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Number Properties and Divisibility
Even, Odd, and the Integers
The integers consist of all whole numbers, their negatives, and zero: $\ldots, -3, -2, -1, 0, 1, 2, 3, \ldots$. Within the positive integers, every number is either even numbers or odd numbers.
$$\text{Even}: n = 2k \quad | \quad \text{Odd}: n = 2k + 1 \quad (k \in \mathbb{Z})$$
Even numbers are multiples of 2; odd numbers leave a remainder of 1 when divided by 2.
$k$=Any integer(integer)
$2k$=Generic even number(integer)
$2k + 1$=Generic odd number(integer)
$k = 0$
→$2k = 0$, confirming that 0 is even.
$n = 1$
→1 is the smallest (and only odd) positive integer that is neither prime numbers nor composite numbers.
Addition Rules: Even + Even = Even, Odd + Odd = Even, Even + Odd = Odd.
Multiplication Rules: Even × Any = Even, Odd × Odd = Odd.
Powers: Odd raised to any power stays odd; even raised to any positive power stays even.
Prime Numbers and Factorization
A prime numbers is a positive integer greater than 1 whose only divisibility factors are 1 and itself. Every integer greater than 1 can be uniquely broken down into a product of primes — this is the prime factorization.
$$N = p_1^{a} \times p_2^{b} \times p_3^{c} \times \cdots$$
Any integer $N > 1$ can be expressed as a unique product of prime powers (Fundamental Theorem of Arithmetic).
$p_1, p_2, p_3, \ldots$=Distinct prime factors(prime numbers)
$a, b, c, \ldots$=Exponents (powers) of each prime(non-negative integers)
First Few Primes: 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47...
2 is Special: 2 is the only even prime and the smallest prime. All other primes are odd.
Composite Numbers: Any integer > 1 that is not prime is composite numbers — it has at least three factors.
Counting Factors: If $N = p_1^{a} \times p_2^{b}$, then the total number of factors is $(a+1)(b+1)$.
To find the prime factorization of any number, repeatedly divide by the smallest prime that divides it. The process stops when every factor at the bottom is prime.
$$84 = 2 \times 2 \times 3 \times 7 = 2^2 \times 3 \times 7$$
Breaking 84 into its prime factors: divide by 2 twice, then by 3, leaving 7.
$84$=The number being factorized(unitless)
Division Ladder: Write $N$, divide by 2 as many times as possible, then 3, then 5, 7, 11... until you reach 1.
Shortcut for Large Numbers: Check divisibility by small primes first (2, 3, 5, 7, 11) before moving to larger ones.
Use in LCM/HCF: Once you have prime factorizations, finding LCM and HCF becomes a matter of comparing exponents.
Divisibility Rules
Divisibility rules let you quickly determine whether one number divides another without performing full division. These are essential tools for working with prime factorization, finding HCF and LCM, and solving remainder-based problems.
$$\text{A number } N \text{ is divisible by } d \iff N \mod d = 0$$
A number is divisible by $d$ if it leaves zero remainder when divided by $d$.
$N \mod d$=Remainder when $N$ is divided by $d$(unitless)
By 2: Last digit is even (0, 2, 4, 6, 8).
By 3: Sum of digits is divisible by 3. For $471$: $4 + 7 + 1 = 12$, and 12 is divisible by 3, so 471 is too.
By 4: Last two digits form a number divisible by 4. For $3{,}516$: check $16 \div 4 = 4$ ✓
By 5: Last digit is 0 or 5.
By 6: Must satisfy both rules for 2 and 3 simultaneously.
By 8: Last three digits form a number divisible by 8.
By 9: Sum of digits is divisible by 9 (similar to rule for 3 but stricter).
By 10: Last digit is 0.
By 11: Alternate sum of digits (sum of odd-position digits minus sum of even-position digits) is 0 or divisible by 11. For $2{,}816$: $(2+1)-(8+6) = 3-14 = -11$ ✓
By 12: Must satisfy both rules for 3 and 4 simultaneously.
