Understanding numbers and their properties is the foundation of all mathematics. On the YÖS exam, number-related problems appear in every section, so a deep knowledge of this topic is essential.
---
Definition: Natural numbers are the numbers used for counting objects.
$$\mathbb{N} = \{1, 2, 3, 4, 5, \ldots\}$$
Sometimes 0 is included in the natural numbers: $\mathbb{N}_0 = \{0, 1, 2, 3, \ldots\}$
Properties of natural numbers:- Every natural number has a successor- There is a smallest natural number (one)- There is no largest natural number
Definition: The integers include the natural numbers, zero, and the numbers opposite to the natural numbers.
$$\mathbb{Z} = \{\ldots, -3, -2, -1, 0, 1, 2, 3, \ldots\}$$
Partition into subsets:- Positive integers: $\{1, 2, 3, \ldots\}$- Negative integers: $\{\ldots, -3, -2, -1\}$- Zero: $\{0\}$
Definition: Rational numbers are numbers that can be represented as the ratio of two integers.
$$\mathbb{Q} = \left\{\frac{p}{q} : p \in \mathbb{Z}, q \in \mathbb{Z}, q \neq 0\right\}$$
Examples: $\frac{1}{2}, -\frac{3}{4}, \frac{7}{1} = 7, 0.\overline{3} = \frac{1}{3}$
Definition: Irrational numbers are real numbers that are not rational. They cannot be represented as a fraction.
Examples:- $\sqrt{2}, \sqrt{3}, \sqrt{5}$ — roots of non-perfect squares- $\pi$ — ratio of a circle's circumference to its diameter- $e$ — base of the natural logarithm- $\phi = \frac{1+\sqrt{5}}{2}$ — the golden ratio
Property: Irrational numbers have an infinite non-repeating decimal representation.
Definition: Real numbers are the union of rational and irrational numbers.
$$\mathbb{R} = \mathbb{Q} \cup (\mathbb{R} \setminus \mathbb{Q})$$
Geometric interpretation: Every real number corresponds to a point on the number line.
For addition and multiplication:$$a + b = b + a$$$$a \cdot b = b \cdot a$$
Example: $5 + 3 = 3 + 5 = 8$ and $4 \cdot 7 = 7 \cdot 4 = 28$
Important: Subtraction and division are not commutative!$$5 - 3 \neq 3 - 5 \quad \text{and} \quad 10 \div 2 \neq 2 \div 10$$
For addition and multiplication:$$(a + b) + c = a + (b + c)$$$$(a \cdot b) \cdot c = a \cdot (b \cdot c)$$
Example: $(2 + 3) + 4 = 2 + (3 + 4) = 9$
$$a \cdot (b + c) = a \cdot b + a \cdot c$$$$a \cdot (b - c) = a \cdot b - a \cdot c$$$$(a + b) \cdot (c + d) = ac + ad + bc + bd$$
Example: $3 \cdot (5 + 2) = 3 \cdot 5 + 3 \cdot 2 = 15 + 6 = 21$
A number $a$ is divisible by a number $b$ (written $b | a$) if there exists an integer $k$ such that:$$a = b \cdot k$$
Here $b$ is called a divisor of $a$, and $a$ is called a multiple of $b$.
#### Divisibility by 2A number is divisible by 2 if its last digit is even (0, 2, 4, 6, 8).
Example: 1346 is divisible by 2, because it ends in 6.
#### Divisibility by 3A number is divisible by 3 if the sum of its digits is divisible by 3.
Example: For 2547: $2 + 5 + 4 + 7 = 18$, which is divisible by 3. Therefore, 2547 is divisible by 3.
#### Divisibility by 4A number is divisible by 4 if the number formed by its last two digits is divisible by 4.
Example: 3524 is divisible by 4, because 24 is divisible by 4.
#### Divisibility by 5A number is divisible by 5 if its last digit is 0 or 5.
Example: 1235 and 1450 are divisible by 5.
#### Divisibility by 6A number is divisible by 6 if it is divisible by both 2 and 3.
Example: 342 is divisible by 6 (divisible by 2 and by 3, since $3 + 4 + 2 = 9$).
#### Divisibility by 8A number is divisible by 8 if the number formed by its last three digits is divisible by 8.
Example: 15648 is divisible by 8, because 648 is divisible by 8.
