Number foundations — SSC CGL Quant Notes

Concepts and rules

Natural numbers are 1, 2, 3, …; whole numbers add 0; integers also include the negatives −1, −2, −3, … . Zero is an even integer and is neither positive nor negative.

A prime has exactly 2 positive divisors: 2, 3, 5, 7, 11, 13, … . The number 1 is neither prime nor composite, 2 is the only even prime, and there are 25 primes below 100.

Divisibility by 2, 5 and 10: the last digit is even, is 0 or 5, is 0. By 4: the number formed by the last 2 digits is a multiple of 4. By 8: the number formed by the last 3 digits is a multiple of 8.

By 3 and 9: the digit sum is a multiple of 3 (or of 9). This works because 10 leaves remainder 1 on division by 3 and by 9, so every digit counts at face value.

By 11: (sum of digits in odd places) − (sum of digits in even places), counted from the right, must be 0 or a multiple of 11. Since 1001 = 7 × 11 × 13, the alternating sum of 3-digit groups from the right tests 7, 11 and 13 together.

Composite divisor: split it into coprime factors. Divisible by 72 means divisible by 8 and by 9; by 24 means by 8 and by 3. Splitting 24 as 4 × 6 is wrong because 4 and 6 are not coprime: 12 passes both tests but is not a multiple of 24.

Division algorithm: for a divisor d > 0, N = d × q + r with 0 ≤ r < d. Remainders can be added and multiplied: the remainder of a × b equals the remainder of (remainder of a) × (remainder of b).

If N = pᵃ × qᵇ × rᶜ with p, q, r distinct primes, the number of positive divisors is (a + 1)(b + 1)(c + 1). Odd divisors ignore the power of 2; only perfect squares have an odd count of divisors.

HCF × LCM = product of the numbers holds for 2 positive numbers only. Any 2 consecutive integers are coprime, and the product of any 3 consecutive integers is divisible by 6.

Unit digits of powers repeat with period 4 for bases ending in 2, 3, 7, 8 (2, 4, 8, 6 / 3, 9, 7, 1 / 7, 9, 3, 1 / 8, 4, 2, 6), with period 2 for 4 and 9, and never change for 0, 1, 5, 6.

Method and shortcuts

Missing digits with a composite divisor: write the condition for each coprime factor separately, list the digits 0 to 9 that pass each condition, then keep only the common ones.

Usually fix the last-digit condition first (for 8 or 4) because it allows few choices, then use the digit-sum condition (for 9 or 3) to find the remaining digit.

Remainder of a large power: find a small power of the base that leaves remainder 1 for that divisor, then cut the exponent. For example 2³ = 8 leaves remainder 1 on division by 7, so 2 raised to any multiple of 3 also leaves 1.

Negative remainder trick: for any d ≥ 2, (d − 1)ⁿ leaves remainder 1 on division by d when n is even and d − 1 when n is odd, because d − 1 behaves like −1.

Least number leaving the same remainder r on division by a, b, c is LCM(a, b, c) + r. If each remainder is short of its divisor by the same k, the least number is LCM − k.

Largest number that divides a, b, c leaving the same remainder is the HCF of the differences (b − a), (c − b) and (c − a), because subtracting removes the common remainder.

Successive division: if N leaves r on division by d and the quotient leaves s on division by e, then N = d(e × k + s) + r, so the remainder of N on division by d × e is d × s + r.

Example 1: least digit for divisibility by 3

Question: Find the least digit x for which the number 4x82 is divisible by 3.

Solution: Digit sum = 4 + x + 8 + 2 = 14 + x. The next multiples of 3 after 14 are 15, 18, 21, so x = 1, 4 or 7. The least value is 1, and 4182 = 3 × 1394 confirms it.

Answer: x = 1

Example 2: counting divisors

Question: How many positive divisors does 360 have, and how many of them are odd?

Solution: 360 = 2³ × 3² × 5. Number of divisors = (3 + 1)(2 + 1)(1 + 1) = 4 × 3 × 2 = 24. Odd divisors cannot contain 2, so count only 3² × 5: (2 + 1)(1 + 1) = 6.

Answer: 24 divisors, of which 6 are odd

Example 3: remainder of a large power

Question: Find the remainder when 2¹⁰⁰ is divided by 7.

Solution: 2³ = 8 = 7 + 1 leaves remainder 1. Write 100 = 3 × 33 + 1, so 2¹⁰⁰ = (2³)³³ × 2. The bracket leaves remainder 1, so the whole leaves 1 × 2 = 2.

Answer: 2

Example 4: unit digit of a difference of powers

Question: The unit digit of 7⁹⁵ − 3⁵⁸ is: (a) 6 (b) 2 (c) 4 (d) 8

Solution: For 7 the cycle is 7, 9, 3, 1. 95 = 4 × 23 + 3, so the unit digit of 7⁹⁵ is the 3rd term, 3. For 3 the cycle is 3, 9, 7, 1. 58 = 4 × 14 + 2, so the unit digit of 3⁵⁸ is 9.

The first number is larger, so borrow 10: unit digit = 13 − 9 = 4.

Answer: (c) 4

Example 5: same remainder plus a divisibility condition

Question: The least number which leaves remainder 8 when divided by 12, 15, 20 and 54, and is itself divisible by 11, is: (a) 1088 (b) 1628 (c) 2168 (d) 548

Solution: LCM(12, 15, 20, 54) = 2² × 3³ × 5 = 540, so the number is 540k + 8 for k = 0, 1, 2, … .

540 = 11 × 49 + 1 leaves remainder 1 on division by 11, so 540k + 8 leaves k + 8. We need k + 8 to be a multiple of 11; the least k is 3.

Number = 540 × 3 + 8 = 1628 = 11 × 148. Option (a) 1088 = 540 × 2 + 8 is not divisible by 11, and (d) 548 = 540 + 8 is not either.

Answer: (b) 1628

Example 6: 2 missing digits for divisibility by 72

Question: The 5-digit number 6x24y is divisible by 72, where x and y are digits. The greatest possible value of x + y is: (a) 6 (b) 12 (c) 9 (d) 15

Solution: 72 = 8 × 9 with 8 and 9 coprime. For 8, the last 3 digits 24y must be a multiple of 8: 240 = 8 × 30 and 248 = 8 × 31 work, so y = 0 or 8.

For 9, the digit sum 6 + x + 2 + 4 + y = 12 + x + y must be a multiple of 9. If y = 0, 12 + x = 18 gives x = 6. If y = 8, 20 + x = 27 gives x = 7.

The numbers are 66240 and 67248, giving x + y = 6 or 15. The greatest value is 15.

Answer: (d) 15

Common mistakes

Calling 1 a prime, or forgetting that 2 is prime. Right: 1 is neither prime nor composite; 2 is the only even prime.

Testing 24 as 4 and 6, or 48 as 6 and 8. Right: use coprime factors only, 24 = 8 × 3 and 48 = 16 × 3.

Using HCF × LCM = product for 3 numbers. Right: the identity is only for 2 numbers; 2, 4, 8 have HCF 2 and LCM 8, but 2 × 8 ≠ 2 × 4 × 8.

Writing a remainder equal to or larger than the divisor, or a negative remainder as the final answer. Right: convert to 0 ≤ r < d, e.g. a remainder of −1 on division by 7 is 6.

In a unit-digit cycle, treating exponent remainder 0 as the 0th term. Right: remainder 0 means the 4th term of the cycle, e.g. 7⁴⁰ ends in 1.

Practice questions

Q1. The largest number which divides 1305, 4665 and 6905 leaving the same remainder in each case is N. The sum of the digits of N is: (a) 4 (b) 5 (c) 6 (d) 8

Q2. A number leaves remainder 3 when divided by 5, and the quotient leaves remainder 4 when divided by 7. The remainder when the number is divided by 35 is: (a) 13 (b) 18 (c) 23 (d) 27

Q3. How many numbers from 100 to 400 are divisible by both 6 and 9? (a) 16 (b) 17 (c) 33 (d) 18

Answers

Q1. (a) Differences: 4665 − 1305 = 3360, 6905 − 4665 = 2240, 6905 − 1305 = 5600. HCF = 1120, digit sum 1 + 1 + 2 + 0 = 4.

Q2. (c) Number = 5(7k + 4) + 3 = 35k + 23, so the remainder is 5 × 4 + 3 = 23.

Q3. (b) Divisible by both means divisible by LCM(6, 9) = 18. The multiples run from 108 to 396: (396 − 108) ÷ 18 + 1 = 17.

Quick revision

Composite divisor ⇒ coprime factors: 72 = 8 × 9, 24 = 8 × 3, 36 = 4 × 9, 88 = 8 × 11.

Divisors of pᵃ × qᵇ: (a + 1)(b + 1). Same remainder r: LCM + r. Same shortfall k: LCM − k.

Largest divisor leaving the same remainder = HCF of the differences.

Big powers: find the power that leaves remainder 1; unit digits repeat every 4.

Concepts and rules

A fraction a/b has b ≠ 0. It is proper if a < b, improper if a ≥ b, and a mixed number such as 2 3/4 means 2 + 3/4 = 11/4. Multiplying or dividing top and bottom by the same non-zero number gives an equal fraction.

Comparing positive fractions: a/b > c/d exactly when a × d > b × c. With equal denominators the larger numerator wins; with equal numerators the smaller denominator wins.

Adding the same positive number to the top and bottom of a positive proper fraction increases it (towards 1); doing it to a fraction greater than 1 decreases it (towards 1).

Add or subtract using the LCM of the denominators. To divide, multiply by the reciprocal: (a/b) ÷ (c/d) = (a × d)/(b × c).

Terminating rule: a fraction in lowest terms gives a terminating decimal if and only if its denominator is of the form 2ᵐ × 5ⁿ. Always reduce first: 21/140 = 3/20 terminates even though 140 contains 7.

Pure recurring decimal: write the repeating block over as many 9s as it has digits. 0.(36), meaning 0.363636…, = 36/99 = 4/11, and 0.(142857) = 142857/999999 = 1/7.

Mixed recurring decimal: (all digits after the point − non-repeating digits) over (one 9 per repeating digit followed by one 0 per non-repeating digit). 0.1(6) = (16 − 1)/90 = 15/90 = 1/6.

For fractions in lowest terms: HCF of fractions = HCF of numerators ÷ LCM of denominators; LCM of fractions = LCM of numerators ÷ HCF of denominators.

Order of operations (VBODMAS): bars (vinculum), brackets, 'of', division, multiplication, addition, subtraction. In SSC simplification, 'of' is done before ÷: 48 ÷ 4 of 3 = 48 ÷ 12 = 4.

Learn these: 1/8 = 0.125, 3/8 = 0.375, 1/6 = 0.1(6), 1/7 = 0.(142857), 1/9 = 0.(1), 1/11 = 0.(09), 1/12 = 0.08(3), 1/16 = 0.0625.

Method and shortcuts

To order several fractions, convert each to a decimal to 3 places, or cross-multiply in pairs. For fractions close to 1, compare the gaps 1 − a/b instead.

To simplify a long expression, convert every decimal and recurring decimal to a fraction first, then apply VBODMAS strictly from left to right within the same level.

Continued fractions: start from the deepest level and work outward, one level at a time: 1 + 1/(1 + 1/2) = 1 + 1/(3/2) = 1 + 2/3 = 5/3.

'Multiplied by a/b instead of b/a' questions: the error equals x × (b/a − a/b), so x = error ÷ (b/a − a/b).

Shortcut limit: the 9s rule needs repetition to start right after the decimal point; if non-repeating digits come first, use the mixed rule with 0s.

Example 1: recurring decimal to fraction

Question: Express 0.(45), that is 0.454545…, as a fraction in lowest terms.

Solution: The block 45 has 2 digits, so 0.(45) = 45/99. Dividing top and bottom by 9 gives 5/11.

Answer: 5/11

Example 2: ordering fractions

Question: Arrange 3/5, 5/8 and 7/11 in ascending order.

Solution: 3/5 = 0.600, 5/8 = 0.625, 7/11 = 0.636… . Cross-check: 5/8 vs 7/11 gives 5 × 11 = 55 < 8 × 7 = 56, so 5/8 < 7/11.

Answer: 3/5 < 5/8 < 7/11

Example 3: VBODMAS with 'of'

Question: Simplify 48 ÷ 4 of 3 + 2 × 5 − 6.

Solution: 'of' first: 4 of 3 = 12. Then division: 48 ÷ 12 = 4. Then multiplication: 2 × 5 = 10. Finally 4 + 10 − 6 = 8.

Answer: 8

Example 4: sum of recurring decimals

Question: 0.(3) + 0.(4) + 0.1(6) equals: (a) 8/9 (b) 17/18 (c) 5/6 (d) 19/18

Solution: 0.(3) = 3/9 = 1/3, 0.(4) = 4/9, 0.1(6) = (16 − 1)/90 = 1/6. Using LCM 18: 6/18 + 8/18 + 3/18 = 17/18.

Answer: (b) 17/18

Example 5: counting terminating decimals

Question: How many of 7/80, 11/96, 13/125, 21/140 and 9/75 have terminating decimal expansions? (a) 2 (b) 3 (c) 5 (d) 4

Solution: Reduce each first. 7/80: 80 = 2⁴ × 5, terminates. 11/96: 96 = 2⁵ × 3, does not. 13/125: 125 = 5³, terminates.

21/140 = 3/20 and 20 = 2² × 5, terminates. 9/75 = 3/25 and 25 = 5², terminates. So 4 fractions terminate; only 11/96 recurs.

Answer: (d) 4

Example 6: LCM of fractions

Question: The LCM of 2/3, 4/9 and 8/15 is: (a) 2/45 (b) 8/45 (c) 8/3 (d) 16/3

Solution: All the fractions are already in lowest terms. LCM of numerators 2, 4, 8 = 8; HCF of denominators 3, 9, 15 = 3. LCM = 8/3.

Check that it is a whole-number multiple of each: (8/3) ÷ (2/3) = 4, (8/3) ÷ (4/9) = 6, (8/3) ÷ (8/15) = 5. Option (a) 2/45 is the HCF, a common trap.

Answer: (c) 8/3

Common mistakes

Judging termination before reducing the fraction. Right: cancel common factors first; 9/75 looks recurring because of the 3, but equals 3/25.

Writing 0.1(6) as 16/99. Right: the 1 does not repeat, so it is (16 − 1)/90 = 1/6.

Doing ÷ before 'of'. Right: 48 ÷ 4 of 3 is 48 ÷ 12 = 4, not 12 × 3 = 36.

Swapping the HCF and LCM rules for fractions. Right: LCM of fractions uses LCM on top and HCF below, so it is at least as large as each fraction.

Comparing fractions by numerators alone when denominators differ. Right: cross-multiply, a/b > c/d when a × d > b × c.

Practice questions

Q1. A student multiplied a number by 3/5 instead of 5/3 and got 32 less than the correct answer. The number is: (a) 45 (b) 30 (c) 25 (d) 60

Q2. 0.2(3), that is 0.2333…, as a fraction is: (a) 7/30 (b) 23/90 (c) 23/99 (d) 2/9

Q3. The value of 1 + 1/(1 + 1/(1 + 1/2)) is: (a) 5/3 (b) 13/8 (c) 3/2 (d) 8/5

Answers

Q1. (b) x × (5/3 − 3/5) = 32, x × 16/15 = 32, x = 30.

Q2. (a) (23 − 2)/90 = 21/90 = 7/30.

Q3. (d) 1 + 1/2 = 3/2; 1 + 2/3 = 5/3; 1 + 3/5 = 8/5.

Quick revision

Terminates ⇔ reduced denominator is 2ᵐ × 5ⁿ.

0.(ab) = ab/99; 0.a(b) = (ab − a)/90.

LCM of fractions = LCM(tops)/HCF(bottoms); HCF of fractions = HCF(tops)/LCM(bottoms).

VBODMAS: 'of' before ÷; continued fractions from the inside out.

Concepts and rules

A surd is an irrational root of a rational number, such as √2, √5 or ∛7. √9 = 3 is not a surd. The symbol √ always means the non-negative root, so √25 = 5, not −5.

For a, b ≥ 0: √(a × b) = √a × √b and √(a/b) = √a/√b (b > 0). But √(a + b) is not √a + √b: √(9 + 16) = 5 while √9 + √16 = 7.

Simplest form: take out the largest square factor, √72 = √(36 × 2) = 6√2. Only like surds (same radical part) can be added: 3√2 + 5√2 = 8√2, but √2 + √3 stays as it is.

Conjugates: (√a + √b)(√a − √b) = a − b and (p + √q)(p − √q) = p² − q. So 1/(√a + √b) = (√a − √b)/(a − b), a rational denominator.

Square root of a + 2√b: find x and y with x + y = a and x × y = b; then √(a + 2√b) = √x + √y and √(a − 2√b) = √x − √y (taking x > y).

If x = p + √q with p² − q = 1, then 1/x = p − √q, so x + 1/x = 2p and x² + 1/x² = (2p)² − 2.

Comparing roots of different orders: raise all of them to the LCM of the orders. √3 and ∛5: the 6th powers are 3³ = 27 and 5² = 25, so √3 > ∛5.

√(n + 1) − √n = 1/(√(n + 1) + √n), so this difference shrinks as n grows: √10 − √9 > √11 − √10 > √12 − √11.

A perfect square ends in 0, 1, 4, 5, 6 or 9, and never in 2, 3, 7 or 8. A number ending in an odd count of zeros is not a perfect square.

Method and shortcuts

Simplify every surd first, then group like surds, then add or subtract their coefficients.

To rationalise, multiply top and bottom by the conjugate of the denominator; the denominator becomes a difference of squares.

For √(a + 2√b), if the inner surd has no 2 in front, create one: √(5 + √24) = √(5 + 2√6) = √3 + √2.

Telescoping sums: each term 1/(√k + √(k + 1)) equals √(k + 1) − √k, so a chain of them collapses to (last root − first root).

Shortcut limit: the x + 1/x = 2p trick needs p² − q = 1. For x = 2 + √5, p² − q = −1, so 1/x = √5 − 2 and it is x − 1/x that equals 4.

Example 1: adding surds

Question: Simplify √50 + √18 − √8.

Solution: √50 = 5√2, √18 = 3√2, √8 = 2√2. Total = (5 + 3 − 2)√2 = 6√2.

Answer: 6√2

Example 2: rationalising a denominator

Question: Rationalise 1/(√7 − √5).

Solution: Multiply top and bottom by √7 + √5. Denominator = 7 − 5 = 2, so the value is (√7 + √5)/2.

Answer: (√7 + √5)/2

Example 3: square root of a + 2√b

Question: Find √(8 + 2√15).

Solution: We need x + y = 8 and x × y = 15, so x = 5 and y = 3. Check: (√5 + √3)² = 5 + 3 + 2√15 = 8 + 2√15.

Answer: √5 + √3

Example 4: x² + 1/x² from a surd

Question: If x = 3 + 2√2, then x² + 1/x² equals: (a) 36 (b) 32 (c) 34 (d) 38

Solution: (3 + 2√2)(3 − 2√2) = 9 − 8 = 1, so 1/x = 3 − 2√2 and x + 1/x = 6.

x² + 1/x² = (x + 1/x)² − 2 = 36 − 2 = 34. Option (a) 36 forgets to subtract 2.

Answer: (c) 34

Example 5: comparing roots of different orders

Question: The smallest of √3, ∛4 and the 4th root of 6 is: (a) √3 (b) ∛4 (c) the 4th root of 6 (d) all are equal

Solution: The orders are 2, 3 and 4 with LCM 12. Raise each to the 12th power: √3 gives 3⁶ = 729, ∛4 gives 4⁴ = 256, the 4th root of 6 gives 6³ = 216.

Since 216 < 256 < 729, the 4th root of 6 is the smallest and √3 is the largest.

Answer: (c) the 4th root of 6

Example 6: telescoping surd sum

Question: The value of 1/(√9 + √10) + 1/(√10 + √11) + … + 1/(√80 + √81) is: (a) 9 (b) 6 (c) 3 (d) 5

Solution: Rationalise a general term: 1/(√k + √(k + 1)) = (√(k + 1) − √k)/((k + 1) − k) = √(k + 1) − √k.

The sum becomes (√10 − √9) + (√11 − √10) + … + (√81 − √80). Every middle root cancels, leaving √81 − √9 = 9 − 3 = 6.

Answer: (b) 6

Common mistakes

Splitting a root over a sum: √(16 + 9) is 5, not 4 + 3 = 7. The root splits over × and ÷ only.

Adding unlike-looking surds wrongly: √2 + √8 is not √10. Right: √8 = 2√2, so the sum is 3√2.

Forgetting the 2 in x² + 1/x² = (x + 1/x)² − 2, or using − 2 where (x − 1/x)² + 2 is needed. Write the identity before substituting.

