Number system

Answer: (d) 200

The product is n raised to the power 2n for n = 1 to 25. Twos are plentiful, so count the fives: only 5, 10, 15, 20 and 25 contribute, giving 10 + 20 + 30 + 40 + 100 = 200.

1. Each zero at the end needs one 2 and one 5; there are far more 2s than 5s here, so count the 5s.
2. Only multiples of 5 carry a 5: 5¹⁰ gives 10 fives, 10²⁰ gives 20, 15³⁰ gives 30, 20⁴⁰ gives 40.
3. 25 = 5², so 25⁵⁰ = 5¹⁰⁰ gives 100 fives.
4. Total fives = 10 + 20 + 30 + 40 + 100 = 200, so there are 200 zeros at the end.
5. Check: forgetting that 25 holds two 5s gives 150, which is not even an option.

Remember · Trailing zeros = number of 5s (2s are always more). Remember 25 = 5², so its power counts double.

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Answer: (b) 8th January, 2023

He saves 1, 3, 5, 7, … rupees, so the total after n days is n². A number that is both a square and a cube is a sixth power, so n must be a perfect cube; n = 8 gives 64 = 8² = 4³.

1. Daily savings are the odd numbers 1, 3, 5, …; the sum of the first n odd numbers is n².
2. n² is also a perfect cube only when it is a sixth power, i.e. when n itself is a perfect cube.
3. The first cube after 1 is 8: total after 8 days = 64 = 8² = 4³.
4. So the date is 8th January, 2023.
5. Check: 7 days give 49 and 9 days give 81, neither a cube. (The ₹1 at the end of 1st January is trivially 1² = 1³, but that date is not among the options.)

Remember · Sum of the first n odd numbers = n². A number that is both a square and a cube is a sixth power.

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Answer: (b) 3 and 37 but not 2

222 = 2 × 3 × 37 and 333 = 3² × 37, so both terms, and hence their sum, are divisible by 3 and 37. But 222³³³ is even and 333²²² is odd, so the sum is odd and not divisible by 2.

1. Factorise the bases: 222 = 2 × 3 × 37 and 333 = 3 × 3 × 37.
2. Both 222³³³ and 333²²² contain the factors 3 and 37, so their sum is divisible by 3 and by 37.
3. 222³³³ is even; 333²²² is odd (a power of an odd number stays odd). Even + odd = odd.
4. So the sum is divisible by 3 and 37 but not by 2.

Remember · Factorise the bases first: a sum is divisible by any factor common to both terms. Even + odd is always odd.

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Answer: (d) 9

30³⁰ = 3³⁰ × 10³⁰, so the digit just before the zeros is the unit digit of 3³⁰. Powers of 3 end in 3, 9, 7, 1 in a cycle of 4, and 30 leaves remainder 2 on division by 4, giving 9.

1. 30³⁰ = 3³⁰ × 10³⁰; the factor 10³⁰ supplies the 30 zeros at the end.
2. The unit digits of 3¹, 3², 3³, 3⁴ are 3, 9, 7, 1, and then they repeat every 4 powers.
3. 30 = 4 × 7 + 2, so 3³⁰ ends like 3², i.e. in 9.
4. So the rightmost digit before the zeros is 9.

Remember · Split off the powers of 10 first; then use the four-step unit-digit cycle (remainder 0 means the fourth digit).

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Answer: (c) 3

If both numbers leave remainder 1, the divisor divides 420 and 426, and hence their HCF, 6. The divisors of 6 that are greater than 1 are 2, 3 and 6, so three numbers work (1 cannot leave a remainder of 1).

1. Remainder 1 means the divisor divides 421 − 1 = 420 and 427 − 1 = 426.
2. HCF(420, 426) = HCF(420, 6) = 6.
3. Divisors of 6 are 1, 2, 3 and 6. A divisor must be larger than the remainder, so 1 is ruled out.
4. Possible divisors: 2, 3 and 6, i.e. 3 numbers.
5. Check: 421 = 6 × 70 + 1 and 427 = 6 × 71 + 1.

