[FORUM DOLLARS] Math questions!

[Nightshade]

#3:
283572 (10) = 4H9C (40)

[Nightshade]

#7:

AO&J42WA^!K = 111,407,776,031,422,260 (assuming no arithmetic errors)

[K73SK]

That's what I got

[Nightshade]

#4:
293859235236333 (10) = 14XM&3UMSD

[meth962]

Shucks! I forgot about this! One left??

6.) IEA!932 = 2,175,074,490

[Nightshade]

I got a different answer:

#6: IEA!932 = 75,189,518,522

2,175,074,490 Mod 40 = 10, so the base-40 representation of that number would have to end in A, not 2. I got L9PLMA as the base 40 representation of that.

Assuming that I haven't also made an arithmetic error, this should be the last of them.

[K73SK]

Yup, Night got it, sorry bud!

I'll try to think of another one now

[meth962]

Hrm, not sure where I got that random number from. I ran it again and got 75,189,518,522. I should have double checked my work! I had all the answers on the first day. I cheated though, I used a base Calculator I wrote in C# years ago

[K73SK]

For people wondering on how others got the answers, here are the methods I use (and I'm sure many people have their own methods).

To go to base-40 from base-10, you would follow the following steps (In this example, we'll use 283572)
283572 = (40*7089) + 12 -> C
7089 = (40*177) + 9 -> 9
177 = (40*4) + 17 -> H
4 = (40*0) + 4 -> 4
And when taking those remainders and going up, you get 4H9C.
So what happen there? We simply divided the main number we started with by 40 and grabbed the remainder to find out what number was used in base-40. So dividing 283572 yielded 7089.3. So we're left with 7089 with a remainder of 12 (40*.3 = 12). We then follow the same procedure with 7089 until we get down to (40*0) + r, where r is the last remainder. At this point, we take that and go backwards for the answers we got.

As for going from base-10 to base-40, this is actually easier (in my opinion). Let's use #6: IEA!932
First count the number of digits we have in IEA!932. There's a total of 7. So, if we include 0, we're going from 0-6. Now we just calculate it this way:
(40^6 * I) + (40^5 * E) + (40^4 * A) + (40^3 * !) + (40^2 * 9) + (40^1 * 3) + (40^0 * 2)
Where we will then substitute the I, E, A, and ! with the decimal digits 18, 14, 10, and 36 respectively.
(40^6 * 18) + (40^5 * 14) + (40^4 * 10) + (40^3 * 36) + (40^2 * 9) + (40^1 * 3) + (40^0 * 2) = 75,189,518,522

At least, this is the way I was taught to do it. There may be easier ways to do it that I'm unaware of... But it's at least an easy way to know how to convert from base-x to base-10 and vice versa.

[Nightshade]

I used a variation of the above algorithms.

In particular, to go from base 40 back to 10, I did the following:

Convert each base 40 digit to a base-10 representation:

4H9C = 4 17 9 12

Now, I do the following:
(((((4 * 40) + 17) * 40) + 9) * 40) + 12