combined the difference was changed by twice that number, which is always even.

Now we know that the key here is the difference between the sums, but how does the shaded square relate to it? Since it appears on both sums, it dictates how the differences will relate to each other.

If you place an odd number in the shaded square, you will have three odd numbers left. If you place two of them or none of them at the top row, you will have one or three of them left for the middle column, respectively. This means that the top row will have an odd sum and the column will have an even sum. Not good.

If you place only one of the remaining odd numbers on the row, you will have two left for the column. Once again, this situation is not good. It creates an even row and an odd column, which do not have an even difference.

But if you place an even number in the shaded square, it's perfect: you will either have both sums even or both sums odd, so you can play with the differences and create a zero difference Magic T.

Here are three examples, each with a "shaded square" of 2, 4, and 6 (they are lying on the side so they could be written here):

7
2 - 1 - 3 - 4 - 5
6

7
4 - 1 - 2 - 3 - 6
5

7
6 - 1 - 2 - 3 - 5
4

Hope you liked it.
Nadav

nadavs