Comparing √3 and ∛5 by their radicands (3 < 5). Right: bring both to the same order first; √3 is larger.

Rationalising with the wrong sign: for 1/(√7 − √5) multiply by √7 + √5, not by √7 − √5.

Practice questions

Q1. Which is the greatest? (a) √11 − √10 (b) √12 − √11 (c) √10 − √9 (d) √13 − √12

Q2. If x = √5 + 2, then x² + 1/x² equals: (a) 18 (b) 20 (c) 16 (d) 22

Q3. (3√2 − 2√3)/(3√2 + 2√3) equals: (a) 5 + 2√6 (b) 5 − 2√6 (c) 2 − √6 (d) 5 − √6

Answers

Q1. (c) √(n + 1) − √n = 1/(√(n + 1) + √n) is largest when the denominator is smallest, i.e. for √10 − √9.

Q2. (a) 1/x = √5 − 2, so x − 1/x = 4 and x² + 1/x² = 4² + 2 = 18.

Q3. (b) Multiply by the conjugate: (3√2 − 2√3)²/(18 − 12) = (30 − 12√6)/6 = 5 − 2√6.

Quick revision

(√a + √b)(√a − √b) = a − b; rationalise with the conjugate.

√(a + 2√b) = √x + √y with x + y = a and x × y = b.

Different orders: raise to the LCM power before comparing.

1/(√k + √(k + 1)) = √(k + 1) − √k; sums telescope to last − first.

Concepts and rules

A 2-digit number with tens digit a and units digit b is 10a + b (a from 1 to 9). Reversed it is 10b + a. Sum = 11(a + b); difference = 9(a − b). A 3-digit number abc is 100a + 10b + c.

Consecutive integers: n, n + 1, n + 2; consecutive even or odd numbers: n, n + 2, n + 4. For an odd count of consecutive terms, the middle term = sum ÷ count.

Standard sums: 1 + 2 + … + n = n(n + 1)/2; 1² + 2² + … + n² = n(n + 1)(2n + 1)/6; 1³ + … + n³ = (n(n + 1)/2)²; first n odd numbers add to n².

Terms of an arithmetic list from a to l with step d: count = (l − a)/d + 1, sum = count × (a + l)/2.

If the HCF of 2 numbers is h, write them as h × a and h × b where a and b are coprime. Then LCM = h × a × b and the product of the numbers = HCF × LCM.

Counting multiples: from 1 to N there are ⌊N/k⌋ multiples of k (⌊ ⌋ means the whole-number part). Multiples of k or m: ⌊N/k⌋ + ⌊N/m⌋ − ⌊N/LCM(k, m)⌋ (inclusion–exclusion).

Trailing zeros of n! = ⌊n/5⌋ + ⌊n/25⌋ + ⌊n/125⌋ + …, because every zero needs a factor 5 and factors of 2 are always more plentiful.

Comparing powers: bring them to a common exponent or a common base. 2³⁰ = 8¹⁰ and 3²⁰ = 9¹⁰, so 3²⁰ > 2³⁰.

For any positive x, x + 1/x ≥ 2, with equality only at x = 1. For a fixed sum, the product of 2 positive numbers is largest when they are equal.

Method and shortcuts

Name the unknowns (digits a, b or the smallest number n), write one equation per sentence of the question, solve, then check digit limits (0 to 9, leading digit not 0).

For consecutive-number sums, take the middle number as n (n − 2, n, n + 2) so the extra terms cancel at once.

Reverse-digit shortcut: if a 2-digit number exceeds its reverse by D, then a − b = D ÷ 9. If D is not a multiple of 9, no such number exists.

HCF and sum (or product) given: divide out the HCF, list coprime pairs that satisfy the reduced condition, and count them.

Options often let you back-solve: test each option against every condition in the question, not only the first one.

Example 1: consecutive even numbers

Question: The sum of 3 consecutive even numbers is 78. Find the largest of them.

Solution: Take them as n − 2, n, n + 2. Sum = 3n = 78, so n = 26. The numbers are 24, 26, 28.

Answer: 28

Example 2: reversed digits

Question: A 2-digit number exceeds the number formed by reversing its digits by 27, and the sum of its digits is 9. Find the number.

Solution: 9(a − b) = 27 gives a − b = 3. With a + b = 9, adding gives 2a = 12, so a = 6 and b = 3. Check: 63 − 36 = 27.