LCM and HCF
The HCF (also called GCD or Greatest Common Divisor) is the largest number that divides two or more given numbers exactly. It represents the greatest shared prime factorization among the numbers.
$$\text{HCF}(12, 18) = \text{HCF}(2^2 \times 3,\; 2 \times 3^2) = 2^1 \times 3^1 = 6$$
Take the prime factorization of each number and pick the LOWEST power of every common prime.
$\text{HCF}$=Highest Common Factor (Greatest Common Divisor)(unitless)
Prime Factorization Method: Express each number as a product of prime powers. For each common prime, take the minimum exponent.
Division Method: Repeatedly divide by common factors until no common factor remains. The product of all common divisors is the HCF.
HCF of Three or More Numbers: Extend the prime factorization method — take the minimum exponent of each prime that appears in ALL numbers.
Relationship to GCD: HCF = GCD = Greatest Common Divisor. These are all the same concept.
The LCM is the smallest positive number that is a multiple of two or more given numbers. It represents the least value that all the given numbers divide into without leaving a remainder.
$$\text{LCM}(12, 18) = \text{LCM}(2^2 \times 3,\; 2 \times 3^2) = 2^2 \times 3^2 = 36$$
Take the prime factorization of each number and pick the HIGHEST power of every prime that appears in any number.
$\text{LCM}$=Least Common Multiple(unitless)
Prime Factorization Method: For each prime appearing in any factorization, take the maximum exponent across all numbers and multiply.
Division Method: Write numbers in a row, divide by common primes, carry down undivided numbers. Multiply all divisors and remaining numbers.
Use Case — Scheduling: If Event A repeats every 12 days and Event B every 18 days, both coincide every $\text{LCM}(12, 18) = 36$ days.
HCF-LCM Relationship: For two numbers $a$ and $b$, the product $a \times b = \text{HCF}(a,b) \times \text{LCM}(a,b)$. This also extends to three numbers only in a modified form.
LCM and HCF of Fractions
When dealing with HCF and LCM of fractions, you can reduce the problem to working with the numerators and denominators separately. The formulas are mirror images of each other — HCF swaps what LCM keeps.
$$\text{HCF}\!\left(\frac{a}{b}, \frac{c}{d}\right) = \frac{\text{HCF}(a, c)}{\text{LCM}(b, d)} \quad | \quad \text{LCM}\!\left(\frac{a}{b}, \frac{c}{d}\right) = \frac{\text{LCM}(a, c)}{\text{HCF}(b, d)}$$
For fractions, HCF uses HCF of numerators and LCM of denominators. LCM uses LCM of numerators and HCF of denominators.
$\frac{a}{b}, \frac{c}{d}$=Two fractions in lowest terms(unitless)
Memory Trick: HCF takes the HCF 'upstairs' and LCM 'downstairs'. LCM takes the LCM 'upstairs' and HCF 'downstairs'. It's a cross-swap.
Must Be in Lowest Terms: Convert fractions to their simplest form before applying these formulas, otherwise the results may be incorrect.
Business Application: If two tasks take $\frac{3}{4}$ hour and $\frac{5}{6}$ hour respectively, and they repeat in cycles, the LCM of the fractions tells you when both cycles align.
Remainder Problems and Number Patterns
When a number $N$ is divided by a divisor $d$, it can be expressed in terms of the remainder $r$ and the quotient $q$. Understanding how remainders behave under addition, subtraction, and multiplication is essential for solving competitive-style problems.
$$N = q \times d + r, \quad \text{where } 0 \le r < d$$
Every integer division produces a quotient and a remainder that are uniquely determined.
$N$=The dividend (number being divided)(unitless)
$q$=Quotient (how many times $d$ fits into $N$)(integer)
$r$=Remainder (what's left over)(non-negative integer)
Addition Rule: $(a + b) \mod d = [(a \mod d) + (b \mod d)] \mod d$.
Multiplication Rule: $(a \times b) \mod d = [(a \mod d) \times (b \mod d)] \mod d$.
Negative Remainders: When $r < 0$, add $d$ to make it positive. A remainder of $-2$ mod $7$ is equivalent to $5$ mod $7$.
