part, and find that 400, the square of 20, subtracted from 2025, gives 1625, the first remainder. I again choose 20, whose square, together with twice itself, multiplied by the preceding part, is 20 × 20 + 2 × 20 × 20, or 1200; which subtracted from 1625, the first remainder, gives 425, the second remainder. I choose 7 for the third part, which appears to be too great, since 7 × 7, increased by 2 × 7 multiplied by the sum of the preceding parts 20 + 20, gives 609, which is more than 425. I therefore choose 5, which closes the process, since 5 × 5, together with 2 × 5 multiplied by 20 + 20, gives exactly 425. The square root of 2025 is therefore 20 + 20 + 5, or 45, which will be found, by trial, to be correct; since 45 × 45 = 2025. Again, I ask if 13340 has, or has not, a square root. Let 100 be the first part, whose square is 10000, and the first remainder is 3340. Let 10 be the second part. Here 10 × 10 + 2 × 10 × 100 is 2100, and the second remainder, or 3340-2100, is 1240. Let 5 be the third part; then 5 × 5 + 2 × 5 × (100 + 10) is 1125, which, subtracted from 1240, leaves 115. There is, then, no square root; for a single additional unit will give a subtraction of 1 × 1 + 2 × 1 × (100 + 10 + 5), or 231, which is greater than 115. But if the number proposed had been less by 115, each of the remainders would have been 115 less, and the last remainder would have been nothing. Therefore 13340-115, or 13225, has the square root 100 + 10 + 5, or 115; and the answer is, that 13340 has no square root, and that 13225 is the next number below it which has one, namely,
- 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,