I have always been interested in numbers and playing with them since I was a small kid. Every time someone has a birthday I am always ready to provide an alternative to have the normal decimal birthday. So e.g. having your 100th birthday when you really have your 49 birthday in decimal numbers.
So here is some number theory for birthdays and IQ tests that you can play around with on your vacations days and prepare for future birthdays and IQ tests. Have fun.
First some short introduction to numbers and the number base. When we use numbers we always assume we're counting with decimal numbers. Decimal numbers means that we are using 10 as the base. Thus when we are saying that someone is 25 years old we really mean that he is 2 * 10^1 + 5 * 10^0 = 2 * 10 + 5 year old. If instead someone has his 25 year birthday in octal number what we are saying is that he has his 2 * 8^1 + 5 * 8^0 = 2 * 8 + 5 = 21 in decimal numbers.
So by varying the number base we can have almost any birthday changed into an even birthday. For example when can we say that we have our 100th birthday. The smallest base is 2, this means that our first 100th birthday happens already at our 4th birthday using base 2. Later in life we can have a 100th birthday when we have 9th birthday, 16th birthday, 25th, 36th, 49th, 64th, 81st and 100th. It is very unlikely that someone will celebrate their 100th birthday in base 11 which would happen at age of 121.
Thus celebrating 100 years happens quite a few times, but not very often still.
Other even numbers are more common. We can have our 20th birthday every second year from our 6th birthday. To be 20, the minimum base is 3 since the number 2 cannot be used in base 2 that only have numbers 0 and 1. Thus 2 * 3 + 0 = 6 is the minimum age to become 20.
However after 6 years old you can always have your 20th birthday at any birthday which is an even number. Thus e.g. at age 38 you will be 20 using base 19, 2 * 19 + 0 = 38.
If you want to search for an appropriate age to celebrate on your next birthday, start by dividing your age into a product of prime numbers. So e.g. 38 is the product of 2 and 19 which both are prime numbers. Thus the most even numbers you can get here is 20 in base 19 and 100110 in base 2. If your age is 18 you have more options, this is divided into prime numbers 2,3 and 3 since 18 = 2 * 3 * 3. So here you can have your 200th birthday in base 3 and 10010 in base 2.
However you stumble into issues with the above approach when the age you have achieved is a prime number itself. So for example when your 37th birthday approaches, how will you divide this into an even number to celebrate. The only obvious even number to reach here is 10 years old which can be achieved with any prime number by using the prime number itself as the base.
Here the age 25 comes to the rescue which is seen as an even birthday by most people. Actually we can prove that every birthday with an odd number of years can become 25 in some base if the odd number is at least 17.
Proof: The proof is fairly simple, first of all an odd number is always written as 2 * k + 1 where k is any number. Second the minimum base to use for an age of 25 is 6 since the number 5 doesn't exist in bases 2,3,4 and 5. Thus the first time to have your 25th birthday happens on your 2 * 6 + 5 = 17th birthday.
So choose any odd number larger than or equal to 17. This number can always be written as 2 * k + 1 where k is at least 8. But it can also be written as 2 * (k - 2) + (1 + 2 * 2) = 2 * (k - 2) + 5 = 25 in base k - 2. Thus to calculate the number base to use one calculates:
(Odd - 5) / 2. Thus with 37 you get (37 - 5) / 2 = 16. Thus at your 37th birthday you have 25th birthday in base 16.
Isn't it nice to know that you can always claim to be 20 or 25 years old after reaching 17 years of age for the rest of your life :)
Have fun on future birthday in figuring what age you want to have this time.
Actually the base 10 was selected in Arabia, in many older cultures the base 12 was used, even some money systems still have the number 12 in them. If you are working with computer programs it is very popular to use hexadecimal numbers using base 16 with digits 1,2,3,4,5,6,7,8,9,A,B,C,D,E and F.
So on to IQ tests. Most of you have seen tests like the below one:
1, 4, 9, 16, ?
This one is fairly easy, it is the square of the index, thus x^2 is the function in this number series. Thus the next number in the series is 5 * 5 = 25.
Let's take a bit more complex number series now.
2, 9, 28, 65, ?
This one is a bit more difficult to see directly, I will give a hint, it is based on the function x^3 + 1. Thus the next number is 5 * 5 * 5 + 1 = 126.
Now let's take another one, we use the function x^2 - 2 * x - 2.
-3, -2, 1, 6, ?
This looks difficult at the outset, but since we know the answer we cheat and simply set it to 5 * 5 - 2 * 5 - 2 = 13.
So how does one solve this type of IQ tests in a quick manner. Well it is fairly simple using difference techniques, a bit like Fibonaccis tree.
So write the difference between the numbers and then the difference of the differences.
In the above calculation we write it up as follows.
-3, -2, 1, 6, 13
1, 3, 5, 7
2, 2, 2
Interesting the difference is simply a linearly increasing function which is very easy to see and the second difference is simply constant so even easier.
We can see that the difference function is simply 2 * x - 2 and the second difference is simply 2.
For those familiar with derivatives, you can see that 2 * x - 2 is the derivative function of x^2 - 2 * x - 2. and 2 is the second derivative of this function.
So now let's try if this works in practice, here is a number series again:
0, 1, 8, 27, ?
We use the difference technique:
0, 1, 8, 27, => 64
1, 7, 19, => 37,
6, 12, => 18
So we write the answer to be 64. Now let's check the answer, the function I used in this case was:
x^3 - 3 * x^2 + 3 * x - 1
Thus using x = 5 we get 5^3 - 3 * 5^5 + 3 * 5 - 1 = 125 - 75 + 15 - 1 = 64
Thus we found the correct answer of a fairly complex IQ test and we can claim to be more intelligent than we really are :)
Have fun in showing off your capabilities in IQ tests.
