Number System

Numbers are grouped step by step. Natural numbers (N): 1, 2, 3... Whole numbers (W): 0, 1, 2, 3... (just add 0 to N). Integers (Z): ...-2, -1, 0, 1, 2... (positive, negative and zero). Rational numbers (Q): any number written as p/q where q is not 0, like 3/4, 0.5, -7. Irrational numbers: cannot be written as p/q, like root2, pi, root3 (non-terminating, non-repeating decimals). Real numbers (R): all rational + irrational together. Prime: exactly two factors (1 and itself), e.g. 2, 3, 5, 7. Composite: more than two factors, e.g. 4, 6, 8, 9. Remember: 1 is NEITHER prime nor composite, and 2 is the only EVEN prime.

Even numbers are divisible by 2 (0, 2, 4, 6...); odd numbers are not (1, 3, 5, 7...). Key memory tricks: even + even = even, odd + odd = even, even + odd = odd. even x anything = even; odd x odd = odd. Zero (0) is an even number, is a whole number, and is neither positive nor negative. The smallest prime is 2; the smallest composite is 4; the smallest odd prime is 3. Co-prime numbers have HCF = 1 (e.g. 8 and 15) even if individually composite. Consecutive integers always alternate even-odd.

Counting primes in a range is one of those SSC CHSL questions that looks like it needs memorisation but actually rewards a quick divisibility test. Master the under-100 prime check once and you will never freeze on questions like "How many primes lie between 10 and 30?" or "How many primes are there between 50 and 70?"

Definition: A prime number is a natural number greater than 1 that has exactly two distinct positive divisors — 1 and itself. The first few primes are 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31...

Definition: A composite number is a natural number greater than 1 that has more than two positive divisors. 1 is neither prime nor composite.

The worked example

Question: How many prime numbers lie between 10 and 30?

Solution:
Step 1: List all integers from 11 to 29 (the strictly-between interpretation): 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29.
Step 2: Skip every even number — they are divisible by 2 and so cannot be prime (except 2 itself, which is not in this range). That eliminates 12, 14, 16, 18, 20, 22, 24, 26, 28.
Step 3: From the odd numbers left — 11, 13, 15, 17, 19, 21, 23, 25, 27, 29 — eliminate multiples of 3: 15 (3×5), 21 (3×7), 27 (3×9).
Step 4: Eliminate multiples of 5: 25 (5×5).
Step 5: What remains: 11, 13, 17, 19, 23, 29. Each is checked against 7 — none is divisible by 7. So all six are prime.
Conclusion: There are 6 primes between 10 and 30.

Answer: 6.

Why this technique works — the √n rule

Why it matters: to test whether a number n is prime, you do NOT have to divide by every integer from 2 to n−1. You only have to try divisors up to √n. The reason is that if n has any divisor greater than √n, it must also have a paired divisor smaller than √n (because divisors come in pairs whose product is n). So if you find no small divisor, there is no large one either.

For n < 100, √n < 10, so the only primes you must check as potential divisors are 2, 3, 5 and 7. That is it. This is the single most useful shortcut in SSC prime-counting questions.

For n < 121 (since 11² = 121), still 2, 3, 5, 7 suffice. For n < 169 (13² = 169), include 11. For n < 289 (17² = 289), include 13. Memorise these tiers.

The under-100 cheat list

Real-world example: every SSC aspirant should be able to recite the 25 primes under 100 from memory:

2, 3, 5, 7, 11, 13, 17, 19, 23, 29,
31, 37, 41, 43, 47,
53, 59, 61, 67,
71, 73, 79,
83, 89,
97.

That is 25 primes — a number worth memorising. Then:

• 1–10: 4 primes (2, 3, 5, 7)
• 11–20: 4 primes (11, 13, 17, 19)
• 21–30: 2 primes (23, 29)
• 31–40: 2 primes (31, 37)
• 41–50: 3 primes (41, 43, 47)
• 51–60: 2 primes (53, 59)
• 61–70: 2 primes (61, 67)
• 71–80: 3 primes (71, 73, 79)
• 81–90: 2 primes (83, 89)
• 91–100: 1 prime (97)

Notice the pattern flips on 91 = 7 × 13. Many students wrongly call 91 prime — it is a notorious SSC trap.

