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Chapter Review · a rapid pre-test review (speedrun)

Foundations of Arithmetic: Chapter Review

A rapid review before the test: the chapter's vocabulary and notation, every formula with the conditions to use it, the standard problem types step by step, and the traps that cost points.

Vocabulary and notation

Base 10 (the decimal system)
Naming every number with the ten digits 00 through 99, letting a digit's column decide its size. Base 1010 means bundling in tens: ten ones make one ten, ten tens one hundred.
Expanded form
A number written as the sum of its digit values: 4,572=4,000+500+70+24{,}572 = 4{,}000 + 500 + 70 + 2. A place holding 00 contributes nothing and is left out.
Placeholder zero
A 00 digit whose job is to hold a column open so the digits around it stay put. Deleting it slides everything right: 3,2053{,}205 becomes 325325.
Period
A group of three digits marked off by commas from the right: ones, thousands, millions. The commas ARE the period boundaries, which is why they fall every three digits.
Property (of an operation)
A statement holding for EVERY choice of numbers, not one computed fact. Proving one needs a general argument; disproving one needs a single failing case.
Commutative versus associative
Commute means move, so commutativity changes the ORDER of two numbers. Associate means group, so associativity changes which pair is combined first. Commutativity needs two numbers, associativity three.
Counterexample
One choice of numbers at which a claimed property fails. A single one disproves it; no number of successful examples can prove it.
Grouping symbols
Parentheses (  )(\;), brackets [  ][\;], and braces {  }\{\;\}, all meaning the same thing: treat what is inside as one quantity. The differing shapes only make matching pairs easy to spot.
Greatest common factor
The largest number dividing every term of a sum, where the terms are above zero. It is what a fully factored answer pulls out front.

Formulas and theorems

  • Value of a digit

    digit value=digit×place value\text{digit value} = \text{digit} \times \text{place value}

    Use when Any digit in any place. The digit alone is only a symbol; multiply it by the worth of the column it sits in. A 00 digit has value 00 and still holds its column.

    e.g. In 6,3086{,}308 the 33 sits in the hundreds column, so its value is 3×100=3003 \times 100 = 300.

  • The ten-times ladder of places

    ,100,000,10,000,1000,100,10,1\ldots, 100{,}000, 10{,}000, 1000, 100, 10, 1
    Every place is ten times the place to its rightFive whole-number places listed downward, ten thousands through ones, each beside its value: 10,000, 1,000, 100, 10 and 1. An upward arrow labelled times ten joins every place to the one above it, so each place is ten times the place below it in the list. The ones place is the bottom rung.ten thousands10,000thousands1,000hundreds100tens10ones1×10×10×10×10
    Text description

    A ladder of the whole-number places from ten thousands down to ones, joined by times-ten arrows, with the ones place at the bottom.

    Use when Every place is exactly 1010 times the place to its right, so a digit one column left is worth 1010 times as much, and one column right a tenth as much. The ones place is the right-hand end of the ladder for the whole numbers this chapter works in.

    e.g. The 55 in 508508 is worth 1010 times the 55 in 5858, since 500÷50=10500 \div 50 = 10.

  • How many of a place a number contains

    tens in n=n10hundreds in n=n100\begin{gathered} \text{tens in } n = \frac{n}{10} \\ \text{hundreds in } n = \frac{n}{100} \end{gathered}

    Use when Divide the WHOLE number by the place value, which is not the digit sitting in that column. Exact only when the division comes out whole.

    e.g. 3,4003{,}400 contains 340340 tens, even though its tens digit is 00.

  • Commutative property

    a+b=b+a,a×b=b×aa + b = b + a, \qquad a \times b = b \times a

    Use when Addition and multiplication only. Subtraction and division are not commutative: swapping the two inputs is NOT guaranteed to preserve the value, so you get no licence to reorder them.

    e.g. 53=25 - 3 = 2, while 353 - 5 takes 55 away from only 33 and cannot give 22.