Power Shortcut: To find $a^n \mod d$, look for a cyclical pattern in $a^1, a^2, a^3, \ldots \mod d$. The pattern always repeats.
Many problems involve identifying repeating patterns in digits, remainders, or unit places of powers. These number series patterns rely on modular arithmetic and cyclical behavior of numbers.
$$\text{Units digit of } 7^n: \quad 7, 9, 3, 1, 7, 9, 3, 1, \ldots \text{ (cycle of 4)}$$
The units digit of powers of any number follows a fixed cycle. For 7, the cycle is 7 → 9 → 3 → 1, repeating every 4 powers.
$n$=The exponent(positive integer)
Units Digit Cycles: 2 and 8 have cycle 4; 3 and 7 have cycle 4; 4 and 9 have cycle 2; 5 and 6 always give 5 and 6; 0 always gives 0.
Finding Position in Cycle: To find the units digit of $a^n$, compute $n \mod \text{cycle length}$, then look up that position in the cycle.
Sum of Digits Patterns: Numbers divisible by 9 have digit sums that are multiples of 9. The remainder of any number divided by 9 equals the remainder of its digit sum divided by 9.
Application Strategy
Worked Examples
Problem Statement
A factory produces batches of 24, 36, and 60 units. Each batch must be repacked into identical smaller boxes with no leftovers. What is the largest possible box size, and what is the smallest total order that uses full boxes of each batch size?
prime factorizationHCFLCM
Strategic Toolkit
$$\text{HCF} = 2^{\min} \times 3^{\min} \qquad \text{LCM} = 2^{\max} \times 3^{\max} \times 5^{\max}$$
Step 1
Prime factorize all three numbers
Break each batch size into its prime factors.
$$24 = 2^3 \times 3, \quad 36 = 2^2 \times 3^2, \quad 60 = 2^2 \times 3 \times 5$$
Step 2
Find HCF using minimum exponents
Take the lowest power of each prime common to ALL three numbers. Only 2 and 3 appear in all three.
$$\text{HCF} = 2^{\min(3,2,2)} \times 3^{\min(1,2,1)} = 2^2 \times 3 = 12$$
Step 3
Find LCM using maximum exponents
Take the highest power of each prime that appears in ANY number.
$$\text{LCM} = 2^{\max(3,2,2)} \times 3^{\max(1,2,1)} \times 5^{\max(0,0,1)} = 2^3 \times 3^2 \times 5 = 360$$
Step 4
Interpret the results
The largest box size is 12 units. The smallest order using full boxes of all three batch sizes is 360 units.
$$\text{Verify: } 360 \div 24 = 15, \quad 360 \div 36 = 10, \quad 360 \div 60 = 6 \quad \checkmark$$
Problem Statement
A store manager needs to check whether a shipment of 4,896 items can be split equally into groups of 6, 8, 9, and 11 for different retail outlets. Which divisors work?
divisibility
Strategic Toolkit
$$d \mid N \iff N \mod d = 0$$
Step 1
Test divisibility by 6
A number is divisible by 6 if it satisfies the rules for both 2 and 3.
$$\text{Rule for 2: } \text{last digit is 6 (even)} \checkmark \quad | \quad \text{Rule for 3: } 4+8+9+6 = 27, \; 27 \div 3 = 9 \checkmark$$
Step 2
Test divisibility by 8
Check if the last three digits form a number divisible by 8.
$$896 \div 8 = 112 \checkmark \quad \Rightarrow \quad 896 \text{ is divisible by } 8$$
Step 3
Test divisibility by 9
Check if the sum of all digits is divisible by 9.
$$4 + 8 + 9 + 6 = 27, \quad 27 \div 9 = 3 \checkmark$$
Step 4
Test divisibility by 11
Use the alternating digit sum rule for 11.
$$(4+9) - (8+6) = 13 - 14 = -1 \quad | \quad -1 \text{ is NOT divisible by } 11 \quad \Rightarrow \quad 4{,}896 \text{ fails}$$
Step 5
Conclusion
4,896 can be split into equal groups of 6, 8, and 9, but NOT 11.
