8.1 Number Sets, Parity, Consecutive Sums, Factorials & Base-n Arithmetic (Taban Aritmetiği)

8.1 Number Sets, Parity, Consecutive Sums, Factorials & Base-$n$ Arithmetic (Taban Aritmetiği)

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1. Turkish Exam Number Set Notation (Temel Kavramlar ve Sayı Kümeleri)

Before solving any equation or inequality, inspect the set membership symbol ($\in$) given at the top of the question. Turkish mathematics conventions use strict definitions that differ slightly from some international textbooks:

• Digits (Rakamlar): The ten base-10 symbols used to write numerals:

$$\text{Rakamlar} = \{0, 1, 2, 3, 4, 5, 6, 7, 8, 9\}$$

Every digit is a number, but multi-digit or negative integers are not digits. If a question specifies that $a, b, c$ are rakamlar, they are automatically bounded by $0 \le a, b, c \le 9$. Read carefully whether the problem says rakamları farklı (distinct digits).

• Natural Numbers (Doğal Sayılar, $\mathbb{N}$): In the Turkish curriculum, zero is a natural number:

$$\mathbb{N} = \{0, 1, 2, 3, 4, \dots\}$$

• Counting Numbers (Sayma Sayıları, $\mathbb{N}^+$ or $\mathbb{Z}^+$): Positive integers starting at $1$:

$$\mathbb{N}^+ = \mathbb{Z}^+ = \{1, 2, 3, 4, \dots\}$$

• Integers (Tam Sayılar, $\mathbb{Z}$): The union of negative integers, zero, and positive integers:

$$\mathbb{Z} = \mathbb{Z}^- \cup \{0\} \cup \mathbb{Z}^+ = \{\dots, -3, -2, -1, 0, 1, 2, 3, \dots\}$$

Crucially, $0$ is neither positive nor negative ($0 \notin \mathbb{Z}^+$ and $0 \notin \mathbb{Z}^-$). Whenever a question asks for the minimum natural number, consider $0$; whenever it asks for the minimum counting number or positive integer, start at $1$.

• Rational (Rasyonel, $\mathbb{Q}$), Irrational (İrrasyonel, $\mathbb{Q}'$ or $\mathbb{I}$), and Real Numbers (Gerçek / Reel Sayılar, $\mathbb{R}$):

$$\mathbb{Q} = \left\{\frac{a}{b} \;\middle|\; a, b \in \mathbb{Z},\; b \ne 0\right\}, \qquad \mathbb{R} = \mathbb{Q} \cup \mathbb{Q}'$$

Sum-Product Extremum Rules (Toplam-Çarpım Ekstremum Kuralları)

1. Fixed Sum, Extremize Product ($a + b = S$ for $a, b \in \mathbb{N}^+$):
   • To maximize $a \cdot b$, choose $a$ and $b$ as close to each other as possible (equal if permitted; otherwise the closest distinct integers compatible with the sum).
   • To minimize $a \cdot b$, choose $a$ and $b$ as far apart as possible (e.g., $1$ and $S - 1$ in $\mathbb{N}^+$, or $0$ and $S$ in $\mathbb{N}$).
2. Fixed Product, Extremize Sum ($a \cdot b = P$ for $a, b \in \mathbb{N}^+$):
   • To maximize $a + b$, choose factors as far apart as possible ($1$ and $P$).
   • To minimize $a + b$, choose factors as close together as possible.
3. The Domain Trap ($\mathbb{Z}$ vs. $\mathbb{R}$):
   • If $a, b \in \mathbb{Z}$ (allowing negatives) and $a \cdot b = 24$, the minimum value of $a + b$ is not $4 + 6 = 10$, but $(-1) + (-24) = -25$!
   • If $a, b \in \mathbb{R}^+$ and $a + b = 9$, you are not restricted to integers $4$ and $5$; the maximum product occurs at $a = b = \frac{9}{2} = 4.5$, yielding $a \cdot b = \frac{81}{4} = 20.25$.

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2. Parity (Tek ve Çift Sayılar), Sign Analysis (İşaret İncelemesi), and Consecutive Sums (Ardışık Sayılar)

Parity Rules (Tek = O, Çift = E)

Parity is a property of integers only ($\mathbb{Z}$); non-integer fractions have neither odd nor even parity. Let $O$ denote an odd integer ($2k - 1$) and $E$ denote an even integer ($2k$, including $0$ and negative evens like $-4$):

$$O \pm O = E, \qquad E \pm E = E, \qquad O \pm E = O$$ $$O \times O = O, \qquad E \times O = E, \qquad E \times E = E$$

A product $x_1 x_2 \cdots x_m$ of integers is odd if and only if every single factor is odd; if at least one factor is even, the entire product is even.

For positive integer exponents ($n \in \mathbb{Z}^+$), exponentiation preserves parity: $O^n = O$ and $E^n = E$. However, watch for two classic TR-YÖS traps:

• Zero or Negative Exponents: If $n \in \mathbb{N}$ or $n \in \mathbb{Z}$, an even base raised to the zero power becomes odd ($2^0 = 1$), and negative exponents yield non-integers ($2^{-2} = \frac{1}{4}$).
• Unconstrained Base ($a \in \mathbb{R}$): If a problem states "$a^2$ is an even integer" without declaring $a \in \mathbb{Z}$, you cannot conclude that $a$ is an even integer (for example, $a = \sqrt{6}$ gives $a^2 = 6$, which is even, yet $a^3 = 6\sqrt{6}$ is not even an integer).

Sign Analysis (İşaret İncelemesi)

When determining signs of $x, y, z \in \mathbb{R} \setminus \{0\}$ from inequalities such as $x^{2k} \cdot y^{2m+1} < 0$:

1. Any even power $x^{2k}$ is strictly positive ($x^{2k} > 0$) regardless of whether $x$ is positive or negative. Cancel $x^{2k}$ from multiplicative sign inequalities immediately (retaining only that $x \ne 0$), without flipping the inequality sign.
2. Any odd power $y^{2m+1}$ has the exact same sign as $y$: replace $y^{2m+1}$ with $y^1 = y$.

Consecutive Numbers & Gauss Summation (Ardışık Sayılar ve Toplam Formülleri)

Consecutive integers advance by $1$ ($n, n+1, n+2$), while consecutive even or consecutive odd integers both advance by step $2$ ($n, n+2, n+4$). Consequently, the difference between any two consecutive terms in an even or odd sequence is $\pm 2$, independent of $n$.

For any arithmetic progression $a_1, a_2, \dots, a_k$ with constant step $d$:

$$\text{Number of Terms (Terim Sayısı)} = \frac{\text{Last Term} - \text{First Term}}{d} + 1$$ $$\text{Sum (Toplam)} = (\text{Number of Terms}) \times \left(\frac{\text{First Term} + \text{Last Term}}{2}\right)$$

Memorize the three foundational Gauss formulas starting at $1$:

• First $n$ counting numbers: $1 + 2 + 3 + \dots + n = \frac{n(n+1)}{2}$
• First $n$ positive even integers: $2 + 4 + 6 + \dots + 2n = n(n+1)$
• First $n$ positive odd integers: $1 + 3 + 5 + \dots + (2n-1) = n^2$

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3. Factorials (Faktöriyel) & Legendre's Formula

For $n \in \mathbb{N}$, the factorial $n!$ is defined by $0! = 1$, $1! = 1$, and $n! = n \cdot (n-1)!$ for $n \ge 1$. Using the elementary factorial definition, $n!$ is defined for non-negative integers, an expression containing $(n - 4)!$ and $(4 - n)!$ simultaneously forces both $n - 4 \ge 0$ and $4 - n \ge 0$, which immediately pins down $n = 4$.

When simplifying fractions with sums or differences of factorials, factor out the smallest factorial:

$$\frac{(n+1)! - n!}{(n-1)!} = \frac{n!(n + 1 - 1)}{(n-1)!} = \frac{n \cdot (n-1)! \cdot n}{(n-1)!} = n^2$$

Legendre's Formula (Faktöriyel İçindeki Asal Çarpan Sayısı)

To find the largest integer exponent $x$ such that $p^x$ divides $n!$ (where $p$ is prime), divide $n$ successively by $p$ and sum the integer quotients:

$$E_p(n!) = \left\lfloor \frac{n}{p} \right\rfloor + \left\lfloor \frac{n}{p^2} \right\rfloor + \left\lfloor \frac{n}{p^3} \right\rfloor + \cdots$$

• Composite Base Case: If asked for the maximum $x$ in $n! = k^x \cdot m$ where $k$ is composite, factor $k$ into primes first. For $k=\prod p_i^{e_i}$, the maximum exponent is $\min_i\lfloor E_{p_i}(n!)/e_i\rfloor$. Check every prime-power constraint; the largest prime alone need not control the result.
• Trailing Zeros (Sondan Kaç Basamağı Sıfırdır): Every trailing zero in base $10$ corresponds to a factor of $10 = 2 \times 5$. Because $E_2(n!) > E_5(n!)$ for all $n \ge 2$, the number of trailing zeros of $n!$ (and likewise the number of trailing $9$s of $n! - 1$) equals $E_5(n!)$.

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4. Positional Digit Decomposition (Basamak Kavramı) & Base-$b$ Arithmetic (Taban Aritmetiği)

A numeral $(d_k \dots d_2 d_1 d_0)_b$ written in integer base $b \ge 2$ represents the polynomial expansion:

$$(d_k \dots d_2 d_1 d_0)_b = d_k b^k + \dots + d_2 b^2 + d_1 b^1 + d_0 b^0$$

subject to the strict digit constraint:

$$0 \le d_i < b \quad \text{for all } i, \qquad d_k \ne 0$$

In base $10$, memorize the fast algebraic identities for two- and three-digit integers $(ab)_{10} = 10a + b$ and $(abc)_{10} = 100a + 10b + c$:

• $(ab)_{10} + (ba)_{10} = 11(a + b)$
• $(ab)_{10} - (ba)_{10} = 9(a - b)$
• $(abc)_{10} - (cba)_{10} = 99(a - c)$ (notice the middle digit $b$ cancels completely!)
• $(abc)_{10} + (bca)_{10} + (cab)_{10} = 111(a + b + c)$

Direct Operations in Base $b$

• Addition in Base $b$: Add column by column from right to left. Whenever a column sum $S \ge b$, write down the remainder $(S \bmod b)$ and carry the quotient $\lfloor S / b \rfloor$ to the next column.
• Subtraction in Base $b$: When the top digit is smaller than the bottom digit in a column, borrow $1$ unit from the next left column, which adds $+b$ (not $+10$!) to the current column.
• Multiplication in Base $b$: Multiply digit by digit from right to left just as in base $10$, but divide every intermediate product $P$ (plus any incoming carry) by the base $b$: write down the remainder $(P \bmod b)$ in the current column and carry the quotient $\lfloor P / b \rfloor$ to the next left column, then add the shifted partial products in base $b$.
• Parity of $(d_k \dots d_1 d_0)_b$: If the base $b$ is even, parity depends solely on the last digit $d_0$. If the base $b$ is odd, parity equals the parity of the sum of the digits $d_k + \dots + d_1 + d_0$.

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Summary Reference Table

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Worked Examples

Worked Example 8.1.1: Factorial Composite Divisors and Trailing Zeros

Problem: Let $63! = 15^x \cdot 12^y \cdot K$, where $x, y, K \in \mathbb{Z}^+$. Find the maximum possible value of $x + y$.

Solution:

1. Decompose the composite bases into primes:

$$15^x \cdot 12^y = (3 \cdot 5)^x \cdot (2^2 \cdot 3)^y = 2^{2y} \cdot 3^{x+y} \cdot 5^x$$

1. Count the available prime factors in $63!$ using Legendre's formula:
   • For $p = 5$: $\lfloor 63/5 \rfloor + \lfloor 12/5 \rfloor = 12 + 2 = 14$. Thus $x \le 14$.
   • For $p = 3$: $\lfloor 63/3 \rfloor + \lfloor 21/3 \rfloor + \lfloor 7/3 \rfloor = 21 + 7 + 2 = 30$. Thus $x + y \le 30$.
   • For $p = 2$: $\lfloor 63/2 \rfloor + \lfloor 31/2 \rfloor + \lfloor 15/2 \rfloor + \lfloor 7/2 \rfloor + \lfloor 3/2 \rfloor = 31 + 15 + 7 + 3 + 1 = 57$. Thus $2y \le 57 \implies y \le 28$.
2. Maximize $x + y$: The prime $3$ directly bounds $x + y \le 30$. Check whether $x + y = 30$ is achievable while satisfying $x \le 14$ and $y \le 28$: choosing $x = 14$ and $y = 16$ satisfies $14 \le 14$, $16 \le 28$, and $x, y \in \mathbb{Z}^+$. Therefore, $\max(x + y) = 30$.

Worked Example 8.1.2: Base-$b$ Subtraction and Domain Constraints

Problem: Given the valid base-$7$ subtraction $(a4c)_7 - (c4a)_7 = (2x4)_7$, determine the digit $x$ and the difference $a - c$.

Solution:

1. Method 1 (Column Borrowing in Base $7$): Since the difference $(2x4)_7$ is positive, $a > c$.
   • Units column: Since $c < a$, we cannot subtract $a$ from $c$ directly. Borrow $1$ from the sevens column, adding $+7$ to the units digit: $(c + 7) - a = 4 \implies a - c = 3$.
   • Sevens column: The top digit $4$ became $3$ after lending $1$. Now $3 < 4$, so we borrow $1$ from the $7^2$ column, adding $+7$: $(3 + 7) - 4 = 6 \implies x = 6$.
2. Method 2 (Polynomial Expansion Verification):

$$(a4c)_7 - (c4a)_7 = (49a + 28 + c) - (49c + 28 + a) = 48(a - c)$$

With $a - c = 3$, the value is $48 \times 3 = 144$ in base $10$. Converting $144$ to base $7$: $144 = 2 \times 49 + 6 \times 7 + 4 = (264)_7$, confirming $x = 6$ and $a - c = 3$.

Review the Method

A positional numeral $(d_k \dots d_1 d_0)_b = \sum_{i=0}^k d_i b^i$ is valid if and only if the base $b \ge 2$, the leading digit $d_k \ne 0$, and every digit satisfies $0 \le d_i < b$.