JEE Mathematics HCF and LCM MCQs Set 01

Answer: (c) 36

Explanation:
1. \( 252 = 2^2 \times 3^2 \times 7 \).
2. \( 792 = 2^3 \times 3^2 \times 11 \).
3. \( 540 = 2^2 \times 3^3 \times 5 \).
4. Take each common prime factor with its lowest power: HCF = 36.

Teacher's Note:
1. HCF is the product of the common prime factors taken with the smallest powers.
2. For large numbers, Euclid's division method is quicker.

Answer: (a) 22

Explanation:
1. \( 990 = 2 \times 3^2 \times 5 \times 11 \).
2. \( 1870 = 2 \times 5 \times 11 \times 17 \).
3. \( 4378 = 2 \times 11 \times 199 \).
4. Take each common prime factor with its lowest power: HCF = 22.

Teacher's Note:
1. HCF is the product of the common prime factors taken with the smallest powers.
2. For large numbers, Euclid's division method is quicker.

Answer: (d) 55

Explanation:
1. \( 605 = 5 \times 11^2 \).
2. \( 1045 = 5 \times 11 \times 19 \).
3. Take each common prime factor with its lowest power: HCF = 55.

Teacher's Note:
1. HCF is the product of the common prime factors taken with the smallest powers.
2. For large numbers, Euclid's division method is quicker.

Answer: (b) 14

Explanation:
1. Euclid's algorithm on 12474 and 21854 gives the remainder chain ending in HCF = 14.

Teacher's Note:
1. HCF is the product of the common prime factors taken with the smallest powers.
2. For large numbers, Euclid's division method is quicker.

Answer: (b) 31

Explanation:
1. Subtract the remainder: the required number is the HCF of \( 20682 - 5 = 20677 \) and \( 31284 - 5 = 31279 \).
2. Euclid's algorithm: \( 31279 = 20677 \times 1 + 10602,\ 20677 = 10602 \times 1 + 10075,\ 10602 = 10075 \times 1 + 527,\ 10075 = 527 \times 19 + 62,\ 527 = 62 \times 8 + 31,\ 62 = 31 \times 2 \).
3. HCF = 31.

Teacher's Note:
1. For equal remainders r, find the HCF of (number - r) for each number.
2. The divisor must also be larger than the remainder; 31 > 5.

Answer: (d) 589

Explanation:
1. The required number is the HCF of \( 13554 - 7 = 13547 \) and \( 17092 - 11 = 17081 \).
2. \( 17081 = 13547 \times 1 + 3534,\ 13547 = 3534 \times 3 + 2945,\ 3534 = 2945 \times 1 + 589,\ 2945 = 589 \times 5 \).
3. HCF = 589.

Teacher's Note:
1. Different remainders: subtract each remainder from its own number first.
2. 589 = 19 x 31, and it is larger than both remainders.

Answer: (a) 31

Explanation:
1. If the same remainder is left, the divisor divides the differences of the numbers.
2. \( 153599 - 142408 = 11191 \) and \( 166402 - 153599 = 12803 \).
3. HCF(11191, 12803): \( 12803 = 11191 \times 1 + 1612,\ 11191 = 1612 \times 6 + 1519,\ 1612 = 1519 + 93,\ 1519 = 93 \times 16 + 31,\ 93 = 31 \times 3 \), so HCF = 31.

Teacher's Note:
1. Equal remainders: use the HCF of the differences.
2. Differences remove the common remainder.

Answer: (c) \( 2a^2bcd \)

Explanation:
1. HCF of the coefficients 12, 16, 18 is 2.
2. Lowest powers: a is 2 (from \( a^2 \)), b is 1, c is 1, d is 1, and e does not occur in the first expression, so it is left out.
3. HCF \( = 2a^2bcd \).

Teacher's Note:
1. For algebraic terms, take the HCF of the coefficients and the lowest power of each common letter.
2. A letter missing from even one term is not part of the HCF.

Answer: (c) \( 144a^5b^3c^5d^2e^2 \)

Explanation:
1. LCM of the coefficients 12, 16, 18 is 144.
2. Highest powers: \( a^5,\ b^3,\ c^5,\ d^2,\ e^2 \).
3. LCM \( = 144a^5b^3c^5d^2e^2 \).

Teacher's Note:
1. For the LCM take the highest power of every letter that appears anywhere.
2. HCF x LCM of two expressions equals their product.

Answer: (a) 7 cm

Explanation:
1. The side of the tile must divide both 224 and 189, and be as large as possible, so it is the HCF.
2. \( 224 = 2^5 \times 7 \) and \( 189 = 3^3 \times 7 \), so HCF = 7.

Teacher's Note:
1. Largest equal pieces from two lengths: think HCF.
2. The tiles needed would be \( 32 \times 27 = 864 \).

Answer: (d) 3780

Explanation:
1. \( 18 = 2 \times 3^2 \).
2. \( 28 = 2^2 \times 7 \).
3. \( 108 = 2^2 \times 3^3 \).
4. \( 105 = 3 \times 5 \times 7 \).
5. Take every prime factor with its highest power: LCM = 3780.

Teacher's Note:
1. LCM is the product of all prime factors taken with the highest powers.
2. Check that every given number divides your LCM.

