by Dudeney, Henry Ernest, 1857-1930
Available in 215 free installments
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The smallest number that can be expressed as the sum of two squares in twelve different ways is 160,225, and this is therefore the smallest army that would answer the Sultan's purpose. The number is composed of the factors 5x5x13x17x29, each of which is of the required form. If they were all different factors, there would be sixteen ways; but as one of the factors is repeated, there are just twelve ways. Here are the sides of the twelve pairs of squares: (400 and 15), (399 and 32), (393 and 76), (392 and 81), (384 and 113), (375 and 140), (360 and 175), (356 and 183), (337 and 216), (329 and 228), (311 and 252), (265 and 300). Square the two numbers in each pair, add them together, and their sum will in every case be 160,225.
137.?A STUDY IN THRIFT.?solution
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 2 + 1 = a square number, x 2 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:?
The successive pairs of numbers are found in this way:?
(1x3) + (3x1)= 6 (8x1) + (3x3) = 17
(1 x 17)+ (3x6) = 35 (8x6) + (3x17) = 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 maybe 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.
