Random Math Thread : Puzzles , Jokes , Problems, Funny Anecdotes, Random Thoughts

[Science Friction]

Use Gauss' formula that I derived above to find the sum of the first 2,015 positive integers!

 

Try it the way he did it in second grade and then verify your answer by using the formula S = n(n+1)/2

[Clyde]

Change the direction a bit.

 

If I asked you to take 15% of 220 and do it in your head you'd probably say you couldn't do it very easily.

 

But if I asked you 10% of 220 you'd quickly tell me 22. You need, though, another 5%. 5% is half of 10%. We know 10% = 22 so half of that is 11 so 15% of 220 = 33. Easy peasy.

 

If I asked you 14% instead of 15% you still take 10% to get 22. You need 4% now. You could say 1% = 10% of 10% which means that's 2.2. 4 times that is 8.8 which you than add to 22 to get 30.8. Easy peasy.

 

Use the 10% calc to break numbers down easily. Impress your friends.

[Science Friction]

Change the direction a bit.

 

If I asked you to take 15% of 220 and do it in your head you'd probably say you couldn't do it very easily.

 

But if I asked you 10% of 220 you'd quickly tell me 22. You need, though, another 5%. 5% is half of 10%. We know 10% = 22 so half of that is 11 so 15% of 220 = 33. Easy peasy.

 

If I asked you 14% instead of 15% you still take 10% to get 22. You need 4% now. You could say 1% = 10% of 10% which means that's 2.2. 4 times that is 8.8 which you than add to 22 to get 30.8. Easy peasy.

 

Use the 10% calc to break numbers down easily. Impress your friends.

 

There was a guy that used to sell videos on infomercials teaching kids how to do rather complex arithmetic calculations very quickly. I forget his name. There are tons of little shortcuts that make calculations easier. Good stuff indeed! :thumb:

[Clyde]

There was a guy that used to sell videos on infomercials teaching kids how to do rather complex arithmetic calculations very quickly. I forget his name. There are tons of little shortcuts that make calculations easier. Good stuff indeed! :thumb:

 

I could see that. We were not taught to THINK as kids. We were taught to memorize and follow a method.

[Science Friction]

I could see that. We were not taught to THINK as kids. We were taught to memorize and follow a method.

 

 

Very true, Clyde, and with today's kids it's even worse. I regularly see college kids punch those calculator buttons to do the most trivial calculations. They have been programmed to punch instead of think. Just the other day, I had a college algebra student punch in "8X 3" on her calculator . After she did it, I could she was a bit embarrassed that she did that without even thinking.

[35Bird]

I'll leave you all with this one. Made this one up a few years ago.

Mismatched Joe is in a pitch dark room picking out socks from a dresser drawer. There is a total of six socks in the drawer. There are only blue and brown socks. He can't tell anything about the color by just the feel of the sock. He chooses two socks at random. If he has a 2/3 chance of picking out a pair of blue socks, what is the chance he chooses a pair of brown socks?

 

0. There is only 1 brown sock present, and there are 5 blue ones.

 

(5/6)*(4/5) = 20/30 = 2/3.

 

If he picks a blue sock first, he has a 1/6 chance of getting a Blue/Brown pair. (5/6)*(1/5) = 1/6

 

If he picks a brown sock first, he has a 1/6 chance of getting a Blue/Brown pair (1/6)*(5/5)= 1/6

 

2/3 + 1/6 + 1/6 = 1

[35Bird]

You have 26 scrabble tiles, one for each letter, and throw them in a bag. You hate the letter Q. You draw 5 tiles. What is the chance that none of the 5 is the letter Q?

[Clyde]

0. There is only 1 brown sock present' date=' and there are 5 blue ones. (5/6)*(4/5) = 20/30 = 2/3. If he picks a blue sock first, he has a 1/6 chance of getting a Blue/Brown pair. (5/6)*(1/5) = 1/6 If he picks a brown sock first, he has a 1/6 chance of getting a Blue/Brown pair (1/6)*(5/5)= 1/6 2/3 + 1/6 + 1/6 = 1[/quote']

 

Stupid me. I had 5 blue and 1 brown.

[UKMustangFan]

You have 26 scrabble tiles, one for each letter, and throw them in a bag. You hate the letter Q. You draw 5 tiles. What is the chance that none of the 5 is the letter Q?

 

If I did this correctly, it's (25/26) * (24/25) * (23/24) * (22/23) * (21/22) = 0.9615 * 0.96 * 0.9583 * 0.9565 * 0.9545 = ~80.8% chance that a Q is not picked.

[35Bird]

That's what I got, Boss.

[ggclfan]

Here is one for you all to ponder and this actually happened. I don't know the answer but my guess is SF can figure it out for me (or someone else on here). A buddy of mine and I were each in a betting pool where each week, we were given 10 NFL/college games and the spread on each game. We could then choose any 7 of the 10 to bet against the spread (assume each pick had a 50/50 chance of winning). We had a side bet on who would do better for the year. We would each decide the games to bet and the winners and then get together to see what we picked. One week we not only picked the same 7 games to bet out of the 10, we also picked the same seven winners! What are the odds of that happening???

[mcpapa]

There was a guy that used to sell videos on infomercials teaching kids how to do rather complex arithmetic calculations very quickly. I forget his name. There are tons of little shortcuts that make calculations easier. Good stuff indeed! :thumb:

 

I owned this book many years ago:

 

[Science Friction]

0. There is only 1 brown sock present, and there are 5 blue ones.

 

(5/6)*(4/5) = 20/30 = 2/3.

 

If he picks a blue sock first, he has a 1/6 chance of getting a Blue/Brown pair. (5/6)*(1/5) = 1/6

 

If he picks a brown sock first, he has a 1/6 chance of getting a Blue/Brown pair (1/6)*(5/5)= 1/6

 

2/3 + 1/6 + 1/6 = 1

 

 

Very nice!!!

[Science Friction]

If I did this correctly, it's (25/26) * (24/25) * (23/24) * (22/23) * (21/22) = 0.9615 * 0.96 * 0.9583 * 0.9565 * 0.9545 = ~80.8% chance that a Q is not picked.

 

Indeed you did. Good job.

[Science Friction]

Here is one for you all to ponder and this actually happened. I don't know the answer but my guess is SF can figure it out for me (or someone else on here). A buddy of mine and I were each in a betting pool where each week, we were given 10 NFL/college games and the spread on each game. We could then choose any 7 of the 10 to bet against the spread (assume each pick had a 50/50 chance of winning). We had a side bet on who would do better for the year. We would each decide the games to bet and the winners and then get together to see what we picked. One week we not only picked the same 7 games to bet out of the 10, we also picked the same seven winners! What are the odds of that happening???

 

There would be 1 chance in 120 of you both picking the same seven games out of ten since there are 10!/(7! 3!)= 120 possible combinations. Also, the chance of both of you picking the same winners in all seven games is 1/(2^7) = 1/128 . Thus, the probability that both of you picked the same seven games AND the same seven winners is 1/(120*128) = 1/15,360.

 

One chance in 15,360.