The-sum-of-two-numbers-are-20-and-their-LCM-is-24-What-are-the-two-numbers-

Question Number 105803 by Anindita last updated on 31/Jul/20

The sum of two

numbers are 20 and

their LCM is 24 .

What are the two

numbers ?

Commented by Anindita last updated on 31/Jul/20

Please help me to solve it .

Answered by floor(10²Eta[1]) last updated on 31/Jul/20

a + b = 20

lcm (a, b) = 24

lcm (a, b) . \gcd (a, b) = ab \Rightarrow 24 . \gcd (a, b) = ab

\gcd (a, b) = d \Rightarrow d ∣ a, d ∣ b \Rightarrow d ∣ a + b ∴ d ∣ 20

a = {da}^{'}, b = {db}^{'}, \gcd (a^{'}, b^{'}) = 1 \Rightarrow ab = d^{2} a^{'} b^{'}

24 d = d^{2} a^{'} b^{'} \Rightarrow a^{'} b^{'} = \frac{24}{d} \Rightarrow d ∣ 24

\Rightarrow d ∣ 20, d ∣ 24 \Rightarrow d ∣ \gcd (20, 24) \Rightarrow d ∣ 4 .

\Rightarrow d \in {1, 2, 4}

d = 1 :

ab = 24

a + b = 20

\Rightarrow a \notin Z

d = 2 :

ab = 48 \Rightarrow a \notin Z

d = 4 :

ab = 96

a + b = 20

\Rightarrow a^{2} - 20 a + 96 = 0 \Rightarrow a = \frac{20 \pm 4}{2} ∴ a = 12, a = 8

b = 8, b = 12

\Rightarrow the two numbers are 12 and 8

Answered by Ar Brandon last updated on 31/Jul/20

a + b = 20

24 = {ak}_{1} = {bk}_{2} \Rightarrow {ak}_{1} + {bk}_{1} = {20 k}_{1}

\Rightarrow {bk}_{2} + {bk}_{1} = {20 k}_{1} \Rightarrow b = \frac{{20 k}_{1}}{k_{1} + k_{2}} \in N

k_{1} = 2 k_{2} = 3 \Rightarrow b = 8, a = 12

Answered by 1549442205PVT last updated on 01/Aug/20

Suppose \gcd (a, b) = d \Rightarrow a = md, b = nd

(where \gcd (m, n) = 1) . Then

LCM (a, b) = 24 = mnd . On the other

hands, 20 = a + b = (m + n) d

\Leftrightarrow \frac{(m + n) d}{mnd} = \frac{20}{24} = \frac{5}{6} \Rightarrow 6 (m + n) = 5 mn

\Rightarrow 25 mn - 30 (m + n) + 36 = 36

\Rightarrow (5 m - 6) (5 n - 6) = 36 .

5 m - 6 \in {1, 2, 3, 4, 6, 9, 18, 36}

It is easy to see that only 5 m - 6 \in {4, 9}

satisfy

i) 5 m - 6 = 4 \Rightarrow m = 2, 5 n - 6 = 9 \Rightarrow n = 3

\Rightarrow d = \frac{20}{m + n} = 4 \Rightarrow a = md = 8, b = nd = 12

ii) 5 m - 6 = 9 \Rightarrow m = 3, n = 2 \Rightarrow d = 4 \Rightarrow a = 12, b = 8

Thus, (m, n) \in {(8, 12), (12, 8)}