Number system

Answer: (c) 8

Dividing N = p + q + r by N turns the condition into 1 = 1/a + 1/b + 1/c with three different whole numbers, and the only such solution is 1/2 + 1/3 + 1/6. So N must be a multiple of 6, and there are 8 multiples of 6 below 50.

1. Divide N = p + q + r by N: 1 = p/N + q/N + r/N = 1/a + 1/b + 1/c, where a = N/p, b = N/q, c = N/r are whole numbers.
2. Distinct factors mean a, b, c are distinct. Take a < b < c. If a ≥ 3, the sum is at most 1/3 + 1/4 + 1/5 < 1, so a = 2.
3. Then 1/b + 1/c = 1/2 with 2 < b < c: b = 3 gives c = 6; b = 4 gives c = 4 (not distinct); b ≥ 5 is too small. So (a, b, c) = (2, 3, 6).
4. Hence N must be divisible by 2, 3 and 6, i.e. a multiple of 6, and then N = N/2 + N/3 + N/6 always works.
5. Multiples of 6 below 50: 6, 12, 18, 24, 30, 36, 42, 48 — that is 8 numbers.
6. Check: 6 = 3 + 2 + 1 and 48 = 24 + 16 + 8.

Remember · When a number equals a sum of its own factors, divide through by it — the problem becomes unit fractions adding to 1.

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Answer: (d) More than 6

Why not the tempting option · UPSC's key is (d). The item says 'three prime numbers p, q and r', not three different primes, and uses 'distinct' only for the sums — so p = q = r is allowed, every prime below 20 can be the middle term and there are 8 sums. Assuming the primes must differ gives 4 sums, option (a), but that assumption is not in the item. In the exam, do not add 'distinct' where the question has not written it.

Since p − q = q − r, the three primes are in arithmetic progression and p + q + r = 3q. The item does not say the primes must be different, so p = q = r is allowed and each of the 8 primes below 20 can be q, giving 8 distinct sums — more than 6.

1. p − q = q − r gives p + r = 2q, so p + q + r = 3q. The sum depends only on the middle prime q.
2. The item does not require the primes to be different, so p = q = r is allowed (both differences are then 0).
3. So every prime q below 20 gives a value: q = 2, 3, 5, 7, 11, 13, 17, 19 gives p + q + r = 6, 9, 15, 21, 33, 39, 51, 57.
4. That is 8 distinct values, which is more than 6.
5. Trap: if one assumed the primes must be different, only (3, 5, 7), (3, 7, 11), (3, 11, 19), (5, 11, 17) and (7, 13, 19) would work, giving 4 sums — but the item says 'distinct' only of the sums, not of the primes, so equal primes count.

Remember · Equal differences mean an arithmetic progression: the sum of three terms is three times the middle one. Apply 'distinct' only where the question states it.

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Answer: (c) Three

Only three unordered triples satisfy 1/p + 1/q + 1/r = 1: (3, 3, 3), (2, 4, 4) and (2, 3, 6). Their sums 9, 10 and 11 are all different, so p + q + r can take three values.

1. Arrange so that p ≤ q ≤ r. Then 1/p is the largest of the three fractions, so 1/p ≥ 1/3, i.e. p ≤ 3; and p ≥ 2 because 1/p must be less than 1.
2. p = 3: 1/q + 1/r = 2/3 with q, r ≥ 3 forces q = r = 3. Sum = 9.
3. p = 2: 1/q + 1/r = 1/2 with q ≤ r gives (q, r) = (3, 6) or (4, 4). Sums = 11 and 10.
4. Possible values of p + q + r: 9, 10 and 11 — three values.
5. Check: 1/2 + 1/3 + 1/6 = 1, 1/2 + 1/4 + 1/4 = 1, 1/3 + 1/3 + 1/3 = 1.

Remember · For unit-fraction equations, sort the variables and bound the smallest one first; the cases then collapse quickly.

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Answer: (c) 1078

Each remainder is 2 less than its divisor, so N + 2 is a multiple of LCM(3, 5, 6, 9) = 90. The smallest 4-digit number of the form 90k − 2 is 1080 − 2 = 1078.

