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.

Q: How many prime numbers lie between 10 and 30? Solution: List numbers and check each: 11 (prime), 13 (prime), 17 (prime), 19 (prime), 23 (prime), 29 (prime). Skip evens and multiples of 3, 5. So primes are 11, 13, 17, 19, 23, 29 = 6 primes. Answer: 6. Tip: to test if a number under 100 is prime, just check divisibility by 2, 3, 5, 7 only.

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).

For divisibility by 11, mark digits from the right as alternate +, −, +, −. Add the plus-positions, add the minus-positions, then take the difference. If that difference is 0 or a multiple of 11, the number is divisible by 11. Example: 4 1 8 0 9 -> from right: 9(+) 0(−) 8(+) 1(−) 4(+) = (9+8+4) − (0+1) = 21 − 1 = 20, not a multiple of 11, so not divisible. This rule is fast and avoids long division in the exam.

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.

Use HCF for 'largest size / greatest number / maximum that divides' problems — e.g. cutting ropes into equal largest pieces, or distributing items equally. Use LCM for 'least / smallest / together again / minimum' problems — e.g. bells ringing together, traffic lights, or the smallest number that leaves the same remainder. Memory trick: HCF = 'splitting into the biggest equal groups', LCM = 'coming back together at the same time'. If a problem asks for the least number leaving remainder r when divided by a, b, c, answer = LCM(a,b,c) + r.

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.