Another worked example

Question: How many prime numbers lie between 50 and 70?

Solution:
Step 1: Numbers 51 through 69.
Step 2: Strip out evens — leaves 51, 53, 55, 57, 59, 61, 63, 65, 67, 69.
Step 3: Strip out multiples of 3 (digit sum divisible by 3): 51 (5+1=6), 57 (5+7=12), 63 (6+3=9), 69 (6+9=15). Removes 51, 57, 63, 69.
Step 4: Strip out multiples of 5: 55, 65. Both end in 5.
Step 5: Remaining: 53, 59, 61, 67. Check divisibility by 7: 53/7 ≈ 7.5, 59/7 ≈ 8.4, 61/7 ≈ 8.7, 67/7 ≈ 9.6 — none divides evenly.
Conclusion: Primes between 50 and 70 are 53, 59, 61, 67 — that is 4 primes.

Common misconception

Common misconception: "1 is a prime number." Wrong. By definition a prime must have exactly two distinct divisors. The number 1 has only one divisor (itself), so it is neither prime nor composite. SSC has tested this.

A second misconception: "All odd numbers greater than 1 are prime." Wrong. 9 = 3×3, 15 = 3×5, 21 = 3×7, 25 = 5×5, 27 = 3×9, 33 = 3×11, 35 = 5×7, 39 = 3×13, 49 = 7×7, 51 = 3×17, 55 = 5×11, 57 = 3×19, 63 = 9×7, 65 = 5×13, 69 = 3×23, 77 = 7×11, 85 = 5×17, 87 = 3×29, 91 = 7×13, 93 = 3×31, 95 = 5×19. All are odd composites.

A third misconception: "2 is not prime because it's even." Wrong. 2 is prime — it is the only even prime, and the smallest prime.

Quick divisibility tests for the under-100 check

To speed up prime testing, drill these one-glance tests:

• By 2 — last digit is 0, 2, 4, 6 or 8.
• By 3 — sum of digits is divisible by 3.
• By 5 — last digit is 0 or 5.
• By 7 — no simple universal rule; just divide. For two-digit numbers, multiples of 7 are 14, 21, 28, 35, 42, 49, 56, 63, 70, 77, 84, 91, 98 — worth memorising.

If a number under 100 escapes all four tests, it is prime.

:::compare

Number	Test trail	Prime?
11	Not even; 1+1=2 not /3; not end in 0/5; 11/7≠int	Yes
91	Not even; 9+1=10 not /3; not end in 0/5; 91/7=13	No (trap!)
51	Not even; 5+1=6 /3 ✓	No
87	Not even; 8+7=15 /3 ✓	No
97	Not even; 9+7=16 not /3; not end in 0/5; 97/7≈13.86	Yes
:::

:::keypoints

• A prime > 1 has exactly two divisors — 1 and itself.
• 1 is neither prime nor composite; 2 is the only even prime.
• To test primality of n < 100, check divisibility by 2, 3, 5, 7 only.
• The √n rule generalises this — divisors come in pairs around √n.
• 25 primes lie under 100; memorise them in tens for instant recall.
• Watch out for 91 = 7 × 13 — a classic SSC composite-pretending-to-be-prime.
• Primes between 10 and 30 = {11, 13, 17, 19, 23, 29} = 6 primes.
• Primes between 50 and 70 = {53, 59, 61, 67} = 4 primes.
  :::

:::memory
"2-3-5-7 BELOW 100" — that is the only divisibility set you need to test any two-digit candidate. Drill the 25 primes under 100 like a multiplication table.
:::

:::recap

• A prime has exactly two divisors; 1 is not prime, 2 is the only even prime.
• For numbers under 100, divisibility by 2, 3, 5 and 7 is sufficient to decide primality.
• There are 6 primes between 10 and 30: 11, 13, 17, 19, 23, 29.
• 91 is composite (7 × 13) — never let it fool you.
  :::

By 2: last digit is even (0,2,4,6,8). By 3: digit sum divisible by 3. By 4: last two digits form a number divisible by 4. By 5: ends in 0 or 5. By 6: divisible by both 2 and 3. By 8: last three digits divisible by 8. By 9: digit sum divisible by 9. By 10: ends in 0. By 11: difference between sum of digits at odd places and even places is 0 or a multiple of 11. By 25: last two digits are 00, 25, 50 or 75. Memory trick: for 6, 12, 15, 18 break into co-prime factors and check each (e.g. 12 = 4 and 3).

Long division in an SSC CHSL exam hall costs you 60 seconds you do not have. The divisibility rule for 11 crunches the same check into 8 seconds, no calculator, no scratch — purely a digit-by-digit dance.

Definition: A number is divisible by 11 if the difference between the sum of its digits in odd positions and the sum of its digits in even positions (counted from the right) is either 0 or a multiple of 11 (such as ±11, ±22, ±33).

Definition: "Position from the right" means the units digit is position 1, tens digit is position 2, hundreds digit is position 3, and so on. Odd positions are 1, 3, 5, ... (units, hundreds, ten-thousands). Even positions are 2, 4, 6, ... (tens, thousands, hundred-thousands).

The mechanical drill — alternate + and − from the right

Take any number. Starting from the rightmost (units) digit, label the digits with alternating signs: +, −, +, −, +, ... Then add together all the digits with the + sign, add together all the digits with the − sign, and find the difference. If the difference is 0 or any multiple of 11, the original number is divisible by 11.

Worked example as in the source: 41809.

Step 1: Label from the right: 9(+), 0(−), 8(+), 1(−), 4(+).
Step 2: Plus-positions sum = 9 + 8 + 4 = 21.
Step 3: Minus-positions sum = 0 + 1 = 1.
Step 4: Difference = 21 − 1 = 20.
Step 5: Is 20 a multiple of 11? No — the multiples are 0, 11, 22, 33, ...
Conclusion: 41809 is not divisible by 11.

Now a second drill, this time with a number that is divisible: 482218.

Step 1: From the right: 8(+), 1(−), 2(+), 2(−), 8(+), 4(−).
Step 2: Plus-positions sum = 8 + 2 + 8 = 18.
Step 3: Minus-positions sum = 1 + 2 + 4 = 7.
Step 4: Difference = 18 − 7 = 11. Yes, 11 is a multiple of 11.
Conclusion: 482218 is divisible by 11. Verify: 482218 ÷ 11 = 43838. Clean.

Why the rule works — the algebra behind the trick

This is one of those magical-looking rules that has a perfectly logical explanation. Consider what each power of 10 does mod 11:

10^0 = 1 ≡ +1 (mod 11)
10^1 = 10 ≡ −1 (mod 11)
10^2 = 100 ≡ +1 (mod 11)
10^3 = 1000 ≡ −1 (mod 11)
10^4 = 10000 ≡ +1 (mod 11)

Notice the pattern: every power of 10 alternates between +1 and −1 (mod 11). So a number d_n d_{n−1} ... d_1 d_0, when reduced mod 11, becomes

(d_0)(+1) + (d_1)(−1) + (d_2)(+1) + (d_3)(−1) + ...

That is exactly the alternating sum from the right. The number is divisible by 11 if and only if this alternating sum is ≡ 0 (mod 11). The "rule" is just modular arithmetic in disguise.

You do not need to remember this proof for the exam, but understanding it once cements the rule for life. You stop fearing it because you know why it works.

A reverse problem — fill in the missing digit

Examiners love the variant: "What digit should replace * in 5*132 to make it divisible by 11?"

Question: Find the digit * in 5*132 such that the number is divisible by 11.

Solution:
Step 1: Label from the right: 2(+), 3(−), 1(+), *(−), 5(+).
Step 2: Plus-sum = 2 + 1 + 5 = 8.
Step 3: Minus-sum = 3 + *.
Step 4: Difference = 8 − (3 + *) = 5 − *.
Step 5: For divisibility, 5 − * must be 0 or ±11 or ±22, ...
Step 6: * is a single digit (0–9). 5 − * = 0 gives * = 5. 5 − * = −11 gives * = 16 (not allowed). 5 − * = 11 gives * = −6 (not allowed).
Conclusion: The missing digit is 5, giving 55132. Verify: 55132 ÷ 11 = 5012, exact.

A second variant — three-digit palindromes

A neat insight: every three-digit palindrome with the same first and last digit and any middle digit has a clean divisibility pattern with 11.

Take aba (e.g., 525). Label from the right: a(+), b(−), a(+). Plus-sum = 2a, minus-sum = b. Difference = 2a − b. So aba is divisible by 11 iff 2a − b ≡ 0 (mod 11). For 525: 2(5) − 2 = 8, not a multiple of 11. For 484: 2(4) − 8 = 0, yes — divisible. 484 ÷ 11 = 44. Clean.

Why it matters: SSC CHSL, SSC CGL, RRB NTPC, and bank PO/clerk papers almost always have a number-system question testing divisibility by 7, 11, or 13. The rule for 11 is the single most common one because the alternating-sum trick is elegant and unique. Add to that 2–3 marks from algebra/quant questions that secretly use divisibility (e.g., "Find the smallest multiple of 11 with...") and you are looking at 4–5 marks per Tier-I that hinge on this one tool.

Real-world parallel

Real-world example: ISBN-10 codes use a divisibility-by-11 checksum. If you weight the first 9 digits of an ISBN by their position and check that the total is divisible by 11 (with X representing 10 if needed), the ISBN is valid. The same modular-arithmetic logic underlies your everyday book barcodes.

Banking sector also uses similar mod-checks. The MICR code on Indian cheques does not specifically use mod-11, but the same family of weighted-sum checks (mod 7 in some cases) catches data-entry errors before they reach the clearing house.

Common misconceptions, cleared

Common misconception: "Start counting from the left." Wrong. The rule pairs each digit with the right power of 10, and 10^0 lives at the rightmost digit. Counting from the left flips the sign pattern and gives you the negative of the correct difference — which, fortunately, is still ±0 or ±11k, so the test still works. But to avoid confusion, always start from the right.

Common misconception: "If the alternating sum is positive, the number is divisible by 11." Sign does not matter — only whether the absolute value is 0 or a multiple of 11. So both +11 and −11 confirm divisibility.

Common misconception: "I have to compute the difference exactly." You only need to know it mod 11. If you spot 22, you can stop and declare "divisible." If you spot 18, you can stop and declare "not divisible." You never need long division.

Common misconception: "The rule works only for small numbers." It works for any number, however large. Try 1331 (which is 11^3): 1(−) + 3(+) + 3(−) + 1(+) (from right) = (1+3) − (3+1) = 0. Divisible. Try 14641 (which is 11^4): from right 1(+), 4(−), 6(+), 4(−), 1(+) = (1+6+1) − (4+4) = 8 − 8 = 0. Divisible. The rule scales effortlessly.

A 30-second drill template

In your scratch sheet, write the number once, draw an arrow over each digit with alternating + and − starting from the right, compute the two sums, subtract, and check against 0, ±11, ±22, ±33. That is the entire workflow. With practice you can do it mentally for 5-digit numbers and on paper for 8–9 digit numbers in under 20 seconds.

:::compare

Step	What you do	Time taken
1	Tag digits from right with +, −, +, −, ...	~3 sec
2	Add the "+" digits	~3 sec
3	Add the "−" digits	~3 sec
4	Subtract and compare to multiples of 11	~2 sec
5	Decide divisibility	instant
:::

:::keypoints

• Number is divisible by 11 iff the alternating-sign sum of digits (from the right) is 0 or a multiple of 11.
• Tag positions from the right: units = +, tens = −, hundreds = +, ...
• Sign of the final difference does not matter — only divisibility by 11 does.
• The rule is exact, not approximate, and works for any digit count.
• Underlying reason: 10 ≡ −1 (mod 11), so powers of 10 alternate ±1 mod 11.
• Reverse problems (find the missing digit) reduce to solving one linear equation.
• Used in real-world checksums like ISBN-10.
  :::

:::memory
"Plus-Minus from the Right, Zero or Eleven brings Delight." Picture writing the number, then underlining each digit alternately with a green + and a red − pen, starting from the rightmost.
:::

:::recap

• The rule of 11 is a fast modular-arithmetic shortcut that skips long division entirely.
• Tag digits from the right alternately +, −, sum each group, subtract.
• Difference of 0 or any multiple of 11 means divisible.
• Drill 10 numbers a day and the rule becomes instinctive within a week.
  :::

Q: For what value of x is the number 34x5 divisible by 9? Solution: A number is divisible by 9 if its digit sum is divisible by 9. Digit sum = 3 + 4 + x + 5 = 12 + x. We need 12 + x to be a multiple of 9. The next multiple of 9 after 12 is 18, so 12 + x = 18, giving x = 6. Check: 3465, digit sum = 18, divisible by 9. Answer: x = 6.

HCF (Highest Common Factor) is the largest number that divides all given numbers exactly. LCM (Lowest Common Multiple) is the smallest number that all given numbers divide into exactly. Find them using prime factorisation: HCF = product of common prime factors with the LOWEST powers; LCM = product of all prime factors with the HIGHEST powers. Most important formula: for any two numbers, HCF × LCM = Product of the two numbers. Also, HCF of fractions = HCF of numerators / LCM of denominators, and LCM of fractions = LCM of numerators / HCF of denominators. HCF is always a factor of LCM.

Two students stare at the same word problem. One reads "largest size" and instantly writes "HCF"; the other reads it and freezes. The difference isn't talent — it's a small set of keyword reflexes that turn HCF and LCM word problems into one-line decisions. This lesson hands you those reflexes for SSC CHSL.

Definition: The HCF (Highest Common Factor, also called GCD) of two or more numbers is the largest positive integer that divides each of them exactly.

Definition: The LCM (Least Common Multiple) of two or more numbers is the smallest positive integer that is divisible by each of them.

The Two Mental Pictures

If you remember just one image per concept, you'll never confuse them again:

• HCF = "splitting into the biggest equal groups". You have several quantities; you want to break each down into equal pieces, and you want those pieces to be as large as possible. Whenever the problem speaks of dividing, cutting, distributing equally, or finding "the largest box / largest piece / largest length" — HCF.

• LCM = "coming back together at the same moment". You have several cycles repeating at different intervals; you want to know when they will all synchronise again. Whenever the problem speaks of bells ringing together, lights blinking together, traffic signals matching, or finding "the smallest / least number that is a multiple of each" — LCM.

Keyword Reflex Table

If the question uses any of these words, your hand should already be writing the answer:

• HCF triggers: "largest", "greatest", "maximum", "biggest", "cuts into equal pieces", "divides", "distributes equally so nothing is left over", "longest tape/rope/length".
• LCM triggers: "least", "smallest", "minimum", "together again", "ring at the same time", "simultaneously", "smallest number divisible by".

Classic HCF Problems

Cutting equal pieces

Question: Three ropes are 60 m, 84 m and 132 m long. They are to be cut into pieces of equal length, each piece being as long as possible. Find the length of each piece.

Solution:

Step 1: "As long as possible, equal pieces from several ropes" → HCF.
Step 2: HCF(60, 84, 132). Prime factorise:
60 = 2² × 3 × 5; 84 = 2² × 3 × 7; 132 = 2² × 3 × 11.
Common factors: 2² × 3 = 12.
Step 3: Each piece = 12 m. Number of pieces = 60/12 + 84/12 + 132/12 = 5 + 7 + 11 = 23 pieces.

