Problems On H_C_F And L_C_M — Part 1 MCQs with Explanations

Required number = H.C.F. of (91 - 43), (183 - 91) and (183 - 43) = H.C.F. of 48, 92 and 140 = 4.

The HCF of a group of numbers will be always a factor of their LCM. HCF is the product of all common prime factors using the least power of each common prime factor. LCM is the product of highest powers of all prime factors. Clearly, the numbers are (23 x 13) and (23 x 14) ∴ Larger number = (23 x 14) = 322

L.C.M. of 2, 4, 6, 8, 10, 12 is 120. So, the bells will toll together after every 120 seconds(2 minutes). In 30 minutes, they will toll together $$\frac{{30}}{2}$$ + 1 = 16 times

Understanding the Question: This question asks us to find the largest number (N) that divides three given numbers (1305, 4665, and 6905) and leaves the same remainder in each division. After finding N, we need to calculate the sum of its digits . Key Concept: Finding N If a number 'N' leaves the same remainder 'R' when dividing numbers 'a', 'b', and 'c', then (a-b), (b-c), and (c-a) are perfectly divisible by N. In other words, 'N' is the Highest Common Factor (HCF) of the differences between the numbers. Steps to Solve: 1. Calculate the differences: * 4665 - 1305 = 3360 * 6905 - 4665 = 2240 * 6905 - 1305 = 5600 2. Find the HCF of these differences (3360, 2240, and 5600): The HCF of these numbers is 1120. Therefore, N = 1120 3. Calculate the sum of the digits of N: * 1 + 1 + 2 + 0 = 4 Conclusion: The sum of the digits in N is 4. Therefore, the correct option is A: 4

Greatest number of 4-digits is 9999. L.C.M. of 15, 25, 40 and 75 is 600. On dividing 9999 by 600, the remainder is 399. ∴ Required number (9999 - 399) = 9600.

Let the numbers be 37 a and 37 b Then, 37 a x 37 b = 4107 ⇒ ab = 3 Now, co-primes with product 3 are (1, 3) So, the required numbers are (37 x 1, 37 x 3) i.e., (37, 111) ∴ Greater number = 111

Let the numbers be 3 x , 4 x and 5 x Then, their L.C.M. = 60 x So, 60 x = 2400 or x = 40 ∴ The numbers are (3 x 40), (4 x 40) and (5 x 40) Hence, required H.C.F. = 40

Given numbers are 1.08, 0.36 and 0.90. H.C.F. of 108, 36 and 90 is 18, ∴ H.C.F. of given numbers = 0.18

Let the numbers 13 a and 13 b Then, 13 a x 13 b = 2028 ⇒ ab = 12 Now, the co-primes with product 12 are (1, 12) and (3, 4) [Note: Two integers a and b are said to be co-prime or relatively prime if they have no common positive factor other than 1 or, equivalently, if their greatest common divisor is 1 ] So, the required numbers are (13 x 1, 13 x 12) and (13 x 3, 13 x 4) Clearly, there are 2 such pairs.

L.C.M. of 6, 9, 15 and 18 is 90 Let required number be 90 k + 4, which is multiple of 7 Least value of k for which (90 k + 4) is divisible by 7 is k = 4 ∴ Required number = (90 x 4) + 4 = 364

$$\eqalign{ & 2\,\,\,|\,\,\,24\,\,\,\,\, - \,\,\,\,\,36\,\,\,\,\, - \,\,\,\,\,40 \cr & - - - - - - - - - - - - - \cr & 2\,\,\,|\,\,\,12\,\,\,\,\, - \,\,\,\,\,18\,\,\,\,\, - \,\,\,\,\,20 \cr & - - - - - - - - - - - - - \cr & 2\,\,\,|\,\,\,\,\,\,6\,\,\,\,\, - \,\,\,\,\,\,\,\,9\,\,\,\,\, - \,\,\,\,\,10 \cr & - - - - - - - - - - - - - \cr & 3\,\,\,|\,\,\,\,\,\,\,3\,\,\,\,\, - \,\,\,\,\,\,\,\,9\,\,\,\,\, - \,\,\,\,\,\,\,\,5 \cr & - - - - - - - - - - - - - \cr & \,\,\,\,\,\,\,|\,\,\,\,\,1\,\,\,\, - \,\,\,\,\,\,\,\,\,\,3\,\,\,\,\, - \,\,\,\,\,\,\,\,5 \cr & L.C.M = 2 \times 2 \times 2 \times 3 \times 3 \times 5 = 360 \cr} $$