Remember · Same remainder r: the divisor divides each number minus r. Count the HCF’s divisors that are larger than r.

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Answer: (b) 4

A bottle size must divide both 399 and 532 exactly, i.e. it must divide their HCF, 133 = 7 × 19. The number 133 has four divisors: 1, 7, 19 and 133.

1. 399 = 3 × 7 × 19 and 532 = 2² × 7 × 19.
2. HCF(399, 532) = 7 × 19 = 133.
3. Every common divisor divides 133: 1, 7, 19 and 133.
4. So 4 different bottle sizes are possible.
5. Check: 532 − 399 = 133, and 399 = 3 × 133, 532 = 4 × 133.

Remember · ‘Equal bottles for both’ means common divisors; their number equals the number of divisors of the HCF.

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Answer: (c) 1 and 3 only

The primes below 10 are 2, 3, 5 and 7, and three distinct ones can be chosen in only four ways, giving sums 10, 12, 14 and 15. Unit digits 0 and 5 occur; 9 does not.

1. Primes less than 10: 2, 3, 5, 7. Choosing 3 distinct ones gives just 4 cases.
2. 2 + 3 + 5 = 10; 2 + 3 + 7 = 12; 2 + 5 + 7 = 14; 3 + 5 + 7 = 15.
3. Possible unit digits: 0, 2, 4, 5.
4. So statements 1 (0) and 3 (5) are correct; statement 2 (9) is not.

• ✓ 1. 2 + 3 + 5 = 10 ends in 0.
• ✗ 2. None of the four possible sums (10, 12, 14, 15) ends in 9.
• ✓ 3. 3 + 5 + 7 = 15 ends in 5.

Remember · When the set is tiny, list every case (four sums here) instead of reasoning in general.

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Answer: (d) 8

X + Y = 11 × (sum of the digits), and it must equal 99, so the digits add up to 9. Both numbers must be two-digit, so neither digit can be 0: X can be 18, 27, 36, 45, 54, 63, 72 or 81, which is 8 values.

1. Let X = 10a + b and Y = 10b + a; then X + Y = 11(a + b).
2. The greatest two-digit number is 99, so 11(a + b) = 99 and a + b = 9.
3. Y must also be a two-digit number, so b cannot be 0; this rules out X = 90 (its reverse is 09).
4. a = 1 to 8 gives X = 18, 27, 36, 45, 54, 63, 72, 81, i.e. 8 values.

Remember · A number plus its reverse is 11 × digit sum. Watch for a 0 digit that makes the reverse a one-digit number.

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Answer: (d) 1, 2 and 3

With p, q odd and r, s even, statement 1 contains the even factor qs and statement 2 the even factor s, so both are even. In statement 3, q + r and p + s are both odd, so the product is odd. All three are correct.

1. Parity rules: odd ± even = odd; any product with an even factor is even; odd × odd = odd.
2. Statement 1: (p − r)² is odd, but qs = odd × even is even, so the product is even. Correct.
3. Statement 2: s is even, so (q − s)q²s is even. Correct.
4. Statement 3: q + r is odd, so (q + r)² is odd; p + s is odd; odd × odd = odd. Correct.
5. Check with p = 1, q = 3, r = 2, s = 4: (−1)² × 12 = 12 (even); (−1) × 9 × 4 = −36 (even); 5² × 5 = 125 (odd).

• ✓ 1. qs is odd × even = even, so the whole product is even whatever (p − r)² is.
• ✓ 2. The factor s is even, so the product is even.
• ✓ 3. q + r and p + s are each odd + even = odd, and a product of odd numbers is odd.

Remember · For parity questions, plug in small numbers of the right parity and test each statement; it is faster than algebra.

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Answer: (b) 56

Count the digit 5 place by place in 1 to 260. It appears 26 times in the units place and 30 times in the tens place (50–59, 150–159, 250–259), a total of 56.

1. Units place: 5, 15, 25, …, 255, which is 26 numbers.
2. Tens place: 50–59, 150–159 and 250–259, 10 each, 30 in all.
3. Hundreds place: there is no page from 500 upwards, so 0.
4. Total fives = 26 + 30 = 56.
5. Check: 55, 155 and 255 each use two fives; they are counted once in the units count and once in the tens count, as they should be.