#### Divisibility by 9A number is divisible by 9 if the sum of its digits is divisible by 9.
Example: For 7263: $7 + 2 + 6 + 3 = 18$, which is divisible by 9. Therefore, 7263 is divisible by 9.
#### Divisibility by 10A number is divisible by 10 if its last digit is 0.
Example: 1250, 4870 are divisible by 10.
#### Divisibility by 11A number is divisible by 11 if the difference between the sum of digits in odd positions and the sum of digits in even positions is divisible by 11.
Example: For 121: $(1 + 1) - 2 = 0$, which is divisible by 11. Therefore, 121 is divisible by 11.
Prime number — a natural number that is divisible only by 1 and by itself.
Composite number — a natural number that has more than two divisors.
Examples of primes: 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, ...
Examples of composites: 4, 6, 8, 9, 10, 12, 14, 15, 16, 18, 20, ...
Important notes:- 1 is neither prime nor composite- 2 is the only even prime number
Theorem: Every natural number greater than 1 can be represented uniquely as a product of prime numbers (up to the order of the factors).
$$n = p_1^{a_1} \cdot p_2^{a_2} \cdot \ldots \cdot p_k^{a_k}$$
where $p_1, p_2, \ldots, p_k$ are distinct primes, and $a_1, a_2, \ldots, a_k$ are natural numbers.
Example: $$360 = 2^3 \cdot 3^2 \cdot 5$$
This is an ancient algorithm for finding all prime numbers up to a given number $n$.
Algorithm:1. Write out all natural numbers from 2 to $n$2. Starting from 2, cross out all multiples of 2 (except 2 itself)3. Find the first number that is not crossed out (it will be 3) and cross out all its multiples4. Repeat step 3 for each next uncrossed number, until that number exceeds $\sqrt{n}$5. All remaining uncrossed numbers are prime
Example for $n = 30$:Primes: 2, 3, 5, 7, 11, 13, 17, 19, 23, 29
Definition: The GCD of two numbers $a$ and $b$ is the largest natural number that divides both.
Notation: $\gcd(a, b)$ or $(a, b)$
To find the GCD, take the product of common factors with the smallest exponents.
Example:- $12 = 2^2 \cdot 3$- $18 = 2 \cdot 3^2$- $\gcd(12, 18) = 2 \cdot 3 = 6$
Algorithm:$$a = b \cdot q + r$$
where $0 \leq r < b$. Then repeat the process with the pair $(b, r)$ until the remainder is zero.
Example: Find $\gcd(144, 60)$
$$144 = 60 \cdot 2 + 24$$$$60 = 24 \cdot 2 + 12$$$$24 = 12 \cdot 2 + 0$$
Answer: $\gcd(144, 60) = 12$
Definition: The LCM of two numbers $a$ and $b$ is the smallest natural number divisible by both.
Notation: $\text{lcm}(a, b)$ or $[a, b]$
To find the LCM, take the product of all factors with the largest exponents.
Example:- $12 = 2^2 \cdot 3$- $18 = 2 \cdot 3^2$- $\text{lcm}(12, 18) = 2^2 \cdot 3^2 = 4 \cdot 9 = 36$
$$\gcd(a, b) \cdot \text{lcm}(a, b) = a \cdot b$$
Application: $\gcd(12, 18) \cdot \text{lcm}(12, 18) = 6 \cdot 36 = 216 = 12 \cdot 18$ ✓
Definition: The factorial of a natural number $n$ (denoted $n!$) is the product of all natural numbers from 1 to $n$.
$$n! = 1 \cdot 2 \cdot 3 \cdot \ldots \cdot n$$
Special case: $0! = 1$ (by definition)
Examples:- $1! = 1$- $2! = 1 \cdot 2 = 2$- $3! = 1 \cdot 2 \cdot 3 = 6$- $4! = 1 \cdot 2 \cdot 3 \cdot 4 = 24$- $5! = 120$- $6! = 720$- $7! = 5040$
Recurrence relation: $n! = n \cdot (n-1)!$
Definition: A permutation of $n$ elements is an ordered arrangement of all $n$ elements.
