a,b,c∈N x = 4(2a+5) = 6(b+9) = 9(c−1) find: min(x+a+b+c) = ?


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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

	$thankyou dear professor$
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