| |
Darkseeker
|
My parents showed me a math riddle that I haven't been able to figure out. Basically, it's the following: Use any operator or arithmetic method to achieve find the results. You may not switch the order of the numbers around, but you can use things like addition, square roots, exponents, factorials etc. 3 7 5 = 1 3 7 5 = 2 3 7 5 = 3 3 7 5 = 4 3 7 5 = 5 3 7 5 = 6 3 7 5 = 7 3 7 5 = 8 3 7 5 = 9 3 7 5 = 10 If you can get all ten correct, then that means you're a genius :D |
|
|
| |
Neutral
|
Do we post the answers? I mean I figured it out ->- |
|
|
| |
Darkseeker
|
You figured it out that fast? O.o I mean, you could if you really wanted, but... probably should put a spoiler on it. You can do that by coloring your text and highlighting it the same color. |
|
|
| |
Neutral
|
Yes...? It may take a while to type it though -_- |
|
|
| |
Darkseeker
|
There's probably multiple answers out there, but you can if you want. lmao |
|
|
| |
Neutral
|
Number 1 is: 3-7 = -4 +5 = 1 will take a few minutes to type all the answers |
|
|
| |
Neutral
|
lol is gcd even a valid arithmetic method #1: 3 - 7 + 5 = 1 #2: (3 + 7) / 5 = 2 #3: gcd (3, (7+5)) = 3 #4: (|3! - 7!|) mod 5 = 4 #5: 3 + 7 - 5 = 5
#6: gcd (3!, (7+5)) = 6
#7: √(|3-7|) + 5 = 7 #8: (3 mod 7) + 5 = 8 #9: |3 - 7| + 5 = 9 #10: gcd ((3+7), 5!) = 10 |
|
|
| |
Neutral
|
(3*7) mod 5=1
(3^7) mod 5=2 (3 + fact(7)) mod 5=3 sqrt(3x7-5)=4 3+(7 mod 5) = 5 3x(7-5)=6
sqrt(abs(3-7))+5=7 fact(3)+7-5=8 3^(7-5)=9 sqrt(abs(3-7))*5=10 |
|
|
| |
Neutral
|
Dire's officially a genius |
|
|
| |
Neutral
|
Life on Mars said:
Dire's officially a genius
Thanks :)
You're doing a nice job too |
|
|
Spring Month: 1 
Night 
Heavy Snow Moon: 
