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I can't figure out how to set up this problem...can anyone help me? A thief steals a number of rare plants from a nursery. On the way out, the thief meets 3 security guards, one after another. To each security guard, the thief is forced to give one-half the plants that he still has, plus 2 more. Finally, the thief leaves the nursery with 1 lone palm. How many plants were originally stolen?

Mathematics
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At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

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OK N = the number of original plants. Let w,x,y,z equal the thief and security guard number of plants in each's possession. The theif has 1 plant so w=1. Now our equation is: 1+ x +y + z = N Let's do the # of plants that the first guard, x, gets: 'To each security guard, the thief is forced to give one-half the plants that he still has, plus 2 more. ' So I read this as... x = N/2 + 2 Now our equation is 1 + (N/2+2) + y + z = N Create a second equation for y...(it get's uglier)... \[y=\Bigg [ \frac {N-\big ( \frac {N} {2} + 2 \big )} {2} + 2\Bigg ]\]
\[z= \bigg (\frac {N-y} {2}+2\bigg )\]
How did it go? Did you get it. Looks ugly doesn't it. BTW I worked backwards to check the answer N=36.

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