TimeFan

does any1 have any math puzzles from like the AMC or somethin bcause this is the pl8ce to share them

I don't know if this is mathematically relevant (maybe Ben can tell me) but I came up with this problem a day ago and it was pretty fun to solve

basically it is this:

using only the numbers 1-9, using each digit only once, come up with an expression that has all the digits 1-9 in its answer.

the rules are: you have to just multiply, divide, add, subtract, or exponent straight numbers. like you could do (2*5*6)+8 to make 68 and use up your 2 5 6 and 8, but you couldn't just use up 6 and 8 and make 68

ALSO the answer can be a decimal, whatever just so long as it has 123456789 in it. so it could be like 95.8476213 if you can get it.


ARE YOU UP TO THE CHALLENGE???

also post your math problems here if you have some

Exploiting the many digits in sevenths:

3*(9*8 + 5*(4+6) + 1) + 2/7 = 369.285714285714285714285714285714285714285714...

Awesome! I think I tried a similar method, on my TI-84 graphing calculator, and failed. But I did get some other from space solutions:

The first one I tried was (((((9^2)^3)*8)/7)/6)*5)=506134.2857, which, using my display, only yielded eight of the nine digits. But then I reworked it, and got (((((9^2)^3)/7)*8)/6)*5), which, now I realize, is the same exact number. Huh.

But I got two other answers. The ^5 part of this was found by guessing: 3((2^7+9*8)/6)^5-1=123456789.1

This was found by working backwards, it uses the least amount of digits (only six!): ((2*5)^9-7)/(3^4)=12345678.93

OK, I got one where the result is an integer, with no repeated digits:

(2*8)^7 - 5*6*9 + 3*4 - 1 = 268435197

My approach was to start by getting a nine-digit number with the first few digits all unique (and nonzero), and then use the remaining numbers to adjust the end of it a bit.


Here's a similar one using just 1 through 8, but still getting all nine digits in the result:

(5+6)^8 + 7*4*3*1 + 2 = 214358967


I'm curious what the smallest n would be such that the numbers from 1 to n are sufficient. Can anyone beat 8?

I am going to beat 8, you wait!

Got one:

(7^(1/2))+6^4=1298.6457513, although the graphing calculator doesn't show the 3 at the end in this form, it does show the 3 in just straight 7^(1/2).

Um, Ben, shouldn't most square roots just have a never-ending pattern of decimals, eventually hitting all 9 digits?

Yeah, 3^(1/2) works that way. It'd take more digits than that for an integer solution, though.

Using 1 to 5, integer result with some repeated digits:

2^(4^3+5-1) = 295147905179352825856

awesome! I didn't say it had to be an integer, but oh well.

also does anyone have any other math puzzles or is this not a thing

Yeah, I just like getting integers.

If you want some interesting maths puzzles, check this out:
http://www.m-a.org.uk/resources/sp/sp_1999.pdf

I'm going to be signing up for my MA membership pretty soon and I'm rather excited about the various publications I will be receiving!