Problem Statement
A restaurant restocks vegetables every 4 days, meat every 6 days, and cleaning supplies every 15 days. If all three are restocked on Monday, after how many days will they all be restocked on the same day again? What day of the week will it be?
LCM
Strategic Toolkit
$$\text{LCM}(4, 6, 15) = \frac{4 \times 6 \times 15}{\text{HCF adjustments}}$$
Step 1
Prime factorize the cycle lengths
Break each restocking interval into prime factors.
$$4 = 2^2, \quad 6 = 2 \times 3, \quad 15 = 3 \times 5$$
Step 2
Compute the LCM
Take the maximum exponent of each prime appearing in any factorization.
$$\text{LCM} = 2^2 \times 3^1 \times 5^1 = 4 \times 3 \times 5 = 60 \text{ days}$$
Step 3
Determine the day of the week
60 days after Monday: divide 60 by 7 to find the weekday offset.
$$60 \div 7 = 8 \text{ weeks remainder } 4 \quad \Rightarrow \quad \text{Monday} + 4 = \text{Friday}$$
Step 4
Verify
Check that 60 is divisible by each cycle length.
$$60 \div 4 = 15 \checkmark \quad | \quad 60 \div 6 = 10 \checkmark \quad | \quad 60 \div 15 = 4 \checkmark$$
Problem Statement
A bank's system generates a security code by computing $3^{87} \mod 7$. What is the remainder?
remaindernumber series patterns
Strategic Toolkit
$$a^n \mod d = a^{(n \mod \text{cycle})} \mod d$$
Step 1
Find the cyclical pattern of powers of 3 mod 7
Compute successive powers of 3 modulo 7 until the pattern repeats.
$$3^1 \equiv 3, \quad 3^2 \equiv 2, \quad 3^3 \equiv 6, \quad 3^4 \equiv 4, \quad 3^5 \equiv 5, \quad 3^6 \equiv 1, \quad 3^7 \equiv 3 \ldots$$
Step 2
Identify the cycle length
The pattern $3, 2, 6, 4, 5, 1$ repeats every 6 powers. This is because $3^6 \equiv 1 \mod 7$.
$$\text{Cycle length} = 6$$
Step 3
Reduce the exponent modulo the cycle length
Find where $3^{87}$ falls in the cycle by computing $87 \mod 6$.
$$87 \div 6 = 14 \text{ remainder } 3 \quad \Rightarrow \quad 87 \mod 6 = 3$$
Step 4
Look up the 3rd position in the cycle
Position 3 in the cycle $\{3, 2, 6, 4, 5, 1\}$ gives the answer.
$$3^{87} \equiv 3^3 \equiv 6 \mod 7 \quad \Rightarrow \quad \text{Remainder} = 6$$
Problem Statement
Two workers complete their tasks in $\frac{4}{9}$ hour and $\frac{8}{15}$ hour respectively. If they work in repeating cycles, find the LCM and HCF of their cycle times.
LCMHCFfraction
Strategic Toolkit
$$\text{HCF}\!\left(\frac{a}{b}, \frac{c}{d}\right) = \frac{\text{HCF}(a,c)}{\text{LCM}(b,d)} \quad | \quad \text{LCM}\!\left(\frac{a}{b}, \frac{c}{d}\right) = \frac{\text{LCM}(a,c)}{\text{HCF}(b,d)}$$
Step 1
Verify fractions are in lowest terms
Check that HCF of numerator and denominator is 1 for each fraction.
$$\frac{4}{9}: \text{HCF}(4,9) = 1 \checkmark \quad | \quad \frac{8}{15}: \text{HCF}(8,15) = 1 \checkmark$$
Step 2
Compute HCF of the fractions
HCF of numerators divided by LCM of denominators (cross-swap).
$$\text{HCF}\!\left(\frac{4}{9}, \frac{8}{15}\right) = \frac{\text{HCF}(4, 8)}{\text{LCM}(9, 15)} = \frac{4}{45}$$
Step 3
Compute LCM of the fractions
LCM of numerators divided by HCF of denominators (cross-swap).
$$\text{LCM}\!\left(\frac{4}{9}, \frac{8}{15}\right) = \frac{\text{LCM}(4, 8)}{\text{HCF}(9, 15)} = \frac{8}{3}$$
Step 4
Interpret the results
The HCF ($\frac{4}{45}$ hour = $\frac{4}{45} \times 60 = 5.33$ minutes) is the largest common sub-cycle. The LCM ($\frac{8}{3}$ hours = 2 hours 40 minutes) is when both cycles align.