1. In every case the remainder is 2 short of the divisor: 3 − 1 = 5 − 3 = 6 − 4 = 9 − 7 = 2.
2. So N + 2 is divisible by 3, 5, 6 and 9, i.e. by their LCM, 90.
3. N = 90k − 2. k = 11 gives 988 (only 3 digits); k = 12 gives 1080 − 2 = 1078.
4. Check: 1078 = 3 × 359 + 1 = 5 × 215 + 3 = 6 × 179 + 4 = 9 × 119 + 7.

Remember · If divisor minus remainder is the same d for every divisor, the number is (LCM × k) − d.

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Answer: (c) 5

The product of all odd numbers up to 999 is odd and contains 5 as a factor. An odd multiple of 5 always ends in 5.

1. The product contains the factor 5, so it is a multiple of 5 and its unit digit is 0 or 5.
2. Every factor is odd, so the product is odd and cannot end in 0.
3. Hence the unit digit is 5.
4. Check: 1 × 3 × 5 = 15 already ends in 5, and multiplying any number ending in 5 by an odd number keeps the last digit 5.

Remember · Odd × 5 ends in 5; an even number times 5 ends in 0. Spot a factor of 5 before multiplying anything.

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Answer: (c) 22

Divisibility by 9 is already covered by 3, so count the numbers from 1 to 100 that are not divisible by 2, 3, 5 or 7. Below 121 these are just 1 and the primes from 11 to 97, which gives 1 + 21 = 22.

1. 9 is a multiple of 3, so only 2, 3, 5 and 7 matter.
2. Multiples of 2, 3, 5, 7 up to 100: 50 + 33 + 20 + 14 = 117.
3. Multiples of 6, 10, 14, 15, 21, 35: 16 + 10 + 7 + 6 + 4 + 2 = 45.
4. Multiples of 30, 42, 70, 105: 3 + 2 + 1 + 0 = 6; multiples of 210: 0.
5. Divisible by at least one = 117 − 45 + 6 = 78, so not divisible by any = 100 − 78 = 22.
6. Check: such numbers are 1 and the 21 primes from 11 to 97 (the smallest composite without the factors 2, 3, 5, 7 is 11² = 121), so 1 + 21 = 22.

Remember · Drop redundant divisors first (9 with 3). Up to 120, numbers free of 2, 3, 5, 7 are just 1 and primes from 11 onwards.

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Answer: (d) None of the above

The largest x + y is 8 + 7 = 15 and the smallest x − y is 4 − 7 = −3, so the ratio is −5. None of the listed values matches; 15/2 comes from wrongly taking the smallest x − y as 4 − 2.

1. Maximum of (x + y): take both at their largest, 8 + 7 = 15.
2. Minimum of (x − y): take the smallest x and the largest y, 4 − 7 = −3.
3. Ratio = 15 ÷ (−3) = −5.
4. −5 is not 6, 15/2 or −15/2, so the answer is 'None of the above'.
5. Check: the trap 15/2 uses 4 − 2 = 2, but y = 7 makes x − y smaller.

Remember · To minimise a difference, take the smallest first term and the largest second term — and keep track of the sign.

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Answer: (b) 5

If p and k were both odd, p² + k would be even and larger than 2, so one of them must be 2. With p = 2, k can be 3, 7, 13 or 19; with k = 2 (and p = 3, giving 11) k = 2 also works, so k has 5 possible values.

1. If p and k are both odd primes, p² + k is odd + odd = even and greater than 2, so it cannot be prime. Hence p = 2 or k = 2.
2. p = 2: 4 + k must be a prime below 30. k = 3, 7, 13, 19 give 7, 11, 17, 23 (prime); k = 2, 5, 11, 17, 23 give 6, 9, 15, 21, 27 (not prime).
3. k = 2: p² + 2 must be a prime below 30. p = 3 gives 11 (prime); p = 5 gives 27 (not prime). So k = 2 is possible.
4. Possible values of k: 2, 3, 7, 13, 19 — five values.