Conclusion: Each piece is 12 m long; you get 23 pieces in total.

Distributing items equally

Question: 84 chocolates, 126 toffees and 168 biscuits are to be packed into identical gift boxes such that each box contains the same number of each item and no item is left over. Find the maximum number of boxes.

Solution:

Step 1: "Maximum number of identical boxes, no item left over" → HCF.
Step 2: HCF(84, 126, 168) = 42 (you can verify: 84 = 42×2, 126 = 42×3, 168 = 42×4).

Conclusion: 42 boxes, each with 2 chocolates, 3 toffees, 4 biscuits.

Classic LCM Problems

Bells, lights and signals

Question: Three bells toll at intervals of 8, 12 and 18 minutes respectively. They tolled together at 9:00 a.m. When do they next toll together?

Solution:

Step 1: "Toll together again" → LCM.
Step 2: LCM(8, 12, 18). Prime: 8 = 2³, 12 = 2² × 3, 18 = 2 × 3². Take the highest power of each prime: 2³ × 3² = 8 × 9 = 72.
Step 3: 72 minutes after 9:00 a.m. = 10:12 a.m.

Conclusion: They next ring together at 10:12 a.m.

Smallest number with the same remainder

Definition: A number that leaves the same remainder r when divided by a, b, c is of the form LCM(a, b, c) × k + r, where k is any non-negative integer.

Question: Find the smallest number that leaves remainder 4 when divided by 6, 8 and 12.

Solution:

Step 1: "Smallest, same remainder" → LCM + remainder.
Step 2: LCM(6, 8, 12) = 24. Add the remainder: 24 + 4 = 28.

Conclusion: 28. Check: 28 ÷ 6 = 4 rem 4; 28 ÷ 8 = 3 rem 4; 28 ÷ 12 = 2 rem 4. ✓

Smallest number with different remainders that share a pattern

If the divisors leave different remainders r₁, r₂, r₃ but the gap d_i = (divisor_i − r_i) is the same for all, the required smallest number is LCM(a, b, c) − d. This pattern is a CHSL favourite — recognise it by checking that (divisor − remainder) is constant.

The Master Formula

For any two positive integers a and b:

HCF(a, b) × LCM(a, b) = a × b

This lets you find one if you have the other. Useful when the problem hands you, say, HCF = 12 and the two numbers as 36 and 60 — instantly LCM = (36×60)/12 = 180.

For three or more numbers this product-rule does not extend directly (be careful — SSC sometimes plants a trap there). It only works for two numbers at a time.

Why It Matters

HCF/LCM word problems appear in every SSC CHSL Tier-I paper, usually as 1–3 questions in the Quant section. They are pure speed: if you spot the keyword in 5 seconds, the rest is arithmetic. Speed here also feeds into ratios, time-and-work, time-distance, and clock problems — every chapter that secretly leans on LCM under the hood.

Why it matters: A confident HCF/LCM aspirant gains 2–4 marks per shift and finishes each problem in under 45 seconds, leaving more time for the trickier Data Interpretation block later.

Real-world example: The Mumbai BEST bus depot uses a slightly more complex LCM idea to schedule four bus routes that share the same junction. If the routes have headways of 6, 9, 15 and 20 minutes, the simultaneous arrival happens once every LCM(6, 9, 15, 20) = 180 minutes, i.e. once every 3 hours. The timetabling team uses exactly this calculation when planning shared-platform usage.

Common Misconceptions

Common misconception: "HCF is for division, LCM is for multiplication." Not quite — both can be computed from prime factorisation. Use them based on the word problem's intent, not on the arithmetic operation.

Common misconception: "Bigger numbers means LCM." No — the biggest piece comes from HCF, the smallest synchronisation interval comes from LCM. The size of the original numbers doesn't tell you which to use.