Answer: (b) \( abc \)

Explanation:
1. The letters present are a, b and c, each with highest power 1.
2. LCM \( = abc \).

Teacher's Note:
1. Each letter appears with power 1, so each appears once in the LCM.
2. Do not multiply the three terms; that would give a common multiple that is not the least.

Answer: (b) \( 20xyz \)

Explanation:
1. LCM of the coefficients 2, 5, 4 is 20.
2. The letters x, y, z each occur with highest power 1, so the literal part is xyz.
3. LCM \( = 20xyz \).

Teacher's Note:
1. Treat the number part and the letter part separately.
2. LCM of 2, 5, 4 is 20, not 40.

Answer: (d) \( 24x^2y^2 \)

Explanation:
1. LCM of 8, 4, 12 is 24.
2. Highest power of x is 2 and of y is 2.
3. LCM \( = 24x^2y^2 \).

Teacher's Note:
1. Highest powers: \( x^2 \) from the first term and \( y^2 \) from the third.
2. The LCM must be divisible by each term.

Answer: (a) 9360

Explanation:
1. \( 60 = 2^2 \times 3 \times 5 \).
2. \( 36 = 2^2 \times 3^2 \).
3. \( 65 = 5 \times 13 \).
4. \( 78 = 2 \times 3 \times 13 \).
5. \( 208 = 2^4 \times 13 \).
6. Take every prime factor with its highest power: LCM = 9360.

Teacher's Note:
1. LCM is the product of all prime factors taken with the highest powers.
2. Check that every given number divides your LCM.

Answer: (c) 167739

Explanation:
1. \( 299 = 13 \times 23 \).
2. \( 221 = 13 \times 17 \).
3. \( 759 = 3 \times 11 \times 23 \).
4. Take every prime factor with its highest power: LCM = 167739.

Teacher's Note:
1. LCM is the product of all prime factors taken with the highest powers.
2. Check that every given number divides your LCM.

Answer: (c) 32

Explanation:
1. \( 3104 = 2^5 \times 97 \).
2. \( 20832 = 2^5 \times 3 \times 7 \times 31 \).
3. \( 34144 = 2^5 \times 11 \times 97 \).
4. Take each common prime factor with its lowest power: HCF = 32.

Teacher's Note:
1. HCF is the product of the common prime factors taken with the smallest powers.
2. For large numbers, Euclid's division method is quicker.

Answer: (a) 39312

Explanation:
1. \( 156 = 2^2 \times 3 \times 13 \).
2. \( 168 = 2^3 \times 3 \times 7 \).
3. \( 208 = 2^4 \times 13 \).
4. \( 432 = 2^4 \times 3^3 \).
5. Take every prime factor with its highest power: LCM = 39312.

Teacher's Note:
1. LCM is the product of all prime factors taken with the highest powers.
2. Check that every given number divides your LCM.

Answer: (d) 729

Explanation:
1. The number is of the form \( \text{LCM} + 9 \).
2. \( 15 = 3 \times 5,\ 20 = 2^2 \times 5,\ 48 = 2^4 \times 3,\ 36 = 2^2 \times 3^2 \); LCM \( = 2^4 \times 3^2 \times 5 = 720 \).
3. The required number \( = 720 + 9 = 729 \).

Teacher's Note:
1. Same remainder: LCM plus that remainder.
2. The remainder 9 is smaller than each divisor, as required.

Answer: (b) 288 seconds

Explanation:
1. They toll together again after the LCM of the intervals.
2. \( 18 = 2 \times 3^2,\ 24 = 2^3 \times 3,\ 32 = 2^5 \); LCM \( = 2^5 \times 3^2 = 288 \).

Teacher's Note:
1. Repeating events that start together meet again after the LCM of their periods.
2. 288 seconds is 4 minutes 48 seconds.

Answer: (b) 1309

Explanation:
1. For two numbers, product \( = \text{HCF} \times \text{LCM} \).
2. Other number \( = \frac{119 \times 11781}{1071} = \frac{1401939}{1071} = 1309 \).

Teacher's Note:
1. Check: HCF(1071, 1309) = 119, since \( 1071 = 119 \times 9 \) and \( 1309 = 119 \times 11 \).
2. 9 and 11 are coprime, as required.

Answer: (d) 175

Explanation:
1. The difference between each divisor and its remainder is the same: \( 12 - 7 = 18 - 13 = 20 - 15 = 5 \).
2. So the number is \( \text{LCM}(12, 18, 20) - 5 \).
3. LCM \( = 180 \), so the number is \( 180 - 5 = 175 \).

Teacher's Note:
1. When divisor minus remainder is constant, subtract that constant from the LCM.
2. Check: 175 = 12 x 14 + 7 = 18 x 9 + 13 = 20 x 8 + 15.

Answer: (a) 301

Explanation:
1. The number is 1 more than a multiple of \( \text{LCM}(2, 3, 4, 5) = 60 \), so it is \( 60k + 1 \): 61, 121, 181, 241, 301, ...
2. It must be a multiple of 7. \( 61, 121, 181, 241 \) leave remainders 5, 2, 6, 3 on division by 7, while \( 301 = 7 \times 43 \).
3. The minimum number is 301.

Teacher's Note:
1. Combine a remainder condition with a divisibility condition by testing the numbers of the form 60k + 1.
2. The next such number would be 301 + 420 = 721.