for x,y,z ∈N, if 38x+40y+41z=520 find x+y+z=?


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Question Number 200321 by mr W last updated on 17/Nov/23

$f o r x, y, z \in N, i f$ $38 x + 40 y + 41 z = 520$ $f i n d x + y + z = ?$
	Answered by Rasheed.Sindhi last updated on 17/Nov/23

	$520 - 38 x - 40 y = 41 z$ $\Rightarrow 41 ∣ (520 - 38 x - 40 y)$ $\Rightarrow 520 - 38 x - 40 y \equiv 0 (m o d 41)$ $28 + 3 x + y \equiv 0 (m o d 41)$ $3 x + y \equiv - 28 (m o d 41)$ $3 x + y \equiv 13 (m o d 41)$ $\land z = \frac{520 - 38 x - 40 y}{41} \land \underset{\Rightarrow 19 x + 20 y ⩽ 260}{38 x + 40 y ⩽ 520}$ $\underset{z}{(x, y)} \underset{=}{=} \underset{2}{(1, 10)} \underset{,}{,} \underset{4}{(2, 7)} \underset{,}{,} \underset{6}{(3, 4)} \underset{,}{,} \underset{8}{(4, 1)}, \underset{10}{\overset{\times}{(5, - 2)}}, . . .$ $x + y + z = 13 ✓$
	Answered by Rasheed.Sindhi last updated on 17/Nov/23

	$38 x + 40 y + 41 z = 520$ $40 x + 40 y + 40 z - 2 x + z = 520$ $40 x + 40 y + 40 z = 520 + 2 x - z$ $x + y + z = \frac{520 + 2 x - z}{40} = 13 + \frac{2 x - z}{40}$ $H e n c e {w e}^{'} v e t o d e t e r m i n e 13 + \frac{2 x - z}{40} \in N$ $13 + \frac{2 x - z}{40} \in N \Rightarrow 40 ∣ 2 x - z$ $2 x - z = 40 k \Rightarrow x = 20 k + \frac{z}{2} \in N \Rightarrow z i s e v e n$ $z = 2 k$ $13 + \frac{2 x - z}{40} = 13 + \frac{2 x - 2 k}{40} = 13 + \frac{x - k}{20}$ $x - k = 0, 20, 40, 60, . . . .$ $x + y + z = 13 + \frac{0}{20}$ $x = k, 20 + k, 40 + k, . . .$ $z = 2 k, 40 + 2 k, 80 + 2 k, . . .$ $C o n t i n u e$ $. . . .$
	Answered by ajfour last updated on 17/Nov/23

	$s = x + y + z$ $z < \frac{520 - 78}{41} \Rightarrow z < 11$ $x < \frac{520 - 81}{38} \Rightarrow x < 12$ $y < \frac{520 - 79}{40} \Rightarrow y < 12$ $38 x + 40 y + 41 (s - x - y) = 520$ $3 x + y = 41 s - 520 ⩾ 4$ $\Rightarrow s > \frac{524}{41} \Rightarrow s = 13, 14, . . .$ $3 x + y = 41 (s - 13) + 13$ $s = 13 + \frac{3 x + y - 13}{41}$ $4 ⩽ 3 x + y ⩽ 36$ $\Rightarrow 3 x + y = 13$ $s = 13$
	Answered by mr W last updated on 17/Nov/23

	$a l t e r n a t i v e :$ $38 x + 40 y + 41 z = 520$ $38 (x + y + z) = 520 - 2 y - 3 z < 520$ $\Rightarrow x + y + z < \frac{520}{38} = 13 \frac{13}{19} ⩽ 13 . . . (i)$ $41 (x + y + z) = 520 + 3 x + y > 520$ $\Rightarrow x + y + z > \frac{520}{41} = 12 \frac{38}{41} ⩾ 13 . . . (i i)$ $\Rightarrow x + y + z = 13$
	Answered by witcher3 last updated on 17/Nov/23

	$38 x + 40 y + 41 z = 520$ $\Leftrightarrow z = 0 [2] \Leftrightarrow$ $z = {2 z}^{'}$ $\Leftrightarrow 19 x + 20 y + {41 z}^{'} = 260$ $19 (x + y + {2 z}^{'}) + {3 z}^{'} + y = 260$ $19 (a) + {3 z}^{'} + y = 260$ ${3 z}^{'} + y = b$ $19 a ⩾ 12 b$ $b ⩽ \frac{19 a}{12}$ $19 a + b ⩽ 19 a . (\frac{13}{12})$ $\Rightarrow a ⩾ \frac{12.260}{19.13}$ $a ⩾ \frac{12.20}{19}$ $a ⩾ 13$ $a = 13 \Rightarrow 19.13 + ({3 z}^{'} + y) = 260$ ${3 z}^{'} + y = 260 - 247 = 13$ $a = 14$ $19.14 + ({3 z}^{'} + y) = 260 \Rightarrow {3 z}^{'} + y = - 6$ $\Rightarrow a = 13$ $x + y + z = 13$ ${3 z}^{'} + y = 13$ $\Rightarrow x - z^{'} = 0 \Rightarrow x = z^{'}$ $2 x = z$
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