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?

6 Answers

Relevance
  • Jon G
    Lv 4
    1 decade ago
    Favorite Answer

    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.

  • 1 decade ago

    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

    That, I believe, is your answer.

    Cheers,

    Himanshu, India.

  • Anonymous
    1 decade ago

    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.

  • 1 decade ago

    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.

  • How do you think about the answers? You can sign in to vote the answer.
  • 1 decade ago

    100*100/2

    =5000

  • Anonymous
    1 decade ago

    100*100-50?

Still have questions? Get your answers by asking now.