Common misconception: "HCF × LCM = product of all numbers, even for three or more." Only for two numbers. For three or more, this product rule fails in general.

Common misconception: "If the problem says 'minimum', use HCF; if 'maximum', use LCM." It's the opposite. Minimum → LCM (smallest common multiple); Maximum → HCF (largest common factor).

:::compare

Clue word in problem	Probably wants	Why
Largest / greatest / maximum length	HCF	Biggest equal piece
Distribute equally, nothing left over	HCF	Largest equal share
Cut into identical pieces	HCF	Largest size that divides each
Smallest / least / minimum number	LCM	Smallest common multiple
Ring / blink / meet together again	LCM	Synchronisation time
Smallest number divisible by a, b, c	LCM	Definition of LCM
Same remainder when divided by a, b, c	LCM + r	LCM gives common multiple
:::

:::keypoints

• HCF answers "biggest equal piece / largest divider".
• LCM answers "smallest common multiple / next time they meet".
• HCF triggers: largest, greatest, maximum, distribute equally, cut.
• LCM triggers: smallest, least, minimum, together again, simultaneous.
• HCF(a,b) × LCM(a,b) = a × b (only for two numbers).
• Number leaving remainder r for divisors a, b, c → LCM(a,b,c) + r.
• If (divisor − remainder) is constant across divisors → LCM(a,b,c) − d.
• Use prime factorisation; pick lowest powers for HCF, highest for LCM.
  :::

:::memory
"H for Huge piece, L for Later meeting." HCF gives the hugest equal piece; LCM gives the next time everything meets again. The first letter of the keyword tells you which to pick.
:::

:::recap

• HCF = largest equal piece / divider (use when keywords are big-sounding).
• LCM = smallest synchronisation / common multiple (use for "together again").
• Master formula HCF × LCM = product works for two numbers only.
• Same-remainder problems: LCM + remainder. Pattern-gap problems: LCM − d.
  :::

Q: Three bells ring at intervals of 9, 12 and 15 minutes. If they ring together at 10:00 am, when will they next ring together? Solution: We need the LCM of 9, 12, 15. Prime factors: 9 = 3², 12 = 2²×3, 15 = 3×5. LCM = 2² × 3² × 5 = 4 × 9 × 5 = 180 minutes = 3 hours. So they ring together again at 10:00 am + 3 hours = 1:00 pm. Answer: 1:00 pm.

The core relation: Dividend = Divisor × Quotient + Remainder. The remainder is always less than the divisor. To find a remainder quickly, you can break the dividend or use remainder properties: the remainder of a sum/product equals the operation on individual remainders. Example: remainder of (17 × 19) ÷ 5 = (remainder of 17÷5) × (remainder of 19÷5) = 2 × 4 = 8, and 8 ÷ 5 leaves 3. So the final remainder is 3. For large powers, find the repeating cycle of remainders. These shortcuts save time versus full long division.

The unit (last) digit of a power repeats in a cycle of at most 4. Digits 0, 1, 5, 6 always keep the same unit digit. For others, the cycle length is: 2 -> (2,4,8,6); 3 -> (3,9,7,1); 7 -> (7,9,3,1); 8 -> (8,4,2,6); these have cycle 4. Digits 4 -> (4,6) and 9 -> (9,1) have cycle 2. Method: divide the power by 4, look at the remainder. If remainder is 1,2,3 take that position in the cycle; if remainder is 0, take the 4th (last) value. Example: unit digit of 7^52 -> 52 ÷ 4 remainder 0 -> 4th value of (7,9,3,1) = 1.

Q: Find the unit digit of 3^47. Solution: The unit digit cycle of 3 is (3, 9, 7, 1) repeating every 4 powers. Divide the exponent 47 by 4: 47 ÷ 4 = 11 remainder 3. A remainder of 3 means we take the 3rd value in the cycle, which is 7. So the unit digit of 3^47 is 7. (Quick check: 3^1=3, 3^2=9, 3^3=27 ends in 7 — matches the 3rd position.) Answer: 7.