  • Associative property

    (a+b)+c=a+(b+c)(a×b)×c=a×(b×c)\begin{gathered} (a + b) + c = a + (b + c) \\ (a \times b) \times c = a \times (b \times c) \end{gathered}

    Use when Addition and multiplication only, which is why a+b+ca + b + c and abcabc need no parentheses. Chains of - or ÷\div are pinned by the left-to-right convention instead.

    e.g. (104)2=4(10 - 4) - 2 = 4 but 10(42)=810 - (4 - 2) = 8, and (16÷4)÷2=2(16 \div 4) \div 2 = 2 but 16÷(4÷2)=816 \div (4 \div 2) = 8.

  • Identity elements and the zero property

    a+0=a,a×1=aa×0=0\begin{gathered} a + 0 = a, \qquad a \times 1 = a \\ a \times 0 = 0 \end{gathered}

    Use when Every number. 00 is the additive identity, 11 the multiplicative identity. The third is a separate fact, not an identity: one zero factor collapses the whole product.

    e.g. (37+58)×(66)=0(37 + 58) \times (6 - 6) = 0, with the first factor never needed.

  • A chain removes the total, or divides by the product

    abc=a(b+c),a÷b÷c=a÷(b×c)a - b - c = a - (b + c), \qquad a \div b \div c = a \div (b \times c)

    Use when Removing bb then cc removes b+cb + c in total; dividing by bb then cc cuts into b×cb \times c parts. Both follow from the left-to-right reading, not associativity. So bb and cc may trade places; aa stays put. Keep b+cab + c \le a, and each division exact.

    e.g. 30128=3020=1030 - 12 - 8 = 30 - 20 = 10, and 720÷12÷3=720÷36=20720 \div 12 \div 3 = 720 \div 36 = 20.

  • The order of operations

    Grouping    Exponents  ×, ÷    +, \begin{gathered} \text{Grouping} \;\to\; \text{Exponents} \;\to \\ \times,\ \div \;\to\; +,\ - \end{gathered}

    Use when Multiplication and division share ONE tier, addition and subtraction share ONE tier, and inside a tier the work runs strictly left to right. Nested groups settle innermost first, with the tiers applying inside each group.

    e.g. 2×(3+1)224÷6=2×164=282 \times (3 + 1)^2 - 24 \div 6 = 2 \times 16 - 4 = 28: group, then power, then the ×,÷\times, \div tier, then the subtraction.

  • The fraction bar as a grouping symbol

    a+bc+d=(a+b)÷(c+d)\frac{a + b}{c + d} = (a + b) \div (c + d)

    Use when The bar wraps invisible parentheses round the WHOLE numerator and the WHOLE denominator, so each is settled before dividing. Needs c+d0c + d \neq 0. Anything outside the fraction stays outside.

    e.g. 8+42+2=124=3\dfrac{8 + 4}{2 + 2} = \dfrac{12}{4} = 3, not the flattened 8+4÷2+2=128 + 4 \div 2 + 2 = 12.

  • Distributive property

    a(b+c)=ab+aca(bc)=abac\begin{gathered} a(b + c) = ab + ac \\ a(b - c) = ab - ac \end{gathered}
    One rectangle, counted either as a times the whole width or as the two parts addedA rectangle with its height marked a on the left. A highlighted vertical line cuts it into two pieces: the left piece has width b marked above it and area ab written inside, the right piece has width c marked above it and area ac written inside. A dimension line below the rectangle spans its entire width and is labelled b plus c.bcaabacb + c
    Text description

    A rectangle of height a, cut into parts of width b and c labelled ab and ac, whose full width is b plus c.

