8.2 Division Algorithm, Divisibility Rules, Prime Factorization, Divisors, EBOB (GCD) & EKOK (LCM)

8.2 Division Algorithm, Divisibility Rules, Prime Factorization, Divisors, EBOB (GCD) & EKOK (LCM)

Number theory questions in the TR-YÖS test structural shortcuts rather than brute-force arithmetic. Whether bounding a variable in a division box, solving for unknown digits in a five-digit number divisible by $36$, counting the non-prime divisors of $720$, or synchronizing remainders via $\text{EKOK}$, mastering these four pillars turns lengthy computations into twenty-second solutions.

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1. The Euclidean Division Algorithm (Bölme Algoritması)

When a dividend (Bölünen, $A$) is divided by a divisor (Bölen, $B$) to produce a quotient (Bölüm, $C$) and a remainder (Kalan, $K$), two fundamental relationships govern the division box:

1. The Division Identity (Bölme Özdeşliği):

$$A = B \cdot C + K$$

1. The Strict Remainder Inequality (Kalan Koşulu):

$$0 \le K < B$$

The remainder $K$ is strictly less than the divisor $B$. If $K = 0$, we say $A$ is exactly divisible by $B$ ($B \mid A$).

Classic TR-YÖS Division Traps

• Swapping Divisor and Quotient: You may interchange the divisor $B$ and the quotient $C$ ($A = C \cdot B + K$) if and only if the remainder is also strictly less than the quotient ($K < C$).
• Algebraic Remainder Bounds: Suppose $A$ divided by $B = 3x + 4$ gives quotient $C = x + 1$ and remainder $K = x^2$ for $x \in \mathbb{N}$. You cannot choose arbitrary values for $x$: the condition $K < B$ forces:

$$x^2 < 3x + 4 \implies x^2 - 3x - 4 < 0 \implies (x - 4)(x + 1) < 0$$

Since $x \in \mathbb{N}$, only $x \in \{0, 1, 2, 3\}$ are valid! Thus the maximum possible value of $A$ occurs at $x = 3$, giving $B = 13$, $C = 4$, $K = 9$, and $A = 13(4) + 9 = 61$.

• Modular Arithmetic of Remainders: The remainder of $A \pm B$, $A \cdot B$, or $A^n$ when divided by $m$ equals the remainder obtained by replacing $A$ and $B$ with their respective remainders modulo $m$.

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2. Divisibility Rules (Bölünebilme Kuralları)

Elementary Rules (By Last Digits vs. By Digit Sums)

• Last-Digit Family ($2, 5, 10, 4, 8$):
  • By $2$: The units digit is even ($\{0, 2, 4, 6, 8\}$).
  • By $5$: The units digit is $0$ or $5$ (remainder mod $5$ is determined entirely by the units digit).
  • By $10$: The units digit is $0$ (remainder mod $10$ IS the units digit).
  • By $4$: The two-digit number formed by the last two digits is divisible by $4$ ($10d_1 + d_0 \equiv 2d_1 + d_0 \pmod 4$).
  • By $8$: The three-digit number formed by the last three digits is divisible by $8$ ($4d_2 + 2d_1 + d_0 \equiv 0 \pmod 8$).
• Digit-Sum Family ($3, 9, 11$):
  • By $3$ (and $9$): The sum of all digits $\sum d_i$ is a multiple of $3$ (or $9$). The remainder of $N$ divided by $3$ or $9$ equals the remainder of $\sum d_i$ divided by $3$ or $9$.
  • By $11$ (Alternating Sign Rule): Starting at the units digit (rightmost digit) with a $+$ sign and alternating $+, -, +, -, \dots$ from right to left:

$$R_{11} \equiv (d_0 - d_1 + d_2 - d_3 + d_4 - \dots) \pmod{11}$$

If this alternating sum is negative, add multiples of $11$ until it falls in $0 \le R_{11} < 11$.

