Question

Determine the sum using method 2 (regroup to next higher place value) $$\begin{array}{r} 159 \\\ +\quad 27 \\\ \hline \end{array}$$ $$\begin{array}{r} 924 \\\ +138 \\\ \hline \end{array}$$ $$\begin{array}{r} 51 \\\ 382 \\\ +\quad 77 \\\ \hline \end{array}$$

Short answer

1. $$159 + 27$$ 2. $$924 + 138$$ 3. $$51 + 382 + 77$$ Answer: 1. The sum for the first problem is $$186$$. 2. The sum for the second problem is $$1062$$. 3. The sum for the third problem is $$510$$.

Step-by-step solution

Problem 1

Add the digits in the ones place: $$9 + 7 = 16$$ Since 16 has two place values (tens and ones), write the 6 in the ones place and carry over the 1 to the tens place. $$\begin{array}{@{}c@{\;}c@{}c} & & \cancelto{1}{\stackrel{1}{}}\\ & 1 & 5 & 9 \\ & & + & 2 & 7\\ \cline{1-4} & & & & \\ \end{array}$$ Next, add the digits in the tens place, as well as the carried over 1: $$5 + 2 + 1 = 8$$ Write the 8 in the tens place. $$\begin{array}{@{}c@{}c@{}c} & & & & \\ & 1 & 5 & 9 \\ & & + & 2 & 7\\ \cline{1-4} & & 1 & 8 & 6\\ \end{array}$$ Finally, bring down the digit in the hundreds place since there is nothing to add: $$\begin{array}{@{}]c@{}c@{}c} & & & & \\ & 1 & 5 & 9 \\ & & + & 2 & 7\\ \cline{1-4} & 1 & 8 & 6\\ \end{array}$$ So the sum of the first problem is $$186.$$

Problem 2

Add the digits in the ones place: $$4 + 8 = 12$$ Since 12 has two place values, write the 2 in the ones place and carry over the 1 to the tens place. $$\begin{array}{@{}c@{}c@{}c} & & & & \\ & 9 & 2 & 4 \\ & & + & 1 & 3 & 8\\ \cline{1-5} & & & & & \\ \end{array}$$ Next, add the digits in the tens place, as well as the carried over 1: $$2 + 3 + 1 = 6$$ Write the 6 in the tens place. $$\begin{array}{@{}c@{}c@{}c} & & & & \\ & 9 & 2 & 4 \\ & & + & 1 & 3 & 8\\ \cline{1-5} & & & 1 & 6 & 2\\ \end{array}$$ Finally, add the digits in the hundreds place: $$9 + 1 = 10$$ Write the 10 in the hundreds and thousands places. $$\begin{array}{@{}c@{}c@{}c} & & & & \\ & 9 & 2 & 4 \\ & & + & 1 & 3 & 8\\ \cline{1-5} & 1 & 0 & 6 & 2\\ \end{array}$$ So the sum of the second problem is $$1062.$$

Problem 3

Add the digits in the ones place: $$1 + 2 + 7 = 10$$ Since 10 has two place values, write the 0 in the ones place and carry over the 1 to the tens place. $$\begin{array}{@{}c@{\;}c@{}c} & & \cancelto{1}{\stackrel{1}{}}\\ & 5 & 1\\ & 3 & 8 & 2\\ & & + & 7 & 7\\ \cline{1-4} & & & & \\ \end{array}$$ Next, add the digits in the tens place, as well as the carried over 1: $$5 + 8 + 7 + 1 = 21$$ Since 21 has two place values, write the 1 in the tens place and carry over the 2 to the hundreds place. $$\begin{array}{@{}c@{}c@{}c} & \cancelto{2}{\stackrel{2}{}}& & & \\ & 5 & 1\\ & 3 & 8 & 2\\ & & + & 7 & 7\\ \cline{1-4} & & & 1 & 0\\ \end{array}$$ Finally, add the 2 carried over to the hundreds place: $$2 + 3 = 5$$ Write the 5 in the hundreds place. $$\begin{array}{@{}c@{}c@{}c} & & & & \\ & 5 & 1\\ & 3 & 8 & 2\\ & & + & 7 & 7\\ \cline{1-4} & 5 & 1 & 0\\ \end{array}$$ So the sum of the third problem is $$510.$$