H.C.F. and L.C.M. of Numbers: Formulas, Shortcuts and Solved Examples — MCQs with Answers (Set 1)

For two or more fractions, the L.C.M. equals the L.C.M. of their numerators divided by the H.C.F. of their denominators. For example, for 2/3 and 4/9: L.C.M.(2,4)=4 and H.C.F.(3,9)=3, so the L.C.M. of the fractions is 4/3.

For two or more fractions, the H.C.F. equals the H.C.F. of their numerators divided by the L.C.M. of their denominators. For example, for 2/3 and 4/9: H.C.F.(2,4)=2 and L.C.M.(3,9)=9, so the H.C.F. of the fractions is 2/9.

The Euclidean method repeatedly divides the larger number by the smaller one. The smaller number is then replaced by that remainder, and the process repeats. This continues until the remainder becomes zero. The last nonzero remainder is the H.C.F. For example, for 18 and 12: 18/12 leaves remainder 6, then 12/6 leaves remainder 0, so H.C.F.(18,12) = 6.

The L.C.M. must be exactly divisible by every one of the given numbers. So for each prime appearing in any number, the L.C.M. must use its highest appearing power. For example, with 2^2 x 3 and 2 x 3^3, the L.C.M. is 2^2 x 3^3, taking the higher exponent of each prime.

When numbers are written as prime factors with exponents, the H.C.F. keeps only the primes common to all of them. For each such prime, it uses the lowest exponent seen among the numbers. For example, with 2^3 x 3^2 and 2^2 x 3^4, the H.C.F. is 2^2 x 3^2, taking the lower exponent of each prime.

For any two positive integers a and b, this identity always holds: H.C.F.(a, b) x L.C.M.(a, b) = a x b. Check with a=4 and b=6: H.C.F.(4,6)=2 and L.C.M.(4,6)=12. 2 x 12 = 24, and 4 x 6 = 24, so the identity checks out.

Co-prime numbers, also called relatively prime numbers, share no common factor other than 1. Their only common divisor is 1. So by definition their H.C.F. equals 1. For example, 8 and 15 share no factor besides 1, so H.C.F.(8, 15) = 1.

Every number divides 0 exactly, since 0 divided by any number leaves no remainder. n also divides itself exactly. So the greatest number dividing both 0 and n is n itself. Thus H.C.F.(0, n) = n.

L.C.M. stands for Least Common Multiple. It is the smallest number that every one of the given numbers divides into exactly. For example, the L.C.M. of 4 and 6 is 12. 12 is the smallest number that both 4 and 6 divide into exactly.

H.C.F. stands for Highest Common Factor. It is the greatest number that divides every one of the given numbers exactly. For example, the H.C.F. of 12 and 18 is 6. 6 is the largest number that divides both exactly.

Let x be the number to be added. 152 + x must be exactly divisible by each of 18, 30, so by their L.C.M. L.C.M.(18, 30) = 90. Dividing 152 by 90: 152/90 = 1 remainder 62. So x = 90 – 62 = 28. Check: 152 + 28 = 180, and 2 x 90 = 180, a whole multiple.

Let x be the number to be added. 5944 + x must be exactly divisible by each of 24, 36, 35, so by their L.C.M. L.C.M.(24, 36, 35) = 2520. Dividing 5944 by 2520: 5944/2520 = 2 remainder 904. So x = 2520 – 904 = 1616. Check: 5944 + 1616 = 7560, and 3 x 2520 = 7560, a whole multiple.

Let x be the number to be added. 389 + x must be exactly divisible by each of 20, 32, so by their L.C.M. L.C.M.(20, 32) = 160. Dividing 389 by 160: 389/160 = 2 remainder 69. So x = 160 – 69 = 91. Check: 389 + 91 = 480, and 3 x 160 = 480, a whole multiple.

Let x be the number to be added. 295 + x must be exactly divisible by each of 24, 30, 20, so by their L.C.M. L.C.M.(24, 30, 20) = 120. Dividing 295 by 120: 295/120 = 2 remainder 55. So x = 120 – 55 = 65. Check: 295 + 65 = 360, and 3 x 120 = 360, a whole multiple.

Let x be the number to be added. 340 + x must be exactly divisible by each of 28, 24, 12, so by their L.C.M. L.C.M.(28, 24, 12) = 168. Dividing 340 by 168: 340/168 = 2 remainder 4. So x = 168 – 4 = 164. Check: 340 + 164 = 504, and 3 x 168 = 504, a whole multiple.

Let x be the number to be added. 114 + x must be exactly divisible by each of 24, 18, 12, so by their L.C.M. L.C.M.(24, 18, 12) = 72. Dividing 114 by 72: 114/72 = 1 remainder 42. So x = 72 – 42 = 30. Check: 114 + 30 = 144, and 2 x 72 = 144, a whole multiple.

Let x be the number to be added. 190 + x must be exactly divisible by each of 20, 36, so by their L.C.M. L.C.M.(20, 36) = 180. Dividing 190 by 180: 190/180 = 1 remainder 10. So x = 180 – 10 = 170. Check: 190 + 170 = 360, and 2 x 180 = 360, a whole multiple.

Let x be the number to be added. 1917 + x must be exactly divisible by each of 24, 28, 40, so by their L.C.M. L.C.M.(24, 28, 40) = 840. Dividing 1917 by 840: 1917/840 = 2 remainder 237. So x = 840 – 237 = 603. Check: 1917 + 603 = 2520, and 3 x 840 = 2520, a whole multiple.

Let the numbers be 57 and 43, with L.C.M. 2451. For any two numbers, H.C.F. x L.C.M. = product of the numbers. Product: 57 x 43 = 2451. So H.C.F. = 2451 / 2451 = 1. Check: 1 x 2451 = 2451, which equals 57 x 43 = 2451.

Let the numbers be 69 and 42, with L.C.M. 966. For any two numbers, H.C.F. x L.C.M. = product of the numbers. Product: 69 x 42 = 2898. So H.C.F. = 2898 / 966 = 3. Check: 3 x 966 = 2898, which equals 69 x 42 = 2898.

Let the numbers be 30 and 17, with L.C.M. 510. For any two numbers, H.C.F. x L.C.M. = product of the numbers. Product: 30 x 17 = 510. So H.C.F. = 510 / 510 = 1. Check: 1 x 510 = 510, which equals 30 x 17 = 510.

Let the numbers be 23 and 26, with H.C.F. 1. For any two numbers, H.C.F. x L.C.M. = product of the numbers. Product: 23 x 26 = 598. So L.C.M. = 598 / 1 = 598. Check: 1 x 598 = 598, which equals 23 x 26 = 598.

Let the numbers be 18 and 50, with H.C.F. 2. For any two numbers, H.C.F. x L.C.M. = product of the numbers. Product: 18 x 50 = 900. So L.C.M. = 900 / 2 = 450. Check: 2 x 450 = 900, which equals 18 x 50 = 900.

Let the numbers be 59 and 54, with L.C.M. 3186. For any two numbers, H.C.F. x L.C.M. = product of the numbers. Product: 59 x 54 = 3186. So H.C.F. = 3186 / 3186 = 1. Check: 1 x 3186 = 3186, which equals 59 x 54 = 3186.

Let the numbers be 78 and 81, with L.C.M. 2106. For any two numbers, H.C.F. x L.C.M. = product of the numbers. Product: 78 x 81 = 6318. So H.C.F. = 6318 / 2106 = 3. Check: 3 x 2106 = 6318, which equals 78 x 81 = 6318.