Composite Divisors (Birleşik Sayılarla Bölünebilme)

A number is divisible by a composite divisor $M$ if and only if it is simultaneously divisible by coprime (aralarında asal) factors $a$ and $b$ such that $M = a \cdot b$ and $\text{EBOB}(a, b) = 1$:

• $6 \iff 2 \text{ and } 3$
• $12 \iff 3 \text{ and } 4$ (never $2 \text{ and } 6$, since $\text{EBOB}(2,6) = 2 \ne 1$!)
• $15 \iff 3 \text{ and } 5$
• $18 \iff 2 \text{ and } 9$
• $30 \iff 3 \text{ and } 10$
• $36 \iff 4 \text{ and } 9$
• $45 \iff 5 \text{ and } 9$
• $90 \iff 9 \text{ and } 10$

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3. Prime Factorization (Asal Çarpanlara Ayırma) & Divisor Counting

A prime number (Asal Sayı) is an integer $p \ge 2$ whose only positive divisors are $1$ and $p$: $\{2, 3, 5, 7, 11, 13, 17, 19, 23, \dots\}$. Note that $2$ is the smallest and the only even prime number ($1$ is not prime!).

Every integer $N \ge 2$ has a unique canonical prime factorization:

$$N = p_1^{e_1} \cdot p_2^{e_2} \cdots p_k^{e_k}$$

where $p_1 < p_2 < \dots < p_k$ are distinct primes and $e_i \in \mathbb{Z}^+$. From the exponents $e_i$, we extract every divisor property of $N$:

1. Number of Positive Divisors (Pozitif Tam Bölen Sayısı — PBS):

$$d^+(N) = (e_1 + 1)(e_2 + 1)\cdots(e_k + 1)$$

1. Number of Negative Divisors (Negatif Tam Bölen Sayısı): $d^-(N) = d^+(N)$.
2. Total Number of Integer Divisors (Tam Bölen Sayısı — TBS):

$$d(N) = 2 \cdot d^+(N) = 2(e_1 + 1)(e_2 + 1)\cdots(e_k + 1)$$

1. Sum of All Integer Divisors: Always $0$, because every positive divisor $d$ cancels with its negative counterpart $-d$.
2. Sum of Positive Divisors:

$$\sigma(N) = \left(\frac{p_1^{e_1+1} - 1}{p_1 - 1}\right)\left(\frac{p_2^{e_2+1} - 1}{p_2 - 1}\right)\cdots\left(\frac{p_k^{e_k+1} - 1}{p_k - 1}\right)$$

1. Sum of Non-Prime Integer Divisors: Since positive and negative divisors cancel except that prime divisors are positive, the sum of all integer divisors that are not prime equals $-(p_1 + p_2 + \dots + p_k)$!
2. Product of Positive Divisors: $P^+(N) = N^{d^+(N)/2}$.

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4. Greatest Common Divisor (EBOB) & Least Common Multiple (EKOK)

For positive integers $a = \prod p_i^{\alpha_i}$ and $b = \prod p_i^{\beta_i}$:

$$\text{EBOB}(a, b) = \prod p_i^{\min(\alpha_i, \beta_i)}, \qquad \text{EKOK}(a, b) = \prod p_i^{\max(\alpha_i, \beta_i)}$$

Fundamental Laws of EBOB and EKOK

• Product Identity (for TWO numbers): $a \cdot b = \text{EBOB}(a, b) \cdot \text{EKOK}(a, b)$.
• Ordering Inequality: $\text{EBOB}(a, b) \le \min(a, b) \le \max(a, b) \le \text{EKOK}(a, b)$.
  • When $a$ and $b$ are allowed to be equal ($a = b$), $\text{EBOB}(a, b) = a = b = \text{EKOK}(a, b)$.
  • When $a \ne b$, to maximize $a + b$ given $\text{EKOK}(a, b) = L$, choose $L$ and its largest proper positive divisor; that divisor is $L/p$ for the smallest prime factor $p$ of $L$ ($L>1$). To minimize $a + b$ given $\text{EBOB}(a, b) = G$, choose $a = G$ and $b = 2G$.
• Coprime Numbers (Aralarında Asal Sayılar): Two positive integers $a$ and $b$ are coprime if $\text{EBOB}(a, b) = 1$, which implies $\text{EKOK}(a, b) = a \cdot b$. Note that $1$ is coprime with every positive integer, and any two consecutive positive integers ($n$ and $n+1$) or consecutive odd integers ($2n-1$ and $2n+1$) are always coprime!
• Constant-Deficit Remainder Problems: To find the smallest positive integer $A$ satisfying