Answer: 63

Example 3: pairs with a given HCF and sum

Question: How many pairs of positive numbers have HCF 12 and sum 96?

Solution: Write the numbers as 12a and 12b with a, b coprime. 12(a + b) = 96 gives a + b = 8. Coprime pairs: (1, 7) and (3, 5); (2, 6) and (4, 4) fail.

So the pairs are (12, 84) and (36, 60).

Answer: 2 pairs

Example 4: counting with inclusion–exclusion

Question: How many numbers from 1 to 300 are divisible by 4 or 6? (a) 125 (b) 100 (c) 105 (d) 75

Solution: Multiples of 4: 300 ÷ 4 = 75. Multiples of 6: 300 ÷ 6 = 50. Both means multiples of LCM 12: 300 ÷ 12 = 25.

Total = 75 + 50 − 25 = 100. Option (a) 125 counts the common multiples twice.

Answer: (b) 100

Example 5: comparing large powers

Question: The largest of 2¹⁵⁰, 3¹⁰⁰, 5⁵⁰ and 6⁷⁵ is: (a) 6⁷⁵ (b) 3¹⁰⁰ (c) 2¹⁵⁰ (d) 5⁵⁰

Solution: The exponents 150, 100, 50, 75 share the factor 25. Write each as a 25th power: 2¹⁵⁰ = (2⁶)²⁵ = 64²⁵, 3¹⁰⁰ = (3⁴)²⁵ = 81²⁵.

5⁵⁰ = (5²)²⁵ = 25²⁵ and 6⁷⁵ = (6³)²⁵ = 216²⁵. Comparing the bases 64, 81, 25, 216, the largest is 216, so 6⁷⁵ is the largest.

Answer: (a) 6⁷⁵

Example 6: sum of numbers with a given remainder

Question: The sum of all 2-digit numbers that leave remainder 3 when divided by 7 is: (a) 672 (b) 680 (c) 702 (d) 676

Solution: Such numbers are 7k + 3. The smallest 2-digit one is 10 (7 × 1 + 3) and the largest is 94 (7 × 13 + 3), since the next, 101, has 3 digits.

Count = (94 − 10) ÷ 7 + 1 = 13. Sum = 13 × (10 + 94) ÷ 2 = 13 × 52 = 676.

Answer: (d) 676

Common mistakes

Allowing the tens digit to be 0, or a digit above 9, when solving digit equations. Always check 1 ≤ a ≤ 9 and 0 ≤ b ≤ 9.

Counting pairs with a given HCF without the coprime check: (24, 72) has sum 96 but HCF 24, not 12.

Adding multiples of 4 and of 6 without removing the multiples of 12. In 'or' questions, subtract the overlap once.

Counting trailing zeros as ⌊n/5⌋ only. 125! has 25 + 5 + 1 = 31 zeros, not 25.

Comparing powers by exponent alone: 5⁵⁰ has a smaller exponent than 3¹⁰⁰ but the bases matter too; bring them to one exponent.

Practice questions

Q1. A 2-digit number is 4 times the sum of its digits. If 27 is added to it, the digits are reversed. The number is: (a) 63 (b) 36 (c) 24 (d) 48

Q2. The number of zeros at the end of 125! is: (a) 25 (b) 30 (c) 31 (d) 28

Q3. 1² + 2² + 3² + … + 20² equals: (a) 2870 (b) 2780 (c) 3080 (d) 2470

Answers

Q1. (b) 10a + b = 4(a + b) gives b = 2a; 9(b − a) = 27 gives b − a = 3; so a = 3, b = 6, number 36.

Q2. (c) ⌊125/5⌋ + ⌊125/25⌋ + ⌊125/125⌋ = 25 + 5 + 1 = 31.

Q3. (a) 20 × 21 × 41 ÷ 6 = 2870.

Quick revision

10a + b and 10b + a: sum 11(a + b), difference 9(a − b).

Σn = n(n + 1)/2, Σn² = n(n + 1)(2n + 1)/6, Σn³ = (n(n + 1)/2)².

HCF h: numbers h × a, h × b with a, b coprime.

k or m: N/k + N/m − N/LCM (whole parts). Zeros in n!: n/5 + n/25 + … (whole parts).