    Use when The outside factor reaches EVERY term inside, for any number of terms: a(b+c+d)=ab+ac+ada(b + c + d) = ab + ac + ad. Over addition and subtraction only, never over a product. With only whole numbers, the difference form needs bcb \ge c.

    e.g. 7(1002)=70014=6867(100 - 2) = 700 - 14 = 686, while 4(3×5)4(3 \times 5) is just 4×15=604 \times 15 = 60.

  • Factoring, the same rule backward

    ab+ac=a(b+c)ab + ac = a(b + c)

    Use when aa must divide every term. Fully factored means aa is the greatest common factor, so what is left inside shares nothing above 11. Distributing back returns the original, which is the check.

    e.g. 45+27=9(5+3)45 + 27 = 9(5 + 3), while 3(15+9)3(15 + 9) is a true equality that is not finished.

Problem types, step by step

Name the place, or the value, of a digit in a large number

  1. Mark the number into periods of three from the right: ones, thousands, millions.
  2. Read the digit's slot within its period as ones, tens, or hundreds, then attach the period name; the hundreds slot of the thousands period is the hundred-thousands place.
  3. For the PLACE, stop at that column name; for the VALUE, multiply the digit by the column's worth.
  4. To compare two digits of one number, write both values and divide the larger by the smaller, or subtract for "exceeds by how much."

e.g. In 7,248,0917{,}248{,}091 the 22 is in the hundred-thousands place, worth 200,000200{,}000; in 2,7172{,}717 the first 77 is worth 700÷7=100700 \div 7 = 100 times the second.

Convert between a number and its expanded form

  1. Going out, multiply each nonzero digit by its place value and join with plus signs.
  2. Coming back, drop each part into its column and write 00 in every column no part mentions.

e.g. 50,409=50,000+400+950{,}409 = 50{,}000 + 400 + 9.

Compare or order whole numbers

  1. Count digits first: with no leading zeros, more digits means the larger number, whatever those digits are.
  2. With equal digit counts, compare place by place from the LEFT.
  3. Stop at the first differing place: it settles the comparison, and nothing to its right can overturn it.

e.g. 4,081>9914{,}081 > 991 on digit count alone, while 3,205>3,1953{,}205 > 3{,}195 is settled at the hundreds.

Build the largest or smallest number from a given set of digits

  1. For the largest, place the digits in decreasing order, biggest in the highest column.
  2. For the smallest, use increasing order, except that the leading digit cannot be 00: lead with the smallest NONZERO digit and send the 00 to the next column.

e.g. From 44, 00, 77, 11 the largest is 7,4107{,}410 and the smallest is 1,0471{,}047.

Evaluate a numerical expression

  1. Scan the whole line for the highest tier present anywhere in it.
  2. Settle the innermost grouping first, counting a fraction bar as grouping its whole top and whole bottom.
  3. Do ONE operation, rewrite the line, repeat; within a tier take the leftmost first.

e.g. 5+2×324=5+2×94=5+184=195 + 2 \times 3^2 - 4 = 5 + 2 \times 9 - 4 = 5 + 18 - 4 = 19.

Place parentheses to force a target value

  1. Evaluate the plain, unbracketed reading and compare it with the target.
  2. Find which lower-tier operation would have to happen early to close the gap.
  3. Bracket exactly that operation and re-evaluate, since the tiers still govern everything outside it.

e.g. 2+3×42 + 3 \times 4 reads 1414; bracketing the sum gives (2+3)×4=20(2 + 3) \times 4 = 20.

Rearrange a sum or product into an easier one

  1. Scan for a pair combining to a round 1010, 100100, or 10001000.
  2. Move that pair together (commutative) and group it (associative).
  3. Combine the round pair first, then finish with what is left.

e.g. 8×125×7=(8×125)×7=1000×7=70008 \times 125 \times 7 = (8 \times 125) \times 7 = 1000 \times 7 = 7000.