$$A = d_1 x + r_1 = d_2 y + r_2 = d_3 z + r_3$$

where $d_1 - r_1 = d_2 - r_2 = d_3 - r_3 = c$, add $c$ to every expression so that $A + c = d_1(x+1) = d_2(y+1) = d_3(z+1)$. Then $A + c = k \cdot \text{EKOK}(d_1, d_2, d_3)$ for $k \in \mathbb{Z}^+$.

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Summary Reference Table

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Worked Examples

Worked Example 8.2.1: Multi-Condition Composite Divisibility

Problem: The five-digit number $(5a38b)_{10}$ with distinct digits yields a remainder of $13$ when divided by $36$. Find the value of the digit $a$.

Solution:

1. Decompose $36$ into coprime factors $4$ and $9$, and reduce the remainder $13$:
   • Remainder when divided by $4$: $13 \equiv 1 \pmod 4$.
   • Remainder when divided by $9$: $13 \equiv 4 \pmod 9$.
2. Apply the mod-$4$ rule to the last two digits $(8b)_{10}$ first: For $(8b)_{10} \equiv 1 \pmod 4$, since $80$ is a multiple of $4$, we need $b \equiv 1 \pmod 4$, which gives $b \in \{1, 5, 9\}$. Because the digits of $(5a38b)_{10}$ are distinct, $b = 5$ is eliminated (since $5$ already appears as the leading digit). Thus $b = 1$ or $b = 9$.
3. Apply the mod-$9$ rule to the digit sum:
   • Case 1 ($b = 1$): Digit sum is $5 + a + 3 + 8 + 1 = 17 + a \equiv 8 + a \pmod 9$. Setting $8 + a \equiv 4 \pmod 9$ yields $a = 5$. However, $5$ is already used in $(5a381)_{10}$, violating distinct digits!
   • Case 2 ($b = 9$): Digit sum is $5 + a + 3 + 8 + 9 = 25 + a \equiv 7 + a \pmod 9$. Setting $7 + a \equiv 4 \pmod 9$ yields $a = 6$. Checking distinctness in $(56389)_{10}$: all five digits $\{5, 6, 3, 8, 9\}$ are distinct. Thus $a = 6$ is the unique valid solution.

Worked Example 8.2.2: Divisor Counting with Embedded Powers of $10$

Problem: Let $N = 18 \cdot 10^n$ be a positive integer ($n \in \mathbb{Z}^+$). If $N$ has exactly $117$ integer divisors that are not prime, find $n$.

Solution:

1. Write the canonical prime factorization of $N$:

$$N = (2 \cdot 3^2) \cdot (2 \cdot 5)^n = 2^{n+1} \cdot 3^2 \cdot 5^n$$

1. Relate total integer divisors (TBS) to non-prime integer divisors: The distinct prime divisors of $N$ are $2, 3, 5$ ($3$ primes in total). Every other integer divisor of $N$ (positive or negative) is non-prime:

$$\text{TBS}(N) - 3 = 117 \implies \text{TBS}(N) = 120$$

Since $\text{TBS}(N) = 2 \cdot d^+(N)$, the number of positive divisors is $d^+(N) = 60$. 3. Use the PBS formula to solve for $n$:

$$d^+(N) = ((n + 1) + 1)(2 + 1)(n + 1) = 3(n + 2)(n + 1) = 60$$ $$(n + 2)(n + 1) = 20 = 5 \times 4 \implies n = 3$$

Review the Method

In simultaneous remainder problems $A = d_1 x + r_1 = d_2 y + r_2 = d_3 z + r_3$ where $d_i - r_i = c$ is constant, adding $c$ to all parts makes $A + c$ a common multiple of $\text{EKOK}(d_1, d_2, d_3)$.