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Question Number 72608 by aliesam last updated on 30/Oct/19

	Answered by mind is power last updated on 30/Oct/19
	$s(x)=n⇒x∈[10^(n−1) ,10^n [ s(y)=m⇒y∈[10^(m−1) ,10^m [ s(z)=p⇒z∈[10^(p−1) ,10^p [ 1&3⇒s(z)=y−x−4≥0⇒y≥x+4⇒n≥m 1&3&2⇒z=2y−4≥0,y≥5 ⇒p≥m 2.y−4≤2.10^m −4<10^(m+1) ⇒p≤m+1 ⇒p∈{m,m+1} p=m⇒s(x)+2s(y)=y−4 s(x)+2s(y)=y−4 s(x)≤s(y)⇒y−4≤3s(y) ⇒10^(m−1) −4≤3m⇒m∈{1,2} m=1 s(x)=x−1 x+y=z−1 s(x)=y−6≥1⇒y≥7 z=2y−4⇒z≥10 but s(z)=1 no solution m=2⇒s(x)=y−8 x+y+2=z=2y−4⇒x=y−2 s(x)+2=x s(x)−x=−2⇒(y−8)−(y−2)=−2⇒−6=−2 no solution ⇒p=m+1 ⇒s(x)+2s(y)+1=y−4≤3s(y)+1 ⇒10^(m−1) −4≤3m+1⇒10^(m−1) ≤3m+5⇒m∈{1,2}m=1 s(x)≤s(y)⇒s(x)=1 m=1⇒s(x)+3=y−4 ⇒y=8 x=2 z=2+8+2=12 (x,y,z)={(2,8,12)} m=2 s(x)≤2 ⇒s(x)+5=y−4⇒s(x)=y−9≤2⇒y≤11 y∈{10,11} s(z)=3⇒ z=2y−4≥100 absurd ⇒{(x,y,z)}={(2,8,12)}$
	$s (x) = n \Rightarrow x \in [10^{n - 1}, 10^{n} [$ $s (y) = m \Rightarrow y \in [10^{m - 1}, 10^{m} [$ $s (z) = p \Rightarrow z \in [10^{p - 1}, 10^{p} [$ $1 & 3 \Rightarrow s (z) = y - x - 4 ⩾ 0 \Rightarrow y ⩾ x + 4 \Rightarrow n ⩾ m$ $1 & 3 & 2 \Rightarrow z = 2 y - 4 ⩾ 0, y ⩾ 5 \Rightarrow p ⩾ m$ $2 . y - 4 ⩽ 2 . 10^{m} - 4 < 10^{m + 1} \Rightarrow p ⩽ m + 1$ $\Rightarrow p \in {m, m + 1}$ $p = m \Rightarrow s (x) + 2 s (y) = y - 4$ $s (x) + 2 s (y) = y - 4$ $s (x) ⩽ s (y) \Rightarrow y - 4 ⩽ 3 s (y)$ $\Rightarrow 10^{m - 1} - 4 ⩽ 3 m \Rightarrow m \in {1, 2}$ $m = 1$ $s (x) = x - 1$ $x + y = z - 1$ $s (x) = y - 6 ⩾ 1 \Rightarrow y ⩾ 7$ $z = 2 y - 4 \Rightarrow z ⩾ 10 but s (z) = 1 no solution$ $m = 2 \Rightarrow s (x) = y - 8$ $x + y + 2 = z = 2 y - 4 \Rightarrow x = y - 2$ $s (x) + 2 = x$ $s (x) - x = - 2 \Rightarrow (y - 8) - (y - 2) = - 2 \Rightarrow - 6 = - 2 no solution$ $