<p>i’m confused: how does one win the game? also, if one wins the game by leaving three numbers that add up to 15, A always loses cause B always erases the 6th number</p>
<p>Sorry, I meant that you win by claiming exactly three numbers that add up to 15. For example,</p>
<p>A takes 7, B takes 9
A takes 3, B takes 5
A takes 2, B takes 4 (this is just an example)
A takes 6.</p>
<p>A wins because 7+2+6 = 15.</p>
<p>Ooh, I have a blackboard problem, too, which I haven’t yet attempted, but I’m going to as soon as I find some time. 
Here it is:</p>
<p>The natural numbers from 1 to 10 are each written on the blackboard 10 times. The students in the class then play the following game: a student deletes 2 of the numbers and instead of them writes down on the blackboard their sum decreased by 1; after that another student deletes 2 of the numbers and instead of them writes down on the blackboard their sum decreased by 1; and so on. The game continues until only one number remains on the blackboard.What is the remaining number?</p>
<p>That’s really easy…solved it in like 1 minute. It’s still harder than an SAT problem, though.</p>
<p>Just did it, too. I’m getting 451.</p>
<p>Another one:
[(2+3)(2^2 + 3^2)(2^4+3^4)…(2^1024 + 3^1024)(2^2048 + 3^2048) + 2^4096] / (3^4096) = ?</p>
<p>Hey guys, needed help on a question I’m having trouble with. </p>
<p>If you can buy c cinnamon rolls for n nickels, how many cinnamon rolls can you buy for d dimes and q quarters? (Answer in terms of c, n, d, q)</p>
<p>Start by converting nickels, dimes, and quarters into a common unit.
1 nickel = 5 cents
1 dime = 10 cents
1 quarter = 25 cents.</p>
<p>Cost of c cinnamon rolls = n x 5 = 5n cents
Cost of 1 cinnamon roll = 5n/c.</p>
<p>d dimes and q quarters = (10d + 25q) cents
No. of cinnamon rolls that can be bought = (10d + 25q)/(5n/c)
= c(2d+5q) = 2dc + 5qc.</p>
<p>Perfect. I think you left the n out from the final answer but thank you!</p>
<p>If that last problem was multiple choice, it would be a perfect time to make up numbers. (I’m assuming it is an sat-like problem – maybe a mistake for this thread…)</p>
<p>The thing about making up numbers for the variables is that it really reduces the level of the problem to the point where it is like a vacation in the middle of the test.</p>
<p>I would have gone with something like c=5 rolls for n=10 nickels, making it 10 cents per roll. Then I’d say something like q= 4 quarters, d= 3 dimes…that’s $1.30 which buys 13 rolls. Then it’s off to the answer choices…plug in my numbers, rule out anything that isn’t 13.</p>
<p>For yellowcat429’s problem (in post #106) multiply the big thingy by (3-2) and then you get 3^4096-2^4096 for that. so the numerator is 3^4096, the answer = 1</p>
<p>@deferred: Yes, I missed the n.
@funsummer: That makes sense. (The answer sheet mentions it 3^2048 though.)</p>
<p>another question guys!
if r > 0. and r^t = 2.56r^ t+2
What is r?</p>
<p>its supposed to be 5/8 but cant figure it out. thanks in advance!</p>
<p>r^t = 2.56 r^t r^2
=> r^2 = 1/2.56 = 25/64
So, r = sqrt(25/64) = 5/8.</p>
<p>bump this cus it’s an awesome thread</p>
<p>A random one:
The number 3^32 - 1 has exactly two divisors which are larger than 75 but smaller than 85. What is the product of these two divisors?</p>
<p>3^32 - 1 = (3^16 - 1)(3^16 + 1) = … = (3 - 1)(3 + 1)(3^2 + 1)(3^3 + 1)…(3^16 + 1)</p>
<p>So right away we know that 3^4 + 1 = 82 is a divisor.</p>
<p>And apparently 3^9 + 1 is divisible by 76, so 76 is a factor. 76*82 = 6232</p>
<p>wait. where did the 3^9+1 come from?</p>
<p>Long time no problem! Here’s one. (:
Are there infinitely many primes p such that (p+2) is also a prime?</p>