Number system

Answer: (b) Wednesday

All three meet again after LCM(5, 24, 9) = 360 days. That is 51 weeks and 3 days, so the meeting falls three days after Sunday — on Wednesday.

1. They meet together again after LCM(5, 24, 9) days.
2. 24 = 2³ × 3 and 9 = 3², so LCM = 2³ × 3² × 5 = 360.
3. 360 = 7 × 51 + 3, so the weekday moves on by 3 days.
4. Sunday + 3 days = Wednesday.

Remember · Next joint meeting = LCM of the intervals; the weekday shifts by (LCM mod 7).

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Answer: (b) 2 only

A 2-digit number and its reverse differ by 9 times the difference of the digits, so 54 means the digits differ by 6 — that is fixed. The digit sum could be 6, 8, 10 or 12; a known product would settle it, but so would other information, so 'only if the product is known' is not true.

1. (10a + b) − (10b + a) = 9(a − b), so 9 × (difference of digits) = 54.
2. The digits differ by 6 — Statement 2 is correct.
3. Possible numbers: 60, 71, 82, 93 (or 17, 28, 39 the other way round), with digit sums 6, 8, 10, 12.
4. Knowing the product (0, 7, 16 or 27) would fix the sum, but so would other clues, such as the larger digit — the product is not the only way, so Statement 1 is incorrect.

• ✗ 1. The product is one way to fix the sum, not the only way, so 'only if' makes the statement false.
• ✓ 2. 9 × (difference of digits) = 54 gives a difference of 6.

Remember · Number − its reverse = 9 × (difference of digits); number + its reverse = 11 × (sum of digits).

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Answer: (d) 15873

The smallest number made only of 1s that 7 divides is 111111 (six ones). 111111 ÷ 7 = 15873.

1. Divide 1, 11, 111, … by 7: the remainders are 1, 4, 6, 5, 2, 0 — the first zero comes at 111111.
2. 111111 ÷ 7 = 15873.
3. Check: 15873 × 7 = 111111.
4. Useful fact: 111111 = 3 × 7 × 11 × 13 × 37.

Remember · Divide repunits (1, 11, 111 …) by 7 until the remainder is 0; 111111 = 7 × 15873.

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Answer: (a) 10

Each trailing zero needs one pair of 2 and 5. Writing the product as 5¹² × 12! shows 14 fives but only 10 twos, so the twos run out first and there are 10 zeroes.

1. The terms after 1 are 5 × 1, 5 × 2, …, 5 × 12, so the product = 5¹² × (1 × 2 × … × 12) = 5¹² × 12!.
2. Fives: 12 from 5¹², plus 2 in 12! (from 5 and 10) = 14.
3. Twos: only from 12! — 6 + 3 + 1 = 10 (multiples of 2, 4 and 8 up to 12).
4. Zeroes = min(10, 14) = 10.
5. Check: 12! = 479001600 has two 5s and ten 2s, as used.

Remember · Trailing zeroes = the smaller of the counts of 2s and 5s. Do not assume 5s are always scarcer.

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Answer: (b) 9

Adding the three rotations puts each digit once in the hundreds, tens and units places, so the sum is 111(X + Y + Z) = 3 × 37 × (X + Y + Z). It is divisible by 3, 37 and (X + Y + Z), but by 9 only if (X + Y + Z) were a multiple of 3 — which it is not.

1. XYZ + YZX + ZXY = (100X + 10Y + Z) + (100Y + 10Z + X) + (100Z + 10X + Y) = 111(X + Y + Z).
2. 111 = 3 × 37, so the sum = 3 × 37 × (X + Y + Z): divisible by 3, 37 and (X + Y + Z).
3. For 9, a second factor 3 must come from (X + Y + Z), which is not a multiple of 3.
4. So the sum is not divisible by 9.
5. Check: 124 + 241 + 412 = 777 = 111 × 7; 777 ÷ 9 is not whole.

Remember · Cyclic rotations of a 3-digit number always add up to 111 × (digit sum) = 3 × 37 × (digit sum).

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Answer: (c) r

Call the common value k. Then each number is k plus or minus a constant, and r = k + 2018 carries the largest addition.

