161. It only remains to put the rule in such a shape as will guide us to those parts which it is most convenient to choose. It is evident (57) that any number which terminates with ciphers, as 4000, has double the number of ciphers in its square. Thus, 4000 × 4000 = 16000000; therefore, any square number,[23] as 49, with an even number of ciphers annexed, as 490000, is a square number. The root[24] of 490000 is 700. This being premised, take any number, for example, 76176; setting out from the right hand towards the left, cut off two figures; then two more, and so on, until one or two figures only are left: thus, 7,61,76. This number is greater than 7,00,00, of which the first figure is not a square number, the nearest square below it being 4. Hence, 4,00,00 is the nearest square number below 7,00,00, which has four ciphers, and its square root is 200. Let this be the first part chosen: its square subtracted from 76176 leaves 36176, the first remainder; and it is evident that we have obtained the highest number of the highest denomination which is to be found in the square root of 76176; for 300 is too great, its square, 9,00,00, being greater than 76176: and any denomination higher than hundreds has a square still greater. It remains, then, to choose a second part, as in the examples of (160), with the remainder 36176. This part cannot be as great as 100, by what has just been said; its highest denomination is therefore a number of tens. Let N stand for a number of tens, which is one of the simple numbers 1, 2, 3, &c.; that is, let the new part be 10N, whose square is 10N × 10N, or 100NN, and whose double multiplied by the former part is 20N × 200, or 4000N; the two together are 4000N + 100NN. Now, N must be so taken that this may not be greater than 36176: still more 4000N must not be greater than 36176. We may therefore try, for N, the number of times which 36176 contains 4000, or that which 36 contains 4. The remark in (80) applies here. Let us try 9 tens or 90. Then, 2 × 90 × 200 + 90 × 90, or 44100, is to be subtracted, which is too great, since the whole remainder is 36176. We then try 8 tens or 80, which gives 2 × 80 × 200 + 80 × 80, or 38400, which is likewise too great. On trying 7 tens, or 70, we find 2 × 70 × 200 + 70 × 70, or 32900, which subtracted from 36176 gives 3276, the second remainder. The rest of the square root can only be units. As before, let N be this number of units. Then, the sum of the preceding parts being 200 + 70, or 270, the number to be subtracted is 270 × 2N + NN, or 540N + NN. Hence, as before, 540N must be less than 3276, or N must not be greater than the number of times which 3276 contains 540, or (80) which 327 contains 54. We therefore try if 6 will do, which gives 2 × 6 × 270 + 6 × 6, or 3276, to be subtracted. This being exactly the second remainder, the third remainder is nothing, and the process is finished. The square root required is therefore 200 + 70 + 6, or 276.