At the end of seventeen days the snail will have climbed 17 ft., and at the end of its eighteenth day-time task it will be at the top. It instantly begins slipping while sleeping, and will be 2 ft. down the other side at the end of the eighteenth day of twenty-four hours. How long will it take over the remaining 18 ft.? If it slips 2 ft. at night it clearly overcomes the tendency to slip 2 ft. during the daytime, in climbing up. In rowing up a river we have the stream against us, but in coming down it is with us and helps us. If the snail can climb 3 ft. and overcome the tendency to slip 2 ft. in twelve hours' ascent, it could with the same exertion crawl 5 ft. a day on the level. Therefore, in going down, the same exertion carries it 7 ft. in twelve hours--that is, 5 ft. by personal exertion and 2 ft. by slip. This, with the night slip, gives it a descending progress of 9 ft. in the twenty-four hours. It can, therefore, do the remaining 18 ft. in exactly two days, and the whole journey, up and down, will take it exactly twenty days. 107.--_The Four Princes._ When Montucla, in his edition of Ozanam's _Recreations in Mathematics_, declared that "No more than three right-angled triangles, equal to each other, can be found in whole numbers, but we may find as many as we choose in fractions," he curiously overlooked the obvious fact that if you give all your sides a common denominator and then cancel that denominator you have the required answer in integers! Every reader should know that if we take any two numbers, _m_ and _n_, then _m_^2 + _n_^2, _m_^2 - _n_^2, and _2mn_ will be the three sides of a rational right-angled triangle. Here _m_ and _n_ are called generating numbers. To form three such triangles of equal area, we use the following simple formula, where _m_ is the greater number:-- _mn_ + _m_^2 + _n_^2 = _a_ _m_^2 - _n_^2 = _b_ 2_mn_ + _n_^2 = _c_ Now, if we form three triangles from the following pairs of generators, _a_ and _b_, _a_ and _c_, _a_ and _b_ + _c_, they will all be of equal area. This is the little problem respecting which Lewis Carroll says in his diary (see his _Life and Letters_ by Collingwood, p. 343), "Sat up last night till 4 a.m., over a tempting problem, sent me from New York, 'to find three equal rational-sided right-angled triangles.' I found two ... but could not find three!" The following is a subtle formula by means of which we may always find a R.A.T. equal in area to any given R.A.T. Let _z_ = hypotenuse, _b_ = base, _h_ = height, _a_ = area of the given triangle; then all we have to do is to form a R.A.T. from the generators _z_^2 and 4_a_, and give each side the denominator 2_z_ (_b_^2 - _h_^2), and we get the required answer in fractions. If we multiply all three sides of the original triangle by the denominator, we shall get at once a solution in whole numbers. The answer to our puzzle in smallest possible numbers is as follows:--