1. Let p − 2016 = q + 2017 = r − 2018 = s + 2019 = k.
2. Then p = k + 2016, q = k − 2017, r = k + 2018, s = k − 2019.
3. The largest is r = k + 2018.

Remember · Set equal expressions to one variable k and compare what is added to k in each.

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Answer: (a) Zero

Every such number uses the digits 1 to 5 once, so its digit sum is 15. A digit sum of 15 makes every one of these numbers divisible by 3, so none can be prime.

1. Digit sum = 1 + 2 + 3 + 4 + 5 = 15, whatever the order.
2. 15 is divisible by 3, so every such five-digit number is divisible by 3.
3. A number greater than 3 that is divisible by 3 is not prime, so the count is zero.

Remember · Before listing permutations for primes, check the digit sum: if it is a multiple of 3, no arrangement is prime.

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Answer: (b) 3

Write the symbol as a digit x and expand each number by place value. The sum becomes a simple linear equation, 24x + 61 = 100 + 11x, which gives x = 3.

1. Let ⊗ = x. Then ⊗ = x, 1⊗ = 10 + x, 5⊗ = 50 + x, ⊗⊗ = 11x, ⊗1 = 10x + 1, and 1⊗⊗ = 100 + 11x.
2. Left side = x + (10 + x) + (50 + x) + 11x + (10x + 1) = 24x + 61.
3. 24x + 61 = 100 + 11x, so 13x = 39 and x = 3.
4. Check: 3 + 13 + 53 + 33 + 31 = 133 = 1⊗⊗ with ⊗ = 3.

Remember · In symbol-for-digit sums, expand every number by place value and solve the resulting equation; then check it back.

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Answer: (b) 0.25 foot

Laying the sticks end to end, forwards or backwards, measures any whole-number combination of 7·5 and 3·25 feet. The smallest positive length such combinations can give is their HCF, 0·25 foot.

1. Convert to hundredths of a foot: 7·5 = 750/100 and 3·25 = 325/100.
2. HCF(750, 325) = 25 (750 = 2 × 3 × 5³ and 325 = 5² × 13), so HCF of the lengths = 0·25 foot.
3. Any length made by laying the sticks forwards and backwards is a multiple of 0·25 foot, and 0·25 itself can be made.
4. Check: 7 × 3·25 = 22·75 and 3 × 7·5 = 22·5; 22·75 − 22·5 = 0·25 foot.

Remember · The smallest length measurable with two rods (by adding and subtracting whole lengths) is the HCF of their lengths.

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Answer: (c) 38

Subtracting 1 from each term gives 13, 17, 19, 23, 29, 31 — consecutive prime numbers. The next prime is 37, so the next term is 37 + 1 = 38.

1. Subtract 1: 14 → 13, 18 → 17, 20 → 19, 24 → 23, 30 → 29, 32 → 31.
2. 13, 17, 19, 23, 29, 31 are consecutive primes.
3. The next prime after 31 is 37 (33, 35 are composite).
4. Next term = 37 + 1 = 38.

Remember · When a series 'relates to primes', try adding or subtracting 1 or 2 to each term and look for consecutive primes.

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Answer: (b) 7, 8

The full booklet's page total must be just above 195. With 20 pages the total is 210, so the torn leaf carries numbers adding to 15 — pages 7 and 8.

1. Sum of pages 1 to n is n(n + 1)/2. For n = 19 it is 190 (too small); for n = 20 it is 210.
2. Missing sum = 210 − 195 = 15.
3. A leaf carries two consecutive numbers, odd then even: 7 + 8 = 15.
4. Check: n = 21 gives 231 − 195 = 36, which cannot be an odd + next-even pair (such pairs add to an odd number).

Remember · Missing leaf: find the first 1 + 2 + … + n above the given total; the gap is the two page numbers' sum.

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Answer: (c) 3

Working column by column from the units, the carries force B = 7 and E = 5, and C and F must be 4 and 8. That leaves 6 and 9 for A and D, whose difference is 3.