Remember · Count a digit place by place (units, tens, hundreds); numbers like 55 are then automatically counted twice.

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Answer: (b) 84

After the first 1, the sequence runs in blocks 1 | 2, 1 | 3, 2, 1 | 4, 3, 2, 1 | …. The first 28 terms are the lone 1, the full blocks up to 6 (21 terms, sum 56) and the first 6 terms of the block 7, 6, …, 1 (sum 27): 1 + 56 + 27 = 84.

1. Group the terms: 1 | 1 | 2, 1 | 3, 2, 1 | 4, 3, 2, 1 | 5, 4, 3, 2, 1 | …; after the first 1, block k is k, k − 1, …, 1.
2. Blocks 1 to 6 have 1 + 2 + 3 + 4 + 5 + 6 = 21 terms; with the first 1 that makes 22 terms.
3. Their sum = 1 + (1 + 3 + 6 + 10 + 15 + 21) = 1 + 56 = 57.
4. Terms 23 to 28 are the first 6 terms of block 7: 7 + 6 + 5 + 4 + 3 + 2 = 27.
5. Sum of the first 28 terms = 57 + 27 = 84.

Remember · In block sequences, count terms block by block (1, 2, 3, … terms) and handle the last, partial block separately.

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Answer: (a) 9

Two two-digit numbers add up to at most 197, so the leading 1 of 1CE is a carry. In the tens column A + C + carry = 10 + C, which forces A = 9 with a carry of 1 from the units column.

1. AB + CD ≤ 99 + 98 = 197, so the sum 1CE lies between 100 and 197.
2. Tens column: A + C + c = C + 10, where c (0 or 1) is the carry from the units column.
3. So A + c = 10. A is a single digit, so c = 1 and A = 9.
4. Check: 97 + 35 = 132 fits, with A = 9, B = 7, C = 3, D = 5, E = 2 all different.

Remember · In digit-sum puzzles, write each column as an equation with its carry; the leftmost column usually fixes a digit at once.

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Answer: (c) 10

32⁵ = (2⁵)⁵ = 2²⁵, so the sum is 2²⁵ + 2²⁷ = 2²⁵ × (1 + 4) = 5 × 2²⁵. It has the factors 2 and 5, so it is divisible by 10.

1. 32 = 2⁵, so 32⁵ = 2²⁵.
2. 32⁵ + 2²⁷ = 2²⁵ + 2²⁷ = 2²⁵ × (1 + 2²) = 5 × 2²⁵.
3. This contains 2 × 5 = 10, so it is divisible by 10.
4. Its only prime factors are 2 and 5, so it is not divisible by 3, 7 or 11.

Remember · Write every term with the same base, take out the common power, and read off the factors.

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Answer: (c) 5

For k = 3 and k = 4 there is only one pair with p < q: (1, 2) and (1, 3). For k = 5 there are two, (1, 4) and (2, 3), so 5 is the smallest k that does not fix p and q.

1. k = 3: only (1, 2). Unique.
2. k = 4: only (1, 3), since (2, 2) breaks p < q. Unique.
3. k = 5: (1, 4) and (2, 3). Not unique.
4. So the smallest such k is 5.

Remember · For a ‘smallest value’ question, test the options in increasing order and stop at the first that works.

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Answer: (b) 155

The terms are n³ + n² + n for n = 1, 2, 3, …: 3, 14, 39, 84, 155, 258. So the missing term is 125 + 25 + 5 = 155.

1. First differences: 11, 25, 45. Second differences: 14, 20, rising by 6, so the next second difference is 26.
2. Next first difference = 45 + 26 = 71, so the missing term = 84 + 71 = 155.
3. Check the last term: next second difference 32, so the next first difference is 71 + 32 = 103, and 155 + 103 = 258.
4. Formula view: n³ + n² + n gives 3, 14, 39, 84, 155, 258.

Remember · If first differences don’t settle a series, take second differences; a steady rise in them points to a cubic pattern.

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