Number of permutations:$$P_n = n!$$
Example: How many ways can 3 people be arranged in a row?$$P_3 = 3! = 6$$
Arrangements: (1,2,3), (1,3,2), (2,1,3), (2,3,1), (3,1,2), (3,2,1)
Definition: An arrangement of $k$ elements from a set of $n$ is an ordered selection of $k$ elements from the $n$.
Number of arrangements:$$A_n^k = \frac{n!}{(n-k)!}$$
Example: How many two-digit numbers can be formed from the digits 1, 2, 3, 4 without repetition?$$A_4^2 = \frac{4!}{(4-2)!} = \frac{4!}{2!} = \frac{24}{2} = 12$$
Definition: A combination of $k$ elements from $n$ is an unordered selection of $k$ elements from the $n$.
Number of combinations:$$C_n^k = \binom{n}{k} = \frac{n!}{k!(n-k)!}$$
Properties:- $C_n^k = C_n^{n-k}$ (symmetry)- $C_n^0 = C_n^n = 1$- $C_n^k + C_n^{k+1} = C_n^{k+1}$ (Pascal's property)
Example: How many ways are there to choose 2 people from 4?$$C_4^2 = \frac{4!}{2! \cdot 2!} = \frac{24}{2 \cdot 2} = 6$$
This method expresses one variable in terms of the other from one equation, then substitutes into the other equation.
Example:$$\begin{cases}x + 2y = 7 \\3x - y = 5\end{cases}$$
From the first equation: $x = 7 - 2y$
Substitute into the second:$$3(7 - 2y) - y = 5$$$$21 - 6y - y = 5$$$$-7y = -16$$$$y = \frac{16}{7}$$
Then: $x = 7 - 2 \cdot \frac{16}{7} = 7 - \frac{32}{7} = \frac{49 - 32}{7} = \frac{17}{7}$
Answer: $\left(\frac{17}{7}, \frac{16}{7}\right)$
This method adds or subtracts the equations to eliminate one of the variables.
Example:$$\begin{cases}2x + 3y = 8 \\x - 2y = 1\end{cases}$$
Multiply the second equation by 2:$$\begin{cases}2x + 3y = 8 \\2x - 4y = 2\end{cases}$$
Subtract the second from the first:$$7y = 6$$$$y = \frac{6}{7}$$
Substitute into the first equation:$$2x + 3 \cdot \frac{6}{7} = 8$$$$2x + \frac{18}{7} = 8$$$$2x = 8 - \frac{18}{7} = \frac{56 - 18}{7} = \frac{38}{7}$$$$x = \frac{19}{7}$$
Answer: $\left(\frac{19}{7}, \frac{6}{7}\right)$
For the system:$$\begin{cases}a_1 x + b_1 y = c_1 \\a_2 x + b_2 y = c_2\end{cases}$$
Determinants:$$\Delta = \begin{vmatrix} a_1 & b_1 \\ a_2 & b_2 \end{vmatrix} = a_1 b_2 - a_2 b_1$$
$$\Delta_x = \begin{vmatrix} c_1 & b_1 \\ c_2 & b_2 \end{vmatrix} = c_1 b_2 - c_2 b_1$$
$$\Delta_y = \begin{vmatrix} a_1 & c_1 \\ a_2 & c_2 \end{vmatrix} = a_1 c_2 - a_2 c_1$$
Solution:$$x = \frac{\Delta_x}{\Delta}, \quad y = \frac{\Delta_y}{\Delta} \quad (\Delta \neq 0)$$
Example:$$\begin{cases}x + 2y = 5 \\3x - y = 4\end{cases}$$
$$\Delta = \begin{vmatrix} 1 & 2 \\ 3 & -1 \end{vmatrix} = 1 \cdot (-1) - 3 \cdot 2 = -1 - 6 = -7$$
$$\Delta_x = \begin{vmatrix} 5 & 2 \\ 4 & -1 \end{vmatrix} = 5 \cdot (-1) - 4 \cdot 2 = -5 - 8 = -13$$
$$\Delta_y = \begin{vmatrix} 1 & 5 \\ 3 & 4 \end{vmatrix} = 1 \cdot 4 - 3 \cdot 5 = 4 - 15 = -11$$
$$x = \frac{-13}{-7} = \frac{13}{7}, \quad y = \frac{-11}{-7} = \frac{11}{7}$$
Statement: Find the largest natural number that gives equal remainders when dividing 127 and 189.