1. Units: C + F ends in 2; both are at least 4, so C + F = 12, carry 1.
2. Tens: B + 2 + 1 ends in 0, so B + 3 = 10 and B = 7, carry 1.
3. Hundreds: 3 + E + 1 ends in 9, so E = 5, carry 0.
4. Thousands: A + D = 15.
5. Digits left from 4–9 after B = 7, E = 5: {4, 6, 8, 9}. C + F = 12 needs {4, 8}; so A and D are 6 and 9 (6 + 9 = 15).
6. Difference = 9 − 6 = 3.
7. Check: 6374 + 9528 = 15902.

Remember · In letter-addition puzzles, start from the units column, track every carry, and use 'different digits' to fix the pairs.

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Answer: (c) 12

There are 19 numbers below 100 containing the digit 4: 4, 14, 24, 34, 40 to 49, and 54, 64, 74, 84, 94. Seven of them are multiples of 4, which leaves 12.

1. Units digit 4: 4, 14, 24, 34, 44, 54, 64, 74, 84, 94 = 10 numbers.
2. Tens digit 4 (other than 44): 40, 41, 42, 43, 45, 46, 47, 48, 49 = 9 numbers. Total = 19.
3. Multiples of 4 among them: 4, 24, 40, 44, 48, 64, 84 = 7.
4. Required count = 19 − 7 = 12.

Remember · Count the whole set first, then subtract the ones to exclude — fewer chances to miss a number.

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Answer: (c) (1/3)⁻⁴

A negative power of a fraction flips it: (1/a)⁻ⁿ = aⁿ. The four values are 2⁶ = 64, 4³ = 64, 3⁴ = 81 and 6² = 36, so (1/3)⁻⁴ is the largest.

1. (1/2)⁻⁶ = 2⁶ = 64.
2. (1/4)⁻³ = 4³ = 64.
3. (1/3)⁻⁴ = 3⁴ = 81.
4. (1/6)⁻² = 6² = 36.
5. The largest is 81, i.e. (1/3)⁻⁴.

Remember · (1/a) raised to −n equals aⁿ; rewrite every option this way before comparing.

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Answer: (d) 1¾ m

The greatest length that divides both is their HCF. In quarters of a metre the lengths are 14/4 and 35/4, and HCF(14, 35) = 7, so x = 7/4 = 1¾ m.

1. 3½ = 14/4 m and 8¾ = 35/4 m.
2. HCF(14, 35) = 7, so the greatest common length = 7/4 m = 1¾ m.
3. Check: 3½ ÷ 1¾ = 2 and 8¾ ÷ 1¾ = 5, both whole numbers.

Remember · HCF of fractions: put them over a common denominator, take the HCF of the numerators, keep the denominator.

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Answer: (b) 14/11

The repeating block 27 has two digits, so 0·2727... = 27/99 = 3/11. Adding the whole part gives 1 + 3/11 = 14/11.

1. Let x = 1·2727...; then 100x = 127·2727...
2. 100x − x = 126, so x = 126/99 = 14/11.
3. Check: 14 ÷ 11 = 1·2727...

Remember · A pure recurring decimal 0·(ab) repeating = ab/99; handle the whole-number part separately.

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Answer: (b) 1022

A number leaving remainder 2 with each divisor is 2 more than a common multiple of 3, 4, 5 and 6. Their LCM is 60, the first four-digit multiple of 60 is 1020, and so the answer is 1022.

1. LCM(3, 4, 5, 6) = 60.
2. Smallest four-digit multiple of 60: 60 × 17 = 1020 (60 × 16 = 960 has three digits).
3. Required number = 1020 + 2 = 1022.
4. Check: 1022 = 3 × 340 + 2 = 4 × 255 + 2 = 5 × 204 + 2 = 6 × 170 + 2.

Remember · Same remainder r for every divisor: the number = k × LCM + r. Find the smallest k that gives the required size.

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Answer: (a) 50

Only the last two digits matter. 62 × 75 = 4650 ends in 50, and multiplying a number ending in 50 by any odd number again ends in 50. The other factors 51, 27 and 35 are all odd, so the remainder is 50.

1. 62 × 75 = 4650, which ends in 50.
2. 50 × (odd number) always ends in 50, since 50 × odd = 100k + 50.
3. 51 × 27 × 35 = 48195 is odd, so the whole product ends in 50.
4. Remainder = 50.