L.C.M. of 5, 6, 4 and 3 = 60 On dividing 2497 by 60, the remainder is 37 ∴ Number to be added = (60 - 37) = 23

$$\eqalign{ & {\text{128352}}\overline {)238368{\text{(}}} 1 \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\text{128352}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\overline {110016{\text{)128352(}}} 1 \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,110016 \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\overline {183336{\text{)110016(}}} {\text{6}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\frac{{110016}}{ \times } \cr & {\text{So, H}}{\text{.C}}{\text{.F of 128352 and 238368}} \cr & {\text{ = 18336}} \cr & \therefore \frac{{128352}}{{238368}} = \frac{{128352 \div 18336}}{{238368 \div 18336}} = \frac{7}{{13}} \cr} $$

L.C.M. of 5, 6, 7, 8 = 840. ∴ Required number is of the form 840 k + 3 Least value of k for which (840 k + 3) is divisible by 9 is k = 2 ∴ Required number = (840 x 2 + 3) = 1683

$$\eqalign{ & {\text{Other number}} \cr & = {\frac{{11 \times 7700}}{{275}}} \cr & = 308 \cr} $$

12 = 2 × 2 × 3 18 = 2 × 3 × 3 21 = 3 × 7 30 = 2 × 3 × 5 L.C.M. of 12, 18, 21 30 = 2 × 3 × 2 × 3 × 7 × 5 = 1260 Required number = $$\frac{{1260}}{2}$$ = 630

Let the numbers be 3 x and 4 x . Then, their H.C.F. = x . So, x = 4 So, the numbers 12 and 16 L.C.M. of 12 and 16 = 48

Required number = (L.C.M. of 12,16, 18, 21, 28) + 7 = 1008 + 7 = 1015

Clearly, 252 = 2 × 2 × 3 × 3 × 7

Required length = H.C.F. of 700 cm, 385 cm and 1295 cm = 35 cm

Since the numbers are co-prime, they contain only 1 as the common factor. Also, the given two products have the middle number in common. So, middle number = H.C.F. of 551 and 1073 = 29 $$\eqalign{ & {\text{First}}\,{\text{number}} \cr & = {\frac{{551}}{{29}}} \cr & = 19 \cr & {\text{Third}}\,{\text{number}} \cr & = {\frac{{1073}}{{29}}} \cr & = 37 \cr & \therefore {\text{Required}}\,{\text{sum}} \cr & = {19 + 29 + 37} \cr & = 85 \cr} $$

36 = 2 2 × 3 2 84 = 2 2 × 3 x 7 ∴ H.C.F. = 2 2 × 3 = 12

$$\eqalign{ & {\text{L}}{\text{.C}}{\text{.M}}{\text{.}}\,{\text{of}}\,8,\,16,\,40\,{\text{and}}\,80 = 80 \cr & \frac{7}{8} = \frac{{70}}{{80}};\,\frac{{13}}{{16}} = \frac{{65}}{{80}};\,\frac{{31}}{{40}} = \frac{{62}}{{80}} \cr & {\text{Since}},\,\frac{{70}}{{80}} > \frac{{65}}{{80}} > \frac{{63}}{{80}} > \frac{{62}}{{80}}, \cr & {\text{So}},\,\frac{7}{8} > \frac{{13}}{{16}} > \frac{{63}}{{80}} > \frac{{31}}{{40}} \cr & {\text{So}},\,\frac{7}{8}\,{\text{is}}\,{\text{the}}\,{\text{largest}}{\text{.}} \cr} $$

Required number = (L.C.M. of 12, 15, 20, 54) + 8 = 540 + 8 = 548

Required number = H.C.F. of (1657 - 6) and (2037 - 5) = H.C.F. of 1651 and 2032 = 127

99 = 1 × 3 × 3 × 11 101 = 1 × 101 176 = 1 × 2 × 2 × 2 × 2 × 11 182 = 1 × 2 × 7 × 13 So, divisors of 99 are 1, 3, 9, 11, 33, 99 Divisors of 101 are 1 and 101 Divisors of 176 are 1, 2, 4, 8, 11, 16, 22, 44, 88 and 176 Divisors of 182 are 1, 2, 7, 13, 14, 26, 91 and 182 Hence, 176 has the most number of divisors.

