fun math puzzles

A:x=lbs of turkey y=lbs of ahm X=times (mathematical funtion) 4x+6y<$72 x=3 12+6y<72, so y<60 y=2, so 2<60 voilá!

A:4 *4 = 16 inversed 61 5 * 5 = 25 inversed 52 6 * 6 = 36 inversed 63 7 * 7 = 49 inversed 94 8 * 8 = 64 inversed 46 9 * 9 = 81 inversed 18

A:jim_thompson5910's answer is right on. The route from any point on the globe to a point diametrically opposite can go through any other point on the globe without having to go 'out of the way' or cover any extra distance. For the really technically-minded puzzle freaks out there, you can go a little deeper into this if you want. Here's my take on it: If the Earth were a perfect sphere, the most you could say is that the man was heading to a point diametrically opposite his current location. Between two such points it is possible to draw a 'great circle' (shortest distance) route that passes through any other point on the globe, i.e. anywhere the young girl was going. However, the Earth isn't a perfect sphere. The Earth's rotation causes it to bulge slightly at the equator; that means the equatorial circumference is 42 miles greater than the meridional (passing through the poles) circumference. To put that in practical terms, if you follow the equator around the earth, you travel 42 miles further than if you go up and over the north pole and then down around the south pole and back to your starting point. Imagine that the man and the girl were starting out somewhere near the equator (say, somewhere in Brazil) and the man was traveling to an opposite point also near the equator (somewhere in Indonesia) - then his shortest route would be over either the north or south pole, rather than along the equator. He'd be saving 42 miles by taking the polar route. If the girl wanted to go somewhere near the equator like Equatorial Guinea, the route along the equator would be about 21 miles longer than the route over the pole (remember - 42 miles longer to go all the way around the equator, so half that, or 21 miles, to go halfway around), so the man's claim about not going more than a few miles out of his way would be a bit of an exaggeration. If we assume the man was telling the truth about not going more than 'a few miles' out of his way, then he must not be so close to the equator that the equatorial bulge would make a difference. This suggests that the two were somewhere nearer one of the poles than the equator, since a pole-to-pole route would not have much variance in its length regardless of where they might be stopping on the way. Their precise location is still difficult to determine, especially without knowing a precise value for 'a few miles', but at least we've narrowed it down a little. Details about the Earth's equatorial bulge were obtained from wikipedia; link below.

A:N(n, i) stands for the number of beans of boy i after run n. We want to know N(0, i), based on the information N(7, i) = 128. If i<>n, boy i did not lose run n, so in that case his amount was doubled, and [1] ... N(n, i) = 2 * N(n-1, i) ... (i =/= n) If i = n, boy i did lose run n, so he had to double the amounts of the others. The total number of beans in the game is 7*128 = 896. If the losing boy had X beans before he lost, the others had 896 - X beans together. Their amounts are doubled, so afterward they have 1792 - 2 X beans, leaving 896 - (1792 - 2 X) = 2 X - 896 beans for the losing boy. Therefore, [2] ... N(n, n) = 2 * N(n-1, n) - 896 Formulas 1 and 2 combined show that every boy doubled his amount of beans, except that in run n, 896 beans were subtracted. In subsequent runs, the term -896 is also doubled. For boy i that doubling takes place 7-i times. Therefore, [3] N(7, i) = 2^7 * N(0,i) - 2^(7-i) * 896 This means that [4] 128 = 128 * N(0,i) - 2^(7-i) * 896 [5] 128 = 128 * (N(0,i) - 896 / 2^i) [6] 1 = N(0,i) - 896 / 2^i [7] N(0,i) = 896 / 2^i + 1 We find Jon: N(0,1) = 896 / 2 + 1 = 449 Tim: N(0,2) = 896 / 4 + 1 = 225 Tom: N(0,3) = 896 / 8 + 1 = 113 Bob: N(0,4) = 896 / 16 + 1 = 57 Bill: N(0,5) = 896 / 32 + 1 = 29 Ben: N(0,6) = 896 / 64 + 1 = 15 Will: N(0,7) = 896 / 128 + 1 = 7 =============================== In general, if there are N boys and they all end up with 2^N beans, their original numbers would be N(0, i) = N * 2^(N-i) + 1 That is, the last boy would start with N+1 beans, the second last boy with one more than double that amount, etc.

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