a-b-c-N-x-4-2a-5-6-b-9-9-c-1-find-min-x-a-b-c-

Question Number 208342 by hardmath last updated on 13/Jun/24

a, b, c \in N

x = 4 (2 a + 5) = 6 (b + 9) = 9 (c - 1)

find : \min (x + a + b + c) = ?

Answered by A5T last updated on 13/Jun/24

c = \frac{6 (b + 9)}{9} + 1 = \frac{2 b}{3} + 7 \Rightarrow b = 3 k

a = \frac{1}{2} [\frac{6 (b + 9)}{4} - 5] = \frac{3 b + 17}{4} \Rightarrow a = \frac{9 k + 17}{4}

\Rightarrow k + 1 \equiv 0 (m o d 4) \Rightarrow k \equiv 3 (m o d 4)

\Rightarrow b = 3 (8 q + 3) \Rightarrow b = 24 q + 9 \Rightarrow m i n (b) = 9

\Rightarrow m i n (a) = 11 \Rightarrow m i n (c) = 13

\Rightarrow m i n (x + a + b + c) = 108 + 11 + 9 + 13 = 141

Commented by hardmath last updated on 13/Jun/24

thankyou dear professor

Answered by Rasheed.Sindhi last updated on 13/Jun/24

a, b, c \in N

x = 4 (2 a + 5) = 6 (b + 9) = 9 (c - 1)

find : \min (x + a + b + c) = ?

lcm (4, 6, 9) ∣ x \Rightarrow 36 ∣ x \Rightarrow x = 36 k

a = \frac{x - 20}{8} = \frac{36 k - 20}{8} = \frac{9 k - 5}{2} \Rightarrow k \in O

b = \frac{x}{6} - 9 = \frac{36 k}{6} - 9 = 6 k - 9 \Rightarrow k ⩾ 2

c = \frac{x}{9} + 1 = \frac{36 k}{9} + 1 = 4 k + 1

k = 3 :

x = 36 k = 36 (3) = 108

a = \frac{9 k - 5}{2} = \frac{9 (3) - 5}{2} = 11

b = 6 k - 9 = 6 (3) - 9 = 9

c = 4 k + 1 = 4 (3) + 1 = 13

\min (x + a + b + c)

= 108 + 11 + 9 + 13 = 141

Commented by hardmath last updated on 13/Jun/24

thankyou dear professor