1.1 Basic Whole Number Arithmetic

1.1 Basic Whole Number Arithmetic

Mastering whole number arithmetic is the foundation for success on the College Board ACCUPLACER Arithmetic test. Whole numbers are non-negative numbers that do not include fractions or decimals: $\mathbb{W} = \{0, 1, 2, 3, 4, \dots\}$. On the ACCUPLACER exam, you will encounter computational items involving place value determination, multi-digit addition with carrying, subtraction requiring regrouping across consecutive zeros, multi-digit multiplication using partial products, and long division with remainders. Because calculators are prohibited or strictly restricted on basic arithmetic portions of placement tests, developing speed, accuracy, and algorithmic fluency is essential.

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The Base-10 Place Value System

The modern number system is a positional base-10 system. The value of any digit depends entirely on its position relative to the ones place. Moving left across a whole number, each position represents a power of $10$ that is ten times greater than the position immediately to its right.

Standard Form vs. Expanded Form

A number written in standard form uses digits ($0$ through $9$) in positional columns: $4,738,215$. In expanded form, the same number expresses the explicit sum of each digit multiplied by its place value: $4,738,215 = (4 \times 1,000,000) + (7 \times 100,000) + (3 \times 10,000) + (8 \times 1,000) + (2 \times 100) + (1 \times 10) + (5 \times 1)$

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Multi-Digit Addition with Carrying (Regrouping)

When adding multi-digit whole numbers, align numbers vertically by place value column (ones under ones, tens under tens). Add digits column-by-column starting from the extreme right (ones place). If the sum of a column is $10$ or greater, write down the ones digit in the sum row and carry (regroup) the tens digit to the next higher place value column to the left.

Worked Example: Multi-Digit Addition

Problem: Compute $48,795 + 63,428$.

Step-by-Step Vertical Addition:

1. Ones Column: $5 + 8 = 13$. Write $3$ under ones, carry $1$ to tens.
2. Tens Column: $1 \text{ (carried)} + 9 + 2 = 12$. Write $2$ under tens, carry $1$ to hundreds.
3. Hundreds Column: $1 \text{ (carried)} + 7 + 4 = 12$. Write $2$ under hundreds, carry $1$ to thousands.
4. Thousands Column: $1 \text{ (carried)} + 8 + 3 = 12$. Write $2$ under thousands, carry $1$ to ten-thousands.
5. Ten-Thousands Column: $1 \text{ (carried)} + 4 + 6 = 11$. Write $11$ across ten-thousands and hundred-thousands.

$$\begin{array}{rccccc} & 1 & 1 & 1 & 1 & \\ & 4 & 8 & 7 & 9 & 5 \\ + & 6 & 3 & 4 & 2 & 8 \\ \hline \mathbf{1} & \mathbf{1} & \mathbf{2} & \mathbf{2} & \mathbf{2} & \mathbf{3} \end{array}$$

Final Sum: $112,223$.

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Multi-Digit Subtraction Across Zeros

Subtraction requires finding the difference between a minuend (top number) and a subtrahend (bottom number). When a digit in the minuend is smaller than the corresponding digit in the subtrahend, you must regroup (borrow) $1$ unit from the adjacent higher place value column to the left, which adds $10$ to the current digit.

When borrowing across one or more zero digits, you must move leftward until you find a non-zero digit, decrement that digit by $1$, convert intermediate zeros to $9$, and add $10$ to the target column.

Worked Example: Subtraction Across Zeros

Problem: Compute $50,004 - 18,367$.

Step-by-Step Borrowing Mechanics:

1. Ones Column: $4 < 7$. We must borrow. Tens, hundreds, and thousands columns are all $0$.
2. Regrouping across zeros: Move to the ten-thousands column ($5$). Decrement $5$ to $4$.
3. The thousands zero becomes $9$, the hundreds zero becomes $9$, and the tens zero becomes $9$.
4. The ones digit $4$ gains $10$, becoming $14$.
5. Column Computations:
   • Ones: $14 - 7 = \mathbf{7}$
   • Tens: $9 - 6 = \mathbf{3}$
   • Hundreds: $9 - 3 = \mathbf{6}$
   • Thousands: $9 - 8 = \mathbf{1}$
   • Ten-Thousands: $4 - 1 = \mathbf{3}$

$$\begin{array}{rccccc} & 4 & 9 & 9 & 9 & 14 \\ & \not{5} & \not{0} & \not{0} & \not{0} & \not{4} \\ - & 1 & 8 & 3 & 6 & 7 \\ \hline & \mathbf{3} & \mathbf{1} & \mathbf{6} & \mathbf{3} & \mathbf{7} \end{array}$$

Final Difference: $31,637$. (Verify by addition: $31,637 + 18,367 = 50,004$).

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Multi-Digit Multiplication Algorithms

Multiplication represents repeated addition. To multiply multi-digit numbers using the standard algorithm:

1. Multiply the top number by the ones digit of the bottom number to obtain the first partial product.
2. Multiply the top number by the tens digit of the bottom number, inserting a placeholder zero in the ones column of the second partial product.
3. Repeat for higher place values, adding an additional placeholder zero for each subsequent shift left.
4. Add all partial products to get the final product.

Worked Example: Multi-Digit Multiplication

Problem: Compute $437 \times 68$.

$$\begin{array}{rccccc} & & & 4 & 3 & 7 \\ \times & & & & 6 & 8 \\ \hline & & 3 & 4 & 9 & 6 \quad \text{(First Partial Product: } 437 \times 8\text{)} \\ + & 2 & 6 & 2 & 2 & 0 \quad \text{(Second Partial Product: } 437 \times 60\text{)} \\ \hline & \mathbf{2} & \mathbf{9} & \mathbf{7} & \mathbf{1} & \mathbf{6} \end{array}$$

• Partial Product 1: $437 \times 8 = (400 \times 8) + (30 \times 8) + (7 \times 8) = 3,200 + 240 + 56 = 3,496$.
• Partial Product 2: $437 \times 60 = 26,220$ (starts with placeholder $0$).
• Sum of Partial Products: $3,496 + 26,220 = 29,716$.

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Long Division Algorithm with Remainders

Long division partitions a dividend into equal groups specified by the divisor. Use the standard DMSB sequence: Divide, Multiply, Subtract, Bring Down.

Worked Example: Long Division

Problem: Divide $8,495$ by $23$.

• Quotient: $369$
• Remainder ($R$): $8$

Verification Formula: $\text{Dividend} = (\text{Divisor} \times \text{Quotient}) + \text{Remainder}$ $8,495 = (23 \times 369) + 8 = 8,487 + 8 = 8,495 \quad \checkmark$

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ACCUPLACER Exam Traps & Common Errors