Remember · Remainder on division by 100 = last two digits; pair factors that quickly give a round ending like 50 or 00.

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Answer: (b) For any whole number n

For any positive integer n, 10ⁿ + 1 is 1 followed by zeros and a final 1, so its digit sum is 2. For n = 0 it is 1 + 1 = 2, whose digit sum is also 2. So it holds for every whole number 0, 1, 2, …, but not for fractional n, where 10ⁿ + 1 is not a whole number.

1. n = 1, 2, 3 …: 11, 101, 1001 … — digit sum 2.
2. n = 0: 10⁰ + 1 = 2 — digit sum 2.
3. n = 1/2: 10ⁿ + 1 = √10 + 1 ≈ 4·16, not a whole number, so 'any real number' fails.
4. So the statement holds for any whole number n.

Remember · Always test the edge value n = 0 before choosing 'positive integers only'.

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Answer: (a) 3

x² − y² = (x − y)(x + y) = 63, so each way of writing 63 as a product of two factors gives one pair. 63 = 1 × 63 = 3 × 21 = 7 × 9, giving (32, 31), (12, 9) and (8, 1).

1. x² − y² = (x − y)(x + y) = 63, with x − y < x + y.
2. Factor pairs of 63: (1, 63), (3, 21), (7, 9) — all odd, so x and y come out whole.
3. (1, 63): x = 32, y = 31. (3, 21): x = 12, y = 9. (7, 9): x = 8, y = 1.
4. 3 pairs.
5. Check: 32² − 31² = 63, 144 − 81 = 63, 64 − 1 = 63.

Remember · Difference of squares: factor the number as (x − y)(x + y) with both factors of the same parity; each factor pair gives one solution.

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Answer: (d) 5/6

Each fraction has the form n/(n + 1). Adding 5 to both parts changes it by 5/[(n + 1)(n + 6)], which shrinks as n grows. So 5/6, the largest, changes least (by 5/66).

1. 2/3 → 7/8: change 7/8 − 2/3 = 5/24 ≈ 0·208.
2. 3/4 → 8/9: change 5/36 ≈ 0·139.
3. 4/5 → 9/10: change 5/50 = 0·1.
4. 5/6 → 10/11: change 5/66 ≈ 0·076 — the smallest.

Remember · Adding the same number to top and bottom pushes a proper fraction towards 1; fractions already closer to 1 move less.

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Answer: (d) 3n + 1

The digits greater than 3 that are divisible by 3 are 6 and 9; 6 is divisible by 6, so n = 9. Then 3n + 1 = 28 is the only option divisible by 4.

1. Digits above 3 divisible by 3: 6 and 9. 6 is divisible by 6, so n = 9.
2. 2n = 18, 3n = 27, 2n + 4 = 22 — none divisible by 4.
3. 3n + 1 = 28 = 4 × 7 — divisible by 4.

Remember · When a condition pins a single digit, find that digit first and then test every option numerically.

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Answer: (c) 300

Written as three-digit strings from 000 to 999, every digit appears equally often in each place, so 5 fills one-tenth of the 3000 digit places: 300. The number 1000 adds no 5.

1. Treat 1 to 999 as three-digit strings 001–999 (leading zeros add no 5s).
2. In each place — units, tens, hundreds — the digit 5 appears in 100 of the 1000 strings 000–999.
3. So 5 appears 3 × 100 = 300 times from 1 to 999.
4. 1000 has no 5, so the total is 300.
5. Check: units place 5, 15, …, 995 = 100; tens place 50–59 in each hundred = 10 × 10 = 100; hundreds place 500–599 = 100.

Remember · From 1 to 10ⁿ − 1, each non-zero digit appears n × 10ⁿ⁻¹ times; check the end number separately.

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Answer: (b) 8

A student plays both games only if the number is divisible by both 4 and 6, that is by their LCM 12. There are 8 multiples of 12 up to 100.

