Mathematics and Problem Solving Course Notes

Arithmetic Sequences and Series Summation Methods

• General Formulas for Finite Arithmetic Sequences:

  • Term Count Formula: $n = \frac{a_n - a_1}{d} + 1$, where $a_n$ is the last term, $a_1$ is the first term, and $d$ is the common difference.

  • Nth Term Formula: $a_n = a_1 + (n - 1)d$.

  • Sum Formula: $S_n = \frac{n}{2}(a_1 + a_n)$.

• Reversal and Pairing Technique:

  • Writing the sum $S$ in ascending order and directly underneath in descending order aligns pairs that each sum to $(a_1 + a_n)$.

  • Adding the two equations produces $2S = n \times (a_1 + a_n)$, which simplifies to $S = \frac{n \times (a_1 + a_n)}{2}$.

• Worked Summation Examples:

  • Sum of Sequence $1 + 2 + 3 + 4 + \dots + 29$:

  • Number of terms: $n = 29 - 1 + 1 = 29$.

  • Reversal alignment:       $S = 1 + 2 + 3 + \dots + 29$       $S = 29 + 28 + 27 + \dots + 1$       $2S = 29 \times (1 + 29) = 29 \times 30 = 870$

  • Final sum calculation: $S = \frac{870}{2} = 435$

  • Sum of Sequence $4 + 5 + 6 + \dots + 56$:

  • Course Section Reference: $601,101.001$ / $2790$

  • Number of terms in sequence: $n = 56 - 4 + 1 = 53$

  • Reversal alignment:       $S = 4 + 5 + 6 + \dots + 55 + 56$       $S = 56 + 55 + 54 + \dots + 5 + 4$       $2S = 53 \times (4 + 56) = 53 \times 60 = 3,180$

  • Final sum calculation: $S = \frac{53 \times 60}{2} = 53 \times 30 = 1,590$

  • Sum of Sequence $1 + 2 + 3 + 4 + \dots + 98$:

  • Number of terms: $n = 98 - 1 + 1 = 98$

  • Pairwise calculation:       $2S = 98 \times (1 + 98) = 98 \times 99 = 9,702$       $S = \frac{9,702}{2} = 4,851$

  • Sum of Sequence $37 + 38 + 39 + 40 + \dots + 104$:

  • Determining term count using $a_n = a_1 + (n - 1)d$:       $104 = 37 + (n - 1) \times 1$       $104 - 37 = n - 1$       $67 = n - 1 \implies n = 68$

  • Pairwise sum: $a_1 + a_n = 37 + 104 = 141$

  • Final sum calculation: $S_{68} = \frac{68}{2}(37 + 104) = 34 \times 141 = 4,794$

  • Sum of Sequence $40 + 41 + 42 + 43 + \dots + 123$:

  • Number of terms: $n = 123 - 40 + 1 = 84$

  • Reversal pair sum: $40 + 123 = 163$

  • Formula calculation: $S_{84} = \frac{84}{2}(40 + 123) = 42 \times 163 = 6,846$

• Comparative Analysis of Series Without Direct Computation:

  • Comparison of Odd Sequence $D = 1 + 3 + 5 + 7 + \dots + 97$ and Even Sequence $E = 4 + 6 + 8 + \dots + 100$:

  • Term count for $D$: $n_D = \frac{97 - 1}{2} + 1 = 48 + 1 = 49$

  • Term count for $E$: $n_E = \frac{100 - 4}{2} + 1 = 48 + 1 = 49$

  • Term-by-term difference comparison:

    • Term 1: $4 - 1 = 3$

    • Term 2: $6 - 3 = 3$

    • Term 3: $8 - 5 = 3$

    • General Term $k$: $E_k - D_k = 3$

  • Total difference accumulation: $49 \times 3 = 147$

  • Conclusion: Sequence $E$ is greater than Sequence $D$ by $147$

Combinatorics and Counting Methods

• Counting Integers with Digit Restrictions:

  • Problem: Determine how many 3-digit positive integers ($100$ through $999$) do not contain the digit $5$

  • Analysis by digit position:

  • Hundreds place: Permissible digits are $\text{\{1, 2, 3, 4, 6, 7, 8, 9\}}$ ($8$ options; $0$ and $5$ excluded)

  • Tens place: Permissible digits are $\text{\{0, 1, 2, 3, 4, 6, 7, 8, 9\}}$ ($9$ options; $5$ excluded)

  • Units place: Permissible digits are $\text{\{0, 1, 2, 3, 4, 6, 7, 8, 9\}}$ ($9$ options; $5$ excluded)

  • Multiplication Principle calculation: $8 \times 9 \times 9 = 648$

  • Answer: $648$ positive integers

• Combinatorial Packaging/Partitioning Problem:

  • Problem: Cookies are sold singly ($1$), in packages of $9$, or packages of $27$. Find how many distinct ways there are to purchase exactly $54$ cookies

  • Case Breakdown by quantity of $27$-cookie packages:

  • Case 1: Two $27$-cookie packages:

    • $2 \times 27 = 54$ cookies

    • Remaining cookies needed: $0$

    • Combination: $(2 \text{ packages of } 27, 0 \text{ packages of } 9, 0 \text{ single cookies})$

    • Ways: $1$

  • Case 2: One $27$-cookie package:

    • $1 \times 27 = 27$ cookies

    • Remaining cookies needed: $27$, using $9$-cookie packages and single cookies:

    • $3$ nine-packs ($3 \times 9 = 27$) + $0$ singles

    • $2$ nine-packs ($2 \times 9 = 18$) + $9$ singles

    • $1$ nine-pack ($1 \times 9 = 9$) + $18$ singles

    • $0$ nine-packs ($0 \times 9 = 0$) + $27$ singles

    • Ways: $4$

  • Case 3: Zero $27$-cookie packages:

    • $0 \times 27 = 0$ cookies

    • Remaining cookies needed: $54$, using $9$-cookie packages and single cookies:

    • $6$ nine-packs ($6 \times 9 = 54$) + $0$ singles

    • $5$ nine-packs ($5 \times 9 = 45$) + $9$ singles

    • $4$ nine-packs ($4 \times 9 = 36$) + $18$ singles

    • $3$ nine-packs ($3 \times 9 = 27$) + $27$ singles

    • $2$ nine-packs ($2 \times 9 = 18$) + $36$ singles

    • $1$ nine-pack ($1 \times 9 = 9$) + $45$ singles

    • $0$ nine-packs ($0 \times 9 = 0$) + $54$ singles

    • Ways: $7$

  • Total combinations sum: $1 + 4 + 7 = 12$ ways

Spatial and Positional Reasoning Problems

• Symmetric Ladder Rung Model (Single Direct Shift):

  • Scenario: Standing on the middle rung of a ladder, moving up $14$ rungs leads to the top rung

  • Structural breakdown:

  • Position $0$ represents the middle rung

  • Moving up $14$ rungs means there are $14$ rungs above the middle rung

  • Middle rung symmetry requires an identical count of $14$ rungs below the middle rung

  • Calculation: $\text{Total Rungs} = \text{Rungs Above} + \text{Middle Rung} + \text{Rungs Below} = 14 + 1 + 14 = 28 + 1 = 29$

  • Total Rungs: $29$ rungs

• Multi-Step Movement Ladder Model (Kathy's Ladder Problem):

  • Scenario: Kathy stands on the middle rung of a ladder and executes the following movement steps:

  1. Climbs up $4$ rungs ($+4$)

  2. Moves down $2$ rungs ($-2$)

  3. Climbs up $7$ rungs ($+7$)

  4. Climbs up the remaining $4$ rungs to reach the top ($+4$)

  • Net Displacement Calculation:

  • Net position above the middle rung: $+4 - 2 + 7 + 4 = 13$ rungs above middle

  • Symmetry analysis:

  • Since the top rung is $13$ rungs above the middle rung, there are $13$ rungs below the middle rung

  • Calculation: $\text{Total Rungs} = 13 + 13 + 1 = 26 + 1 = 27$

  • Total Rungs: $27$ rungs

Logical Deduction and Grid Optimization

• Ordering Deduction Puzzle (Horse Race Problem):

  • Competitors: Evan, Marlena, Mona, Soolin

  • Constraints:

  1. Marlena is the slowest competitor (4th place)

  2. Mona is faster than Soolin but slower than Evan

  • Step-by-step order construction:

  • Relationship between Evan, Mona, and Soolin: $\text{Evan} > \text{Mona} > \text{Soolin}$

  • Incorporating Marlena in 4th place: $\text{Evan} > \text{Mona} > \text{Soolin} > \text{Marlena}$

  • Final Finishing Order: Evan (1st), Mona (2nd), Soolin (3rd), Marlena (4th)

• Magic Square Solutions:

  • Property: Every row, column, and diagonal in a magic square sums to the same constant $S_{\text{magic}}$

  • Constant Sum Value: $S_{\text{magic}} = 48$

  • Variable Evaluations:

  • Solving for $b$: $b + 18 + 26 = 48 \implies b + 44 = 48 \implies b = 4$

  • Solving for $c$: $16 + b + c + 3 = 48 \implies 16 + 4 + c + 3 = 48 \implies 23 + c = 48 \implies 34 + c = 48 \implies c = 14$

  • Solving for $a$: $6 + a + 18 = 48 \implies a + 24 = 48 \implies a = 24$

  • Solving for $d$: $d = 22$

  • Final Variable Values: $a = 24$, $b = 4$, $c = 14$, $d = 22$

Rate and Proportion Problems

• Comparative Reading Rates (Audrey and Lauren Novel Problem):

  • Reading Rates:

  • Audrey reads $6\,\text{pages/day}$

  • Lauren reads $4\,\text{pages/day}$

  • Conditions: Both start reading the novel on the same day. Audrey is currently on page $18$

  • Step-by-Step Solution:

  • Elapsed time for Audrey: $\frac{18}{6} = 3\,\text{days}$

  • Page reached by Lauren after $3\,\text{days}$: $3 \times 4 = 12$

  • Answer: Lauren is on page $12$