Multiply mentally by splitting one factor

  1. Pick the round number nearest the awkward factor.
  2. Rewrite that factor as the round number plus or minus a small correction, leaving the OTHER factor whole.
  3. Distribute across the two parts and combine, keeping the sign that joined them.

e.g. 9×97=9(1003)=90027=8739 \times 97 = 9(100 - 3) = 900 - 27 = 873.

Factor a sum completely

  1. Find the greatest number dividing every term.
  2. Divide each term by it to fill the parentheses, and write it outside.
  3. Check by distributing back, and confirm what is left inside shares no factor above 11.

e.g. 48+80=16(3+5)48 + 80 = 16(3 + 5), and 16×8=12816 \times 8 = 128 matches.

Name the property that justifies a step

  1. Compare the two sides and ask what actually moved.
  2. Numbers changed places, parentheses unchanged: commutative. Parentheses moved, order unchanged: associative.
  3. A 00 or 11 appearing or vanishing: an identity, or the zero property.
  4. An outside factor reaching inside a parenthesis, or a shared factor coming out: distributive.

e.g. 8×7×5=8×5×78 \times 7 \times 5 = 8 \times 5 \times 7 is commutative, then 8×5×7=(8×5)×78 \times 5 \times 7 = (8 \times 5) \times 7 is associative.

Exam traps

  • Trap Reading the tiers as a strict ranking, so multiplication always beats division and addition always beats subtraction.

    Fix Each pair shares one tier and runs left to right. 16÷4×2=816 \div 4 \times 2 = 8, not 22, and 104+1=710 - 4 + 1 = 7, not 55.

  • Trap Finishing a parenthesis and then sweeping left to right through whatever is left.

    Fix A settled group is only a number, and the tiers resume around it: 302×(3+4)=3014=1630 - 2 \times (3 + 4) = 30 - 14 = 16, not 28×7=19628 \times 7 = 196.

  • Trap Flattening a stacked fraction onto one line without writing the parentheses the bar was supplying.

    Fix The bar groups the whole top over the whole bottom: 32+61+2×2=153×2=10\dfrac{3^2 + 6}{1 + 2} \times 2 = \dfrac{15}{3} \times 2 = 10, while 32+6÷1+2×2=193^2 + 6 \div 1 + 2 \times 2 = 19 divides one term instead.

  • Trap Calling a regrouping "commutative" because the expression merely looks rearranged.

    Fix (5×8)×2=5×(8×2)(5 \times 8) \times 2 = 5 \times (8 \times 2) leaves the order 55, 88, 22 untouched and only moves the parentheses, so it is associative. Commutative would need two numbers to trade places.

  • Trap Letting the outside factor reach only the first term inside the parentheses.

    Fix Every term gets multiplied: 9×53=9(50+3)=450+27=4779 \times 53 = 9(50 + 3) = 450 + 27 = 477, never 450+3=453450 + 3 = 453. With three terms inside, none may be skipped either.

  • Trap Answering "hundreds" when asked for the VALUE of the 33 in 6,3086{,}308, or answering 300300 when asked for its PLACE.

    Fix The place is the column's name and the value is the digit times that column's worth, so the place is the hundreds and the value is 300300.

  • Trap Reassembling 40,000+2,000+90+540{,}000 + 2{,}000 + 90 + 5 as 42,90542{,}905.

    Fix A column no part mentions gets a 00. There is no hundreds term, so the hundreds digit is 00 and the number is 42,09542{,}095.

  • Trap Answering "how many tens are in 9,0609{,}060" with the tens digit, 66.

    Fix The question counts groups of ten in the whole number, so divide: 9,060÷10=9069{,}060 \div 10 = 906 tens.

  • Trap Reading "a subtraction chain may be reordered" as licence to move ANY of its numbers.

    Fix Only the numbers being REMOVED may trade places: 80251580 - 25 - 15 and 80152580 - 15 - 25 are both 4040. The first number never moves, and 80(2515)=7080 - (25 - 15) = 70 shows regrouping stays barred.

Chapter test Questions from across the chapter