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1.1 Basic Whole Number Arithmetic

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ACCUPLACER Arithmetic Exam

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1 Chapter 1: Whole Number Operations

1.1 Basic Whole Number Arithmetic 1.2 Order of Operations & Grouping Symbols 1.3 Number Properties, Factors & Prime Factorization 1.4 Whole Number Estimation, Rounding & Applied Word Problems

2 Chapter 2: Fraction Operations

2.1 Fraction Fundamentals & Simplifying 2.2 Addition & Subtraction of Fractions and Mixed Numbers 2.3 Multiplication & Division of Fractions and Mixed Numbers 2.4 Real-World Fraction Applications & Word Problems

3 Chapter 3: Decimal Operations

3.1 Decimal Place Value, Rounding & Comparing Decimals 3.2 Addition & Subtraction of Decimals 3.3 Multiplication & Division of Decimals 3.4 Decimal Word Problems & Financial Calculations

4 Chapter 4: Percent

4.1 Percent Concepts & Conversions 4.2 Percent of a Quantity & Percent Equations 4.3 Percent Increase, Decrease & Percent Change 4.4 Practical Percent Applications

5 Chapter 5: Number Comparisons & Equivalents

5.1 Ordering Mixed Numerical Forms 5.2 Equivalent Numerical Representations & Conversion Strategies 5.3 Ratios, Unit Rates & Proportional Reasoning

1.1 Basic Whole Number Arithmetic

Key Takeaways Explain This Topic AI

The Base-10 place value system increases by a power of 10 for each position moving left from the ones place up to millions.

Column addition requires carrying (regrouping) tens values to the adjacent left place value whenever a column sum equals 10 or greater.

Subtraction across zeros requires regrouping from the nearest non-zero place value on the left, turning intermediate zeros into 9s.

Multi-digit multiplication relies on partial products, requiring a placeholder zero for each place value shift leftward.

The standard long division algorithm (DMSB) calculates quotients and remainders, satisfying Dividend = (Divisor × Quotient) + Remainder.

Last updated: August 2026

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: W = { 0 , 1 , 2 , 3 , 4 , … } \mathbb{W} = \{0, 1, 2, 3, 4, \dots\} W = { 0 , 1 , 2 , 3 , 4 , … } . 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.

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 10 10 that is ten times greater than the position immediately to its right.

Place Value Position Exponential Notation Value in Base-10 Example Digit Position ( 4 , 738 , 215 4,738,215 4 , 738 , 215 )

  • Millions 10 6 10^6 1 0 6 1 , 000 , 000 1,000,000 1 , 000 , 000 4 4 4
  • Hundred Thousands 10 5 10^5 1 0 5 100 , 000 100,000 100 , 000 7 7 7
  • Ten Thousands 10 4 10^4 1 0 4 10 , 000 10,000 10 , 000 3 3 3
  • Thousands 10 3 10^3 1 0 3 1 , 000 1,000 1 , 000 8 8 8
  • Hundreds 10 2 10^2 1 0 2 100 100 100 2 2 2
  • Tens 10 1 10^1 1 0 1 10 10 10 1 1 1
  • Ones 10 0 10^0 1 0 0 1 1 1 5 5 5
  • Standard Form vs. Expanded Form

A number written in standard form uses digits ( 0 0 0 through 9 9 9 ) in positional columns: 4 , 738 , 215 4,738,215 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 × 1 , 000 , 000 ) + ( 7 × 100 , 000 ) + ( 3 × 10 , 000 ) + ( 8 × 1 , 000 ) + ( 2 × 100 ) + ( 1 × 10 ) + ( 5 × 1 ) 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) 4 , 738 , 215 = ( 4 × 1 , 000 , 000 ) + ( 7 × 100 , 000 ) + ( 3 × 10 , 000 ) + ( 8 × 1 , 000 ) + ( 2 × 100 ) + ( 1 × 10 ) + ( 5 × 1 )

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 10 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 48,795 + 63,428 48 , 795 + 63 , 428 .

Step-by-Step Vertical Addition:

Ones Column: 5 + 8 = 13 5 + 8 = 13 5 + 8 = 13 . Write 3 3 3 under ones, carry 1 1 1 to tens.

Tens Column: 1 (carried) + 9 + 2 = 12 1 \text{ (carried)} + 9 + 2 = 12 1 (carried) + 9 + 2 = 12 . Write 2 2 2 under tens, carry 1 1 1 to hundreds.

Hundreds Column: 1 (carried) + 7 + 4 = 12 1 \text{ (carried)} + 7 + 4 = 12 1 (carried) + 7 + 4 = 12 . Write 2 2 2 under hundreds, carry 1 1 1 to thousands.

Thousands Column: 1 (carried) + 8 + 3 = 12 1 \text{ (carried)} + 8 + 3 = 12 1 (carried) + 8 + 3 = 12 . Write 2 2 2 under thousands, carry 1 1 1 to ten-thousands.

Ten-Thousands Column: 1 (carried) + 4 + 6 = 11 1 \text{ (carried)} + 4 + 6 = 11 1 (carried) + 4 + 6 = 11 . Write 11 11 11 across ten-thousands and hundred-thousands. 1 1 1 1 4 8 7 9 5 + 6 3 4 2 8 1 1 2 2 2 3 \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} + 1 ​ 1 4 6 1 ​ 1 8 3 2 ​ 1 7 4 2 ​ 1 9 2 2 ​ 5 8 3 ​ ​

Final Sum: 112 , 223 112,223 112 , 223 .

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 1 1 unit from the adjacent higher place value column to the left, which adds 10 10 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 1 1 , convert intermediate zeros to 9 9 9 , and add 10 10 10 to the target column.

Worked Example: Subtraction Across Zeros

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

Step-by-Step Borrowing Mechanics:

Ones Column: 4 < 7 4 < 7 4 < 7 . We must borrow. Tens, hundreds, and thousands columns are all 0 0 0 .

Regrouping across zeros: Move to the ten-thousands column ( 5 5 5 ). Decrement 5 5 5 to 4 4 4 .

The thousands zero becomes 9 9 9 , the hundreds zero becomes 9 9 9 , and the tens zero becomes 9 9 9 .

The ones digit 4 4 4 gains 10 10 10 , becoming 14 14 14 .

Column Computations:

  • Ones: 14 − 7 = 7 14 - 7 = \mathbf{7} 14 − 7 = 7
  • Tens: 9 − 6 = 3 9 - 6 = \mathbf{3} 9 − 6 = 3
  • Hundreds: 9 − 3 = 6 9 - 3 = \mathbf{6} 9 − 3 = 6
  • Thousands: 9 − 8 = 1 9 - 8 = \mathbf{1} 9 − 8 = 1

Ten-Thousands: 4 − 1 = 3 4 - 1 = \mathbf{3} 4 − 1 = 3 4 9 9 9 14 5̸ 0̸ 0̸ 0̸ 4̸ − 1 8 3 6 7 3 1 6 3 7 \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} − ​ 4  5 1 3 ​ 9  0 8 1 ​ 9  0 3 6 ​ 9  0 6 3 ​ 14  4 7 7 ​ ​

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

Multi-Digit Multiplication Algorithms

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

Multiply the top number by the ones digit of the bottom number to obtain the first partial product .

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 .

Repeat for higher place values, adding an additional placeholder zero for each subsequent shift left.

Add all partial products to get the final product.

Worked Example: Multi-Digit Multiplication

Problem: Compute 437 × 68 437 \times 68 437 × 68 . 4 3 7 × 6 8 3 4 9 6 (First Partial Product: 437 × 8 ) + 2 6 2 2 0 (Second Partial Product: 437 × 60 ) 2 9 7 1 6 \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} × + ​ 2 2 ​ 3 6 9 ​ 4 4 2 7 ​ 3 6 9 2 1 ​ 7 8 6 (First Partial Product: 437 × 8 ) 0 (Second Partial Product: 437 × 60 ) 6 ​ ​

Partial Product 1: 437 × 8 = ( 400 × 8 ) + ( 30 × 8 ) + ( 7 × 8 ) = 3 , 200 + 240 + 56 = 3 , 496 437 \times 8 = (400 \times 8) + (30 \times 8) + (7 \times 8) = 3,200 + 240 + 56 = 3,496 437 × 8 = ( 400 × 8 ) + ( 30 × 8 ) + ( 7 × 8 ) = 3 , 200 + 240 + 56 = 3 , 496 .

Partial Product 2: 437 × 60 = 26 , 220 437 \times 60 = 26,220 437 × 60 = 26 , 220 (starts with placeholder 0 0 0 ).

Sum of Partial Products: 3 , 496 + 26 , 220 = 29 , 716 3,496 + 26,220 = 29,716 3 , 496 + 26 , 220 = 29 , 716 .

Long Division Algorithm with Remainders

Long division partitions a dividend into equal groups specified by the divisor. Use the standard DMSB sequence: D ivide, M ultiply, S ubtract, B ring Down.

Worked Example: Long Division

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

Step Operation Working Calculations Intermediate Quotient Digit

1. Hundreds 84 ÷ 23 84 \div 23 84 ÷ 23 23 × 3 = 69 23 \times 3 = 69 23 × 3 = 69 ; 84 − 69 = 15 84 - 69 = 15 84 − 69 = 15 3 3 3 in hundreds place

2. Tens Bring down 9 → 159 9 \to 159 9 → 159 ; 159 ÷ 23 159 \div 23 159 ÷ 23 23 × 6 = 138 23 \times 6 = 138 23 × 6 = 138 ; 159 − 138 = 21 159 - 138 = 21 159 − 138 = 21 6 6 6 in tens place

3. Ones Bring down 5 → 215 5 \to 215 5 → 215 ; 215 ÷ 23 215 \div 23 215 ÷ 23 23 × 9 = 207 23 \times 9 = 207 23 × 9 = 207 ; 215 − 207 = 8 215 - 207 = 8 215 − 207 = 8 9 9 9 in ones place

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

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

ACCUPLACER Exam Traps & Common Errors

Caution

Trap 1: Misaligning Digits by Place Value When adding or subtracting numbers with unequal numbers of digits (e.g., 12 , 450 + 385 12,450 + 385 12 , 450 + 385 ), never align numbers to the left! Always align right-justified along the ones column.

Warning

Trap 2: Forgetting Placeholder Zeros in Partial Products When multiplying by a tens digit, failing to write down a zero ( 0 0 0 ) in the ones column shifts your partial product by a factor of 10, causing massive calculation errors.

Note

Trap 3: Improper Regrouping Across Zeros A frequent mistake in subtraction like 700 − 243 700 - 243 700 − 243 is turning all zeros into 10 10 10 . Remember: only the final target ones place becomes 10 10 10 ; intermediate zeros turn into 9 9 9 because 1 1 1 ten was borrowed to supply the ones column.