Solution:
If dividing both numbers by $d$ gives the same remainder:$$127 = d \cdot q_1 + r$$$$189 = d \cdot q_2 + r$$
Subtracting:$$189 - 127 = d(q_2 - q_1)$$$$62 = d(q_2 - q_1)$$
Therefore, $d$ is a divisor of 62. The divisors of 62 are 1, 2, 31, 62.
Check $d = 62$:- $127 = 62 \cdot 2 + 3$ (remainder 3)- $189 = 62 \cdot 3 + 3$ (remainder 3) ✓
Answer: 62
Statement: How many natural divisors does 360 have?
Factor 360 into primes:$$360 = 8 \cdot 45 = 8 \cdot 9 \cdot 5 = 2^3 \cdot 3^2 \cdot 5^1$$
The number of divisors is given by:$$\tau(n) = (a_1 + 1)(a_2 + 1) \cdots (a_k + 1)$$
where $n = p_1^{a_1} \cdot p_2^{a_2} \cdot \ldots \cdot p_k^{a_k}$
For 360:$$\tau(360) = (3 + 1)(2 + 1)(1 + 1) = 4 \cdot 3 \cdot 2 = 24$$
Answer: 24 divisors
Statement: Find LCM(12, 18, 24).
Factor each number:- $12 = 2^2 \cdot 3$- $18 = 2 \cdot 3^2$- $24 = 2^3 \cdot 3$
LCM — product of the largest exponents of all factors:$$\text{lcm}(12, 18, 24) = 2^3 \cdot 3^2 = 8 \cdot 9 = 72$$
Check:- $72 \div 12 = 6$ ✓- $72 \div 18 = 4$ ✓- $72 \div 24 = 3$ ✓
Answer: 72
Statement: In how many ways can 3 people be chosen from 5 to participate in a contest?
Order does not matter, so we use the combinations formula:$$C_5^3 = \frac{5!}{3!(5-3)!} = \frac{5!}{3! \cdot 2!} = \frac{5 \cdot 4 \cdot 3!}{3! \cdot 2 \cdot 1} = \frac{5 \cdot 4}{2} = 10$$
Answer: 10 ways
Statement: Solve the system:$$\begin{cases}2x + y = 11 \\x - y = 1\end{cases}$$
Solution by elimination:
Add both equations:$$2x + y + x - y = 11 + 1$$$$3x = 12$$$$x = 4$$
Substitute into the first equation:$$2(4) + y = 11$$$$8 + y = 11$$$$y = 3$$
Check:- $2(4) + 3 = 8 + 3 = 11$ ✓- $4 - 3 = 1$ ✓
Answer: $(4, 3)$
Wrong: "123 is divisible by 5 because it contains 2 and 3"
Right: A number is divisible by 5 only if it ends in 0 or 5. 123 is not divisible by 5.
Wrong: LCM(12, 18) = 6 (that's the GCD!)
Right: LCM(12, 18) = 36, and GCD(12, 18) = 6
Wrong: When computing the GCD of three numbers, find the GCD of two and ignore the third.
Right: GCD(12, 18, 24) = 6 — all numbers must be considered.
Wrong: Substituting an un-simplified expression.
Right: Simplify the expression first, then substitute.
Wrong: $(n!)! = (n!)!$ without understanding what it means. For example, $3! = 6$, but $(3!)! = 6! = 720$
Right: Remember that the factorial is a single operation: $3! = 6$, not $3 \cdot ! = ?$
### Tip 1: Quick factorial valuesOn the exam, you often need to compute factorials up to 7–8 quickly. Memorize:- $1! = 1$- $2! = 2$- $3! = 6$- $4! = 24$- $5! = 120$- $6! = 720$- $7! = 5040$
### Tip 2: Divisibility rules save timeInstead of long division, use divisibility rules for a fast check. This saves 20–30 seconds per problem!
### Tip 3: Euclidean algorithm is more efficient than factorizationFor finding the GCD of large numbers, the Euclidean algorithm is faster than factoring into primes.
### Tip 4: Cramer's rule for larger systemsIf the coefficients are small integers, Cramer's rule is often faster than substitution.
### Tip 5: Verify the answer in the original systemAfter solving a system, always substitute the answer back into both original equations. This reduces errors by 50%.
All course topics · Practice problems
Please enable JavaScript to use the practice app.