Math puzzles!

Hi guys. I'm here to challenge all the mathy/puzzle loving people of SP. Even if you don't consider yourself a part of this category, give these a shot! You never know, you might see something they don't



Let's begin!

1) What two numbers, that between them contain the digits 1 through 9, have the largest product when multiplied?

ex: 12345*6789 are two numbers that multiply to 83810205, but you can do better than that! As you can see I used each of the nine digits only once!



2) Use the numbers 987654321 in that order. Add in any + - signs where ever is needed to make an expression that totals to 100.

ex 9+8+7+654-321=(not 100). Your job is to find an expression that does. But you MUST keep the numbers in that order. Multiple solutions.



(I'll add in more later. Try not to cheat please :O It ruins the fun of the people trying these.)

1) 9765*84321 = 823394565
2) (9*8)+7+6+5+4+3+2+1

1) 9765*84321 = 823394565
Nope. It's possible to go about 20 million higher than that. Give others a chance before you answer again
2) (9*8)+7+6+5+4+3+2+1
Good Good. People can keep submitting more.

9654 * 87321 = 842996934

1) 94321*8765 = 826723565

Little bit higher, by 3 million.

ninja'd with a better answer.

97531*8462=842,862,902

Little bit of trying around Excel:

2) 98-76+54+3+21 (thought it was just + or - signs, what Jon did was cheating sorta).

9654 * 87321 = 842996934

Closest answer so far.

Give up yet?

My second try here:

1) 9643*87521 = 843 965 003

Closest answer so far.

Give up yet?

8754 * 96321 = 843194034

1) 9642*87531=843 973 902

Slightly better

1) 9642*87531=843 973 902

Slightly better

Ding ding ding.

Okay next up:

3) Find three three-digit numbers of the form n, 2n, and 3n such that, collectively they contain the numbers from 1 through 9 exactly once each.

ex: Here's one: 192 384 576. There's three more such trios.

4)
(ignore the ~, I just need it so the problem lines up)
~SEND
+MORE
MONEY

Each different letter represents a unique digit. Which ones are they?

Little bit of trying around Excel:

2) 98-76+54+3+21 (thought it was just + or - signs, what Jon did was cheating sorta).

Yeah, I thought about it while afk after I posted it and remembered it was supposed to be +/-. :/ I'll do a legit one, here:

2) 98-7+6-5+4+3+2-1


3) 219 438 657
4) I don't understand what I'm trying to achieve here. Ah. After Googling, an image of the equation cleared it up for me. But the page also gave the answer away.

273 546 819
327 654 981
219 438 657
192 384 576

I don't think number 4 is fair because I've done it before and is fairly common.

 Spoiler

~9567
+1085
------
10652

273 546 819
327 654 981
219 438 657
192 384 576

I don't think number 4 is fair because I've done it before and is fairly common.

 Spoiler

~9567
+1085
------
10652

Well the math puzzles I'll give out will be a mix of common with... wait wut? style problems (as seen in the n, 2n, 3n problem).

5) This is a gambling game problem:

Players bet on a number between 1 and 6 (including 1 and 6). The operator (think dealer) tosses three dice. If all three dice show the same face value, say three 6s, those who bet on 6 win three times their bet, and everyone else loses. If the dice came up with two 6s and a 3, those who backed 6 win twice their bet, and those who bet on 3 win the amount of their bet (aka, they get their money back AND they get that much more back).

"Since each face has a one in six chance of coming up, when three dice are tossed there are therefore 3 in 6 chances that the number I bet on will appear... and if my number comes up more than once... I win even more! So it must be a good game to play."

ex: you bet $1 on 6.
Triple sixes occur. you get your dollar back and win 3 dollars on top of that.
double sixes occur. you get your dollar back and win 2 dollars on top of that.
single six occurs. you get your dollar back and win 1 dollar on top of that.
no sixes occur. You lose your dollar.


Is it? Your job is to work out the true odds of this game!

Mathematical Hope for the value returned for this game is 0.5 of the initial bet:

(5/6)^3 chance to get zero - ~58%
3 * 1/6 * (5/6)^2 chance to get what you bet - ~35%
3 * (1/6)^2 * 5/6 chance to get double your bet ~7%
(1/6)^3 chance to get triple your bet ~0.5%

It adds to exactly 100% and multiplying it by the returns will give you 0.5 =D

It means that on average you will walk off with half of what you invested.
The statement that "Since each face has a one in six chance of coming up, when three dice are tossed there are therefore 3 in 6 chances that the number I bet on will appear..." is already wrong. The chance that at least one face will be the one you chose is (1 - (5/6)^3) = ~42%.

Whoops, I realized that I misworded it, so I added in an example bet with payouts. Sorry shidoshi D:

P® = C(n,r)*(p^r)*[q^(n-r)]
p = 0.167
q = 0.833
n = 3
r = # times a given face of a die comes up

P(r=0) = 0.579
P(r=1) = 0.348
P(r=2) = 0.0697
P(r=3) = 0.00466

assuming win = p(r>1):

p = 0.348 + 0.0697 + 0.00466
= 0.42

avg. gross gain given win (where bet = x) [μ(gain|p(r>1)]: (2x+3x+4x)/3 = 3x

μ = 3x(0.421) - -x(0.579)
= 0.684; you'll walk away with 0.684 times what you bet on average

6 x 6 x 6 = 216 total combos
5 x 5 x 5 = 125 combos without the number YOU guessed.

125/216 = 57.9% chance that you lose.

6 x 6 x 6 = 216 total combos
5 x 5 x 5 = 125 combos without the number YOU guessed.

125/216 = 57.9% chance that you lose.

That's only the first part. You have the losses part right, now you have to split up the winning sections based off expected gain. Do the gain amounts outweigh the loss chance?

Also, not yet Chrome D: