Mrs. Sandy McAllister will have to save a tremendous sum out of her housekeeping allowance if she is to win that sixth present that her canny husband promised her. And the allowance must be a very liberal one if it is to admit of such savings. The problem required that we should find five numbers higher than 36 the units of which may be displayed so as to form a square, a triangle, two triangles, and three triangles, using the complete number in every one of the four cases. Every triangular number is such that if we multiply it by 8 and add 1 the result is an odd square number. For example, multiply 1, 3, 6, 10, 15 respectively by 8 and add 1, and we get 9, 25, 49, 81, 121, which are the squares of the odd numbers 3, 5, 7, 9, 11. Therefore in every case where 8x squared + 1 = a square number, x squared is also a triangular. This point is dealt with in our puzzle, "The Battle of Hastings." I will now merely show again how, when the first solution is found, the others may be discovered without any difficulty. First of all, here are the figures:-- 8 x 1 squared + 1 = 3 squared 8 x 6 squared + 1 = 17 squared 8 x 35 squared + 1 = 99 squared 8 x 204 squared + 1 = 577 squared 8 x 1189 squared + 1 = 3363 squared 8 x 6930 squared + 1 = 19601 squared 8 x 40391 squared + 1 = 114243 squared The successive pairs of numbers are found in this way:-- (1 x 3) + (3 x 1) = 6 (8 x 1) + (3 x 3) = 17 (1 x 17) + (3 x 6) = 35 (8 x 6) + (3 x 17) = 99 (1 x 99) + (3 x 35) = 204 (8 x 35) + (3 x 99) = 577 and so on. Look for the numbers in the table above, and the method will explain itself. Thus we find that the numbers 36, 1225, 41616, 1413721, 48024900, and 1631432881 will form squares with sides of 6, 35, 204, 1189, 6930, and 40391; and they will also form single triangles with sides of 8, 49, 288, 1681, 9800, and 57121. These numbers may be obtained from the last column in the first table above in this way: simply divide the numbers by 2 and reject the remainder. Thus the integral halves of 17, 99, and 577 are 8, 49, and 288. All the numbers we have found will form either two or three triangles at will. The following little diagram will show you graphically at a glance that every square number must necessarily be the sum of two triangulars, and that the side of one triangle will be the same as the side of the corresponding square, while the other will be just 1 less. [Illustration +-----------+ +---------+ |. . . . ./.| |. . . ./.| |. . . ./. .| |. . ./. .| |. . ./. . .| |. ./. . .| |. ./. . . .| |./. . . .| |./. . . . .| /. . . . .| /. . . . . .| +---------+ +-----------+ ]