Square root of 5 and 6 digit numbers by long division

This lesson finds the square root of larger perfect squares (five and six digits) using the long division method. The digits are grouped in pairs from the right, and the root is built up one digit at a time. Three perfect squares are worked in full.

The grouping rule

To find a square root by long division, split the number into groups of two digits, starting from the right. A six digit number gives three groups; a five digit number gives a leftmost group of a single digit. For example, $50625$ splits as $5\,\mid\,06\,\mid\,25$.

You then work left to right: find the largest digit whose square fits the first group, subtract, bring down the next group, and at each later step double the current quotient to form the new divisor.

Example 1: $\sqrt{10816}$

Group the digits: $1\,\mid\,08\,\mid\,16$.

First group. The leftmost group is $1$. The largest square not exceeding it is $1^2 = 1$, so the first digit of the root is $1$. Subtracting leaves $0$, and we bring down $08$ to get $8$.

Second group. Double the quotient $1$ to get $2$ as the start of the divisor. We need a digit $d$ with $\overline{2d} \times d \le 8$. Even $21 \times 1 = 21$ is too big, so this group contributes a $0$ to the root. The quotient is now $10$, and we bring down $16$ to get $816$.

Third group. Double the quotient $10$ to get $20$. We need a digit $d$ with $\overline{20d}\times d = 816$. Since the number ends in $6$, the last digit of the root is $4$ or $6$; testing $204 \times 4 = 816$ works exactly. So the final digit is $4$.

$\sqrt{10816} = 104$

Check: $104^2 = 10816$.

Example 2: $\sqrt{50625}$

Group the digits: $5\,\mid\,06\,\mid\,25$.

First group. The leftmost group is $5$. The largest square not exceeding $5$ is $2^2 = 4$, so the first digit is $2$. Subtracting gives $1$, and bringing down $06$ gives $106$.

Second group. Double the quotient $2$ to get $4$. We need a digit $d$ with $\overline{4d}\times d \le 106$. Testing $42 \times 2 = 84 \le 106$, while $43 \times 3 = 129$ is too big, so the next digit is $2$. Subtracting $84$ from $106$ leaves $22$, and bringing down $25$ gives $2225$.

Third group. Double the quotient $22$ to get $44$. The number ends in $5$, so the final digit of the root must be $5$. Testing $445 \times 5 = 2225$ works exactly.

$\sqrt{50625} = 225$

Check: $225^2 = 50625$.

Example 3: $\sqrt{143641}$

Group the digits: $14\,\mid\,36\,\mid\,41$.

First group. The leftmost group is $14$. The largest square not exceeding $14$ is $3^2 = 9$, so the first digit is $3$. Subtracting gives $5$, and bringing down $36$ gives $536$.

Second group. Double the quotient $3$ to get $6$. We need a digit $d$ with $\overline{6d}\times d \le 536$. Testing $68 \times 8 = 544$ is too big, so we step back to $67 \times 7 = 469 \le 536$. The next digit is $7$. Subtracting $469$ from $536$ leaves $67$, and bringing down $41$ gives $6741$.

Third group. Double the quotient $37$ to get $74$. The number ends in $1$, so the last digit of the root is $1$ or $9$. Testing $749 \times 9 = 6741$ works exactly, so the final digit is $9$.

$\sqrt{143641} = 379$

Check: $379^2 = 143641$.

Key takeaways

• Split the number into pairs of digits from the right; a five digit number leaves a single leftmost digit as its first group.
• At each later step, double the current quotient and append a trial digit to form the divisor, choosing the largest digit that keeps the product within the running remainder.
• If no nonzero digit fits a group, that group contributes a $0$ to the root.
• The units digit of a perfect square narrows down the final digit of its square root.