1. Divisible by 4 and by 6 ⇔ divisible by LCM(4, 6) = 12.
2. Multiples of 12 from 1 to 100: 12, 24, 36, 48, 60, 72, 84, 96.
3. Count = 8 (100 ÷ 12 = 8 remainder 4).
4. Check: using 24 (= 4 × 6) would wrongly drop 12, 36, 60 and 84.

Remember · 'Divisible by a and by b' means divisible by LCM(a, b), not a × b.

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Answer: (b) 4

Writing the number as 10a + b, the ratio 4 : 7 simplifies to b = 2a. The digit a can be 1 to 4, giving 12, 24, 36 and 48 — four pairs.

1. Let the number be 10a + b; reversed, it is 10b + a.
2. 7(10a + b) = 4(10b + a) → 70a + 7b = 40b + 4a → 66a = 33b → b = 2a.
3. b must be a digit, so a = 1, 2, 3, 4.
4. Pairs: 12 & 21, 24 & 42, 36 & 63, 48 & 84 — each in the ratio 4 : 7.
5. Number of pairs = 4.

Remember · Turn digit-reversal conditions into 10a + b form; the equation usually gives a simple digit relation to list.

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Answer: (d) 13th January

A gap of 2 days means Seeta swims every 3rd day; a gap of 3 days means Geeta swims every 4th day. They meet again after LCM(3, 4) = 12 days, on 13th January.

1. Gap of 2 days: Seeta swims on 1, 4, 7, 10, 13 … January (every 3rd day).
2. Gap of 3 days: Geeta swims on 1, 5, 9, 13 … January (every 4th day).
3. LCM(3, 4) = 12, so the next common day is 1 + 12 = 13 January.
4. Check: 13 is in both lists; 7 is only Seeta's and 12 is neither's.

Remember · 'After a gap of n days' means the event repeats every (n + 1) days; then take the LCM.

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Answer: (c) (y − x) is less than or equal to 15

y − x is largest when y is as big as allowed (40) and x as small as allowed (25), giving 15. So y − x can never exceed 15; the other options fail for suitable values.

1. Maximum of y − x = 40 − 25 = 15, so y − x ≤ 15 always.
2. (a) fails: x = 25, y = 40 gives x < y.
3. (b) fails: y − x never goes above 15.
4. (d) fails: x = 25, y = 10 gives x + y = 35.

Remember · For 'always correct' with bounds, test the extreme values; a single counter-example removes an option.

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Answer: (c) 76

The terms fall by 3, 5, 7, 11 and 13 — consecutive primes. The next prime is 17, so X = 93 − 17 = 76.

1. Differences: 132 − 129 = 3, 129 − 124 = 5, 124 − 117 = 7, 117 − 106 = 11, 106 − 93 = 13.
2. These are consecutive primes; the next one is 17.
3. X = 93 − 17 = 76.

Remember · When differences grow irregularly, test primes, squares and cubes first.

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Answer: (b) always greater than the original fraction

For a proper fraction a/b with 0 < a < b, adding the same k > 0 to both parts gives a difference of k(b − a)/[b(b + k)], which is positive. The fraction always increases.

1. Let the fraction be a/b with 0 < a < b, and add k > 0 to both.
2. (a + k)/(b + k) − a/b = [b(a + k) − a(b + k)]/[b(b + k)] = k(b − a)/[b(b + k)].
3. k > 0 and b − a > 0, so the difference is positive: the new fraction is larger.
4. Example: 1/2 becomes (1 + 1)/(2 + 1) = 2/3, which is greater than 1/2.

Remember · Adding the same positive number to both parts pulls a fraction towards 1: proper fractions rise, improper ones fall.

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Answer: (b) 64

Every term is a square: 2², 14², 4², 12², 6², 10², 8². Bases in odd places rise 2, 4, 6, 8 and in even places fall 14, 12, 10, so the next even-place base is 8 and X = 64.

1. 4 = 2², 196 = 14², 16 = 4², 144 = 12², 36 = 6², 100 = 10², 64 = 8².
2. Odd positions: bases 2, 4, 6, 8 (rising by 2). Even positions: 14, 12, 10, … (falling by 2).
3. X is in an even position, so its base is 8 and X = 8² = 64.

Remember · In mixed sequences, split the odd and even positions and look for squares or cubes in each.

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