Let the numbers be 2 x and 3 x Then, their L.C.M. = 6 x So, 6 x = 48 or x = 8 ∴ The numbers are 16 and 24 Hence, required sum = (16 + 24) = 40

$$\eqalign{ & {\text{Required}}\,{\text{H}}{\text{.C}}{\text{.F}}{\text{.}} \cr & = \frac{{{\text{H}}{\text{.C}}{\text{.F}}{\text{.}}\,{\text{of}}\,9,\,12,\,18,\,21}}{{{\text{L}}{\text{.C}}{\text{.M}}{\text{.}}\,{\text{of}}\,10,\,25,\,35,\,40}} \cr & = \frac{3}{{1400}} \cr} $$

$$\eqalign{ & {\text{Let}}\,{\text{the}}\,{\text{numbers}}\,{\text{be}}\,a\,{\text{and}}\,b \cr & {\text{Then}},\,a + b = 55\,{\text{and}} \cr & ab = 5 \times 120 = 600 \cr & \therefore {\text{The}}\,{\text{required}}\,{\text{sum}} \cr & = \frac{1}{a} + \frac{1}{b} = \frac{{a + b}}{{ab}} = \frac{{55}}{{600}} = \frac{{11}}{{120}} \cr} $$

HCF = 8 LCM = 48 One number = 24 Let other number be = y ∴ 24y = 48 × 8 ⇔ y = 16

LCM of (4, 6, 8, 12, 16) ⇒ 16 × 3 = 48 ∴ The number when divided by (4, 6, 8, 12, 16) leaves remainder 2 is = 48 + 2 = 50

LCM = 495 HCF = 5 ( given ) ∴ Let numbers are = 5x and 5y ∴ LCM = 5xy ⇒ 5xy = 495 ⇒ xy = 99 ∴ Possible co-prime factors are 1, 99 9, 11 ∴ Possible numbers are 5x, 5y 45, 55 5, 495 Now given that sum of numbers = 100 So, required numbers are = (45, 55) ∴ Difference of numbers = 55 - 45 = 10

99 = 1 × 3 × 3 × 11 101 = 1 × 101 176 = 1 × 2 × 2 × 2 × 2 × 11 182 = 1 × 2 × 7 × 13 So, divisor of 99 are 1, 3, 9, 11, 33 and 99 Divisor of 101 are 1 and 101 Divisor of 176 are 1, 2, 4, 8, 11, 16, 22, 44, 88 and 176 Divisor of 182 are 1, 2, 7, 13, 14, 26, 91 and 182 Hence, 176 has the most number of divisors.

HCF = 8 Now LCM should have a factors 8 So check also the option we have only 60 which does not have a factor 8. So it will never be the LCM.

Numbers x = 28, y = 42 HCF (28 , 42) Difference = 42 - 28 = 14 For HCF of any numbers take their difference. HCF will be either the factor of that difference or the difference itself. Now, LCM of 28, 42 ∴ 14 × 2 × 3 = 84 ⇒ LCM : HCF ⇒ 84 : 14 ⇒ 6 : 1

$$\eqalign{ & {\text{116690151}}\overline {){\text{427863887(}}} {\text{3}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\text{350070453}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\overline {{\text{77793434)116690151(}}} 1 \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\text{77793434}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\overline {{\text{38896717)77793434(}}} {\text{2}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\frac{{77793434}}{ \times } \cr & {\text{So, H}}{\text{.C}}{\text{.F of 116690151 and 427863887}} \cr & {\text{ = 38896717}} \cr & \therefore \frac{{116690151}}{{427863887}} = \frac{{116690151 \div 38896717}}{{427863887 \div 38896717}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{3}{{11}} \cr} $$

$$\eqalign{ & {\text{LCM = 1820}} \cr & {\text{HCF = 26}} \cr & {{\text{1}}^{{\text{st}}}}{\text{ number = 130}} \cr & {\text{LCM}} \times {\text{HCF }} \cr & {\text{ = Product of numbers}} \cr & \Rightarrow {\text{Let the other number is }}x{\text{ }} \cr & 130 \times x = 1820 \times 26 \cr & x = \frac{{1820 \times 26}}{{130}} = 364 \cr} $$

HCF = 478 Number are = 12906 and 14818 LCM × HCF = 12906 × 14818 LCM × 478 = 12906 × 14818 LCM = 400086

LCM of (3, 5, 8, 12) ⇒ 3 × 5 × 4 × 2 = 120 ⇒ Now greatest five digits number is 99999 On dividing 99999 by = 120 (LCM) We get remainder = $$\frac{{99999}}{{120}},$$ remainder = 39 ⇒ By subtracting remainder from 99999 we get the greatest five digits number which is completely divisible by given numbers (3, 5, 8, 12) ∴ 99999 - 39 = 99960 ⇒ Now, we required greatest five digit number which when divided by (3, 5, 8, 12) leaves remainder 2 in each case. ⇒ add 2 in the 99960 ⇒ 99960 + 2 ⇒ 99962

H.C.F. = Product of lowest powers of common factors = 3 × 5 × 7 = 105