Memorize formulas
HCF × LCM Identity
For exactly two numbers $a$ and $b$: use this to find one when the other is known.
Formula
$$a \times b = \text{HCF}(a,b) \times \text{LCM}(a,b)$$
HCF of Fractions
Cross-swap: HCF upstairs, LCM downstairs.
Formula
$$\text{HCF}\!\left(\frac{a}{b}, \frac{c}{d}\right) = \frac{\text{HCF}(a,c)}{\text{LCM}(b,d)}$$
LCM of Fractions
Cross-swap: LCM upstairs, HCF downstairs.
Formula
$$\text{LCM}\!\left(\frac{a}{b}, \frac{c}{d}\right) = \frac{\text{LCM}(a,c)}{\text{HCF}(b,d)}$$
Total Number of Factors
If $N = p_1^{a} \times p_2^{b} \times \cdots$, the total count of factors is the product of (exponent + 1) for each prime.
Formula
$$\text{Factors}(N) = (a+1)(b+1)(c+1) \cdots$$
Remainder Multiplication Rule
To find the remainder of a product, multiply the individual remainders and mod again.
Formula
$$(a \times b) \mod d = [(a \mod d) \times (b \mod d)] \mod d$$
Review
1
Which of the following numbers is composite (not prime)?
Ans
51
Review
Tap to reveal
Test divisibility by small primes
Check each number for divisibility by 2, 3, 5, and 7.
$$51: \text{ digit sum } = 6, \; 6 \div 3 = 2 \implies 51 = 3 \times 17$$
2
If $a$ is an odd integer, which of the following expressions is always even?
Ans
$a(a + 1)$
Review
Tap to reveal
Determine the parity of each factor
Since $a$ is odd, $a + 1$ must be even.
$$a \text{ is odd} \implies a + 1 \text{ is even}$$
Evaluate the product
The product of an odd and an even integer is always even.
$$a(a+1) = \text{odd} \times \text{even} = \text{even}$$
3
A warehouse stores 180 identical crates. The manager wants to arrange them in equal rows for a display. How many different arrangements (different possible numbers of rows) are there?
Ans
18
Review
Tap to reveal
Prime factorize 180
Break 180 into its prime factorization.
$$180 = 2^2 \times 3^2 \times 5^1$$
Apply the factor-counting formula
For $N = p_1^{a} p_2^{b} p_3^{c}$, the number of factors is $(a+1)(b+1)(c+1)$.
$$(2+1)(2+1)(1+1) = 3 \times 3 \times 2 = 18$$
4
Three delivery trucks depart from a warehouse. Truck A leaves every 12 minutes, Truck B every 18 minutes, and Truck C every 30 minutes. If all three depart together at 8:00 AM, when will they next depart simultaneously?
Ans
11:00 AM
Review
Tap to reveal
Find the LCM of the three intervals
Prime factorize each interval and take the highest power of every prime.
$$\text{LCM}(12, 18, 30) = 2^2 \times 3^2 \times 5 = 180 \text{ minutes}$$
Convert to hours and add to the start time
$180 \div 60 = 3$ hours. Add to 8:00 AM.
$$8{:}00 \text{ AM} + 3 \text{ hours} = 11{:}00 \text{ AM}$$
5
Two positive integers have an HCF of 12 and an LCM of 180. If one number is 36, what is the other number?
