# If I memorized all math multiplication tables from 1 to 100 then how many unique calculations would I have?

For example, 2 * 3 is the same as 3 * 2. Hence you would be memorizing certain things in double, right? So 1 to 100 would be how many unique calculations?

Relevance

This is basically asking for a Combination of 100 numbers with a subset of 2, or C(100,2). Without getting into why, this is represented by

100! / (2! * 98!)

Or 100*99 / 2 = 4950

However, this does not take into account for squares, or numbers like 88*88. There are 100 of these, so add 100 to that answer to get:

5050 distinct calculations

Note: the response of 100! is a permutation, and that adds all the doubles you don't want, for instance 2*3 and 3*2 are both counted.

• Assuming you memorize tables in the following way:

1 x 1 = 1 ... 1 x 100 = 100

...

...

...

100 x 1 = 100 ... 100 x 100 = 10000

This makes 100 x 100 = 10000 calculations in all (doubles included). Now, we follow a policy of elimination of the doubles.

Consider this simple 4 x 4 calculation matrix.

1 2 3 4

1 1 2 3 4

2 2 4 6 8

3 3 6 9 12

4 4 8 12 16

Here, double calculation is done only on the upper-right (or left) diagonal elements and all others are unique. This essentially makes a (different) Pascal's triangle as folows:

1

2 4

3 6 9

4 8 12 16

and so on....

Essentially, the number of elements in this triangle is the number of unique calculations you want. Which is:

1 + 2 + 3 + 4 + ...... + 100 = 100 x 101 / 2 = 5050

Cheers,

Himanshu, India.

• Anonymous

you want to how many *unique* combinations of 2 you can get out of a set of 100. This function is called a "combination" (a permutation is the total number of possible sets).

k = 2

n = 100

combination(k,n) = n! / (k! * (n - k)!)

solve:

100! / (2! * 98!) = 4950

however this doesn't account for identical numbers (1*1, 2*2, etc) there are 100 of these, so the total would be 5050.

• You can figure this out as a series. For each number n, you have x possible multipliers. x = 100 - n. So, calculate that formula as a series to get sum(x), and you get 4950.

If you're going to allow for a number times itself (2 x 2) you have to modify your formula slightly to x = 100 - (n - 1) and you get 5050.