Remember · Odd + odd is even: when a sum of primes must be prime, one of them is usually 2.

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Answer: (d) More than 8

Every number in such a set is a divisor of 1001 = 7 × 11 × 13. The sets that include 1 alone number 12, already more than 8; a full count gives 32 sets of three different numbers.

1. 1001 = 7 × 11 × 13, so every number must be one of its divisors: 1, 7, 11, 13, 77, 91, 143, 1001.
2. The LCM must contain 7, 11 and 13, and the HCF must be 1, so no prime may divide all three numbers.
3. Sets containing 1 automatically have HCF 1. Those with LCM 1001: {1, 7, 143}, {1, 11, 91}, {1, 13, 77}, {1, 77, 91}, {1, 77, 143}, {1, 91, 143}, and {1, x, 1001} for x = 7, 11, 13, 77, 91, 143 — 12 sets.
4. 12 is already more than 8; sets such as {7, 11, 13} and {7, 11, 143} add still more (32 in all with three different numbers).
5. So n is more than 8.

Remember · With a 'more than' option, you only need to cross the threshold: count the easiest family of cases and stop.

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Answer: (b) 3

Writing the numbers in place-value form gives 30 + R − S − T = 20P, which forces P = 1 and S + T = R + 10. With distinct digits, Q = 3 and T below 6, only (R, S) = (2, 8), (2, 7) and (4, 9) work.

1. PQR − PS = PPT means (100P + 10Q + R) − (10P + S) = 100P + 10P + T.
2. Simplify: 90P + 10Q + R − S = 110P + T, so 10Q + R − S − T = 20P. With Q = 3: 30 + R − S − T = 20P.
3. The left side is at most 30 + 9 − 1 − 1 = 37, so 20P = 20 and P = 1. Then S + T = R + 10.
4. P = 1 and Q = 3 are used, so R, S, T are different digits from 2, 4, 5, 6, 7, 8, 9, with T = 2, 4 or 5.
5. T = 2: S = R + 8 needs R = 1 (not allowed). T = 4: S = R + 6 gives R = 2, S = 8. T = 5: S = R + 5 gives (R, S) = (2, 7) or (4, 9).
6. So (R, S) has 3 possible values.
7. Check: 132 − 18 = 114, 132 − 17 = 115, 134 − 19 = 115.

Remember · For letter-digit equations, expand in place values, simplify, and use digit limits to fix the leading digit first.

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Answer: (b) 4

35ⁿ needs n factors of 5 and n factors of 7. The product has only four 5s (one from 385, three from 1000) but ten 7s, so the maximum n is 4.

1. 35 = 5 × 7, so count the 5s and the 7s in the product; n is the smaller count.
2. Fives: 385 = 5 × 7 × 11 gives one, and 1000 = 2³ × 5³ gives three; 7, 343, 2401 and 77777 have none. Total = 4.
3. Sevens: 7 (one), 343 = 7³ (three), 385 (one), 2401 = 7⁴ (four), 77777 = 7 × 11111 (one). Total = 10.
4. Each 35 needs one 5 and one 7, so n = the smaller of 4 and 10 = 4.

Remember · For divisibility by a power of a composite number, count each prime factor separately; the scarcer prime sets the limit.

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Answer: (b) 9

A perfect square with 17 digits has a square root of (17 + 1)/2 = 9 digits. Indeed 111111111² = 12345678987654321, and 111111111 has 9 digits.

1. 12345678987654321 has 17 digits.
2. A square with an odd number of digits d has a root with (d + 1)/2 digits: (17 + 1)/2 = 9.
3. Check: 111111111² = 12345678987654321, and 111111111 has 9 digits.

Remember · Digits in √N: (d + 1)/2 if N has d digits with d odd, d/2 if d is even. Also, 11…1² gives 12…n…21.

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Answer: (c) 2

1ⁿ is always 1, and 2ⁿ is a multiple of 4 once n ≥ 2. So the remainder is 3 when n = 1 and 1 for every larger n — two distinct remainders.

1. n = 1: 1 + 2 = 3, remainder 3.
2. n ≥ 2: 2ⁿ is divisible by 4 and 1ⁿ = 1, so 1ⁿ + 2ⁿ leaves remainder 1.
3. Only two remainders occur: 3 and 1.
4. Check: n = 2 gives 5 (remainder 1), n = 3 gives 9 (remainder 1).

Remember · Powers of 2 are multiples of 4 from 2² onwards; always check the first case separately.

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Answer: (c) 37

Every number of the form QQQ equals Q × 111 = Q × 3 × 37, and 481 = 13 × 37. They always share the factor 37, and 13 can never divide 3Q for a single digit Q, so the HCF is 37.

1. QQQ = Q × 111 = Q × 3 × 37.
2. 481 = 13 × 37.
3. HCF = 37 × HCF(3Q, 13). For a single digit Q, 3Q is at most 27 and is never 13 or 26, so HCF(3Q, 13) = 1.
4. So the HCF of P and 481 is 37, whatever the digit Q.
5. Check: HCF(222, 481) = 37, since 222 = 6 × 37 and 481 = 13 × 37.

Remember · Repeated-digit numbers: aaa = a × 3 × 37. Knowing 111 = 3 × 37 settles such items at sight.

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Answer: (d) 9

The numbers 1 to 99 use 9 + 180 = 189 digits. The remaining 300 digits cover exactly 100 three-digit numbers, 100 to 199, so the 489th digit is the final 9 of 199.

1. Digits from 1–9: 9. Digits from 10–99: 90 × 2 = 180. Total so far: 189.
2. Digits still needed: 489 − 189 = 300, all from three-digit numbers.
3. 300 ÷ 3 = 100 exactly, so the 489th digit is the last digit of the 100th three-digit number.
4. The 100th three-digit number is 100 + 99 = 199; its last digit is 9.
5. Check: digits 487–489 are 1, 9, 9 (from 199), and digit 490 is the 2 of 200.

Remember · Count digit blocks: 9 one-digit, 180 two-digit, 2700 three-digit. An exact division means the digit is the last one of that number.

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Answer: (b) 4

P = 3T allows (T, P) = (1, 3), (2, 6) or (3, 9), and S = R + 7 allows (R, Q, S) = (0, 3, 7), (1, 4, 8) or (2, 5, 9). Keeping all five digits different leaves 35291, 63072, 64182 and 94183 — four numbers.

1. P = 3T with T ≠ 0 and P a single digit: (T, P) = (1, 3), (2, 6) or (3, 9).
2. Q = R + 3 and S = Q + 4 = R + 7 ≤ 9, so R = 0, 1 or 2: (R, Q, S) = (0, 3, 7), (1, 4, 8) or (2, 5, 9).
3. T = 1, P = 3: only (2, 5, 9) avoids repeats → 35291.
4. T = 2, P = 6: (0, 3, 7) and (1, 4, 8) work → 63072 and 64182.
5. T = 3, P = 9: only (1, 4, 8) works → 94183.
6. Total: 4 numbers.

Remember · Tie the chained conditions to one variable (here R), list the few cases for each part, then strike out repeated digits.

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Answer: (c) Both I and II

'There exists' needs only one example. The number 2 becomes 3 after a 50% increase and 5 after a 150% increase; all three are primes with exactly two factors, so both statements are correct.

1. Statement I: 2 has 2 factors. Increased by 50%, it becomes 3, which also has 2 factors. So I is correct.
2. Statement II: 2 increased by 150% becomes 2 × 2.5 = 5, which also has 2 factors. So II is correct.
3. Both statements are correct.
4. Check: other examples exist too, e.g. 10 → 15 (4 factors each) for I and 6 → 15 (4 factors each) for II.

• ✓ I 2 → 3: both are primes with exactly 2 factors.
• ✓ II 2 → 5: both are primes with exactly 2 factors.

Remember · 'There exists' statements need just one example — try small primes first.

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Answer: (a) 0

Every power of 9 leaves remainder 3 when divided by 6, and there are 98 terms. 98 × 3 = 294 is a multiple of 6, so the remainder is 0.

1. 9 leaves remainder 3 when divided by 6, and 3 × 3 = 9 again leaves 3. So every power 9ᵏ leaves remainder 3.
2. Number of terms from 9³ to 9¹⁰⁰: 100 − 3 + 1 = 98.
3. Sum of remainders = 98 × 3 = 294 = 6 × 49, so the remainder is 0.
4. Check: each term is odd and a multiple of 3; 98 odd terms add to an even number, and an even multiple of 3 is divisible by 6.

Remember · For the remainder of a sum, reduce each term first; count terms carefully (3 to 100 is 98 terms).

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Answer: (c) Value-I = Value-II

For a fixed sum of 10, the product p × q is largest when p = q = 5, giving 25. Allowing some negative values only adds negative products, so both maxima are 25 and the two values are equal.

1. Value-I: with p + q = 10 and both positive, p × q = p(10 − p) is largest at p = q = 5, giving 25.
2. Value-II: p ≥ −6 and q ≥ −4 allow p from −6 to 14. The product p(10 − p) still peaks at p = 5 (25).
3. The extra cases give negative products, e.g. (−6) × 16 = −96 and 14 × (−4) = −56.
4. So Value-II = 25 = Value-I.

Remember · For a fixed sum, the product is greatest when the two numbers are equal (or as close as possible).

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Answer: (c) Value-I = Value-II

The lowest allowed set, −5 to 5, is balanced around 0, so its average is 0. The product of 0, 1, …, 10 is 0, the smallest possible for non-negative integers. Both values are 0.

1. Value-I: the lowest possible set is −5, −4, …, 4, 5. Its average is 0; any later set has a larger average. So Value-I = 0.
2. Value-II: the set 0, 1, …, 10 has product 0; any set starting above 0 has a positive product. So Value-II = 0.
3. Value-I = Value-II.

Remember · Consecutive integers balanced around 0 average 0; a product of non-negative numbers is smallest (0) when it includes 0.

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Answer: (a) I only

For a number between 0 and 1, higher powers get smaller and square roots get larger. So x² > x³ is true, but x > √x is false.

1. Take x = 0.25 as a test value.
2. I: x² = 0.0625 and x³ = 0.015625, so x² > x³. In general x² − x³ = x²(1 − x) > 0 for 0 < x < 1. True.
3. II: √0.25 = 0.5, which is larger than 0.25, so x > √x is false. For 0 < x < 1, √x is always larger than x.
4. Only statement I is correct.

• ✓ I x² − x³ = x²(1 − x) is positive when 0 < x < 1, so x² > x³.
• ✗ II For a fraction between 0 and 1 the square root is larger: √0.25 = 0.5 > 0.25.

Remember · Between 0 and 1: x³ < x² < x < √x. Test with x = 0.25 to confirm quickly.

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Answer: (c) There can be more than one such number.

Two natural numbers 10 apart have nine numbers strictly between them. These contain one multiple of 5 when both ends are multiples of 5 (5 and 15) and two otherwise (3 and 13). So there can be more than one such number, as (c) says.

1. Take 5 and 15: the only multiple of 5 strictly between them is 10 — one number.
2. Take 3 and 13: the multiples of 5 between them are 5 and 10 — two numbers.
3. The nine numbers between the two always include at least one multiple of 5, so the count is 1 or 2, never 0.
4. Only (c), 'There can be more than one such number', holds in general.

• ✗ (a) True only when both numbers are multiples of 5 (e.g. 5 and 15); 3 and 13 have two such numbers between them.
• ✗ (b) True only when neither number is a multiple of 5; 5 and 15 have just one (10) between them.
• ✓ (c) The count can be 2 (3 and 13 give 5 and 10), so more than one such number is possible.
• ✗ (d) Nine consecutive numbers always include a multiple of 5, so at least one always exists.

Remember · When a count depends on the case, test a boundary case and a general case; pick the option true in every case.

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