Ans
60
Review
Tap to reveal
Apply the HCF $\times$ LCM identity
Set the product of the two numbers equal to the product of their HCF and LCM.
$$36 \times b = 12 \times 180$$
Solve for the unknown number
Divide both sides by 36 to isolate $b$.
$$b = \frac{12 \times 180}{36} = \frac{2160}{36} = 60$$
6
What is the HCF of the fractions $\dfrac{2}{3}$ and $\dfrac{4}{9}$?
Ans
$\dfrac{2}{9}$
Review
Tap to reveal
Apply the HCF of fractions formula
Compute the HCF of numerators and the LCM of denominators separately.
$$\text{HCF}\!\left(\frac{2}{3},\, \frac{4}{9}\right) = \frac{\text{HCF}(2, 4)}{\text{LCM}(3, 9)} = \frac{2}{9}$$
7
What is the remainder when $3^{25}$ is divided by 7?
Ans
3
Review
Tap to reveal
Find the repeating cycle of $3^n \bmod 7$
Compute successive powers until the remainder returns to 1.
$$3,\; 2,\; 6,\; 4,\; 5,\; 1 \pmod{7} \quad \text{(cycle length = 6)}$$
Reduce the exponent using the cycle
Write $25 = 6 \times 4 + 1$ and use $3^{6k + r} \equiv 3^r \pmod{7}$.
$$3^{25} \equiv 3^{6 \times 4 + 1} \equiv (3^6)^4 \times 3 \equiv 1 \times 3 \equiv 3 \pmod{7}$$
8
A merchant has 120 kg of rice, 180 kg of wheat, and 300 kg of sugar. He wants to pack these into bags of equal weight, each bag holding only one type of grain. What is the maximum possible weight of each bag so that no grain is left over?
Ans
60 kg
Review
Tap to reveal
Prime factorize all three quantities
Express each weight as a product of prime factors to find common divisors.
$$120 = 2^3 \!\times\! 3 \!\times\! 5, \quad 180 = 2^2 \!\times\! 3^2 \!\times\! 5, \quad 300 = 2^2 \!\times\! 3 \!\times\! 5^2$$
Compute the HCF
Take the minimum exponent for each prime common to all three factorizations.
$$\text{HCF} = 2^2 \times 3 \times 5 = 60 \text{ kg}$$
9
A game show calculates a contestant's score as $17 \times 23 \times 31$. The total is then divided into groups of 8 to award bonus prizes. How many points are left over after forming complete groups of 8?
Ans
1
Review
Tap to reveal
Find each factor's remainder modulo 8
Compute $17 \bmod 8$, $23 \bmod 8$, and $31 \bmod 8$ separately.
$$17 \bmod 8 = 1, \quad 23 \bmod 8 = 7, \quad 31 \bmod 8 = 7$$
Multiply remainders and reduce modulo 8
Apply the remainder multiplication rule: multiply the individual remainders, then take the result mod 8.
$$(1 \times 7 \times 7) \bmod 8 = 49 \bmod 8 = 1$$
10
A product serial number is the 5-digit number 43,27$d$ where $d$ is the units digit. The serial number must be divisible by 6. What is the largest possible value of $d$?
Ans
8
Review
Tap to reveal
Apply the divisibility rules for 2 and 3
$d$ must be even, and the digit sum $16 + d$ must be divisible by 3.
$$d \in \{0, 2, 4, 6, 8\}, \quad 16 + d \equiv 0 \pmod{3}$$
Test candidates and select the largest
Check each even digit against the digit-sum condition.
$$d = 2: 18 \checkmark \qquad d = 8: 24 \checkmark \qquad \text{Largest} = 8$$
11
What is the LCM of the fractions $\dfrac{1}{4}$ and $\dfrac{1}{6}$?
Ans
$\dfrac{1}{2}$
Review
Tap to reveal
Apply the LCM of fractions formula
Compute the LCM of numerators and the HCF of denominators separately.
$$\text{LCM}\!\left(\frac{1}{4},\, \frac{1}{6}\right) = \frac{\text{LCM}(1, 1)}{\text{HCF}(4, 6)} = \frac{1}{2}$$
Question 1 of 11
0:00
1
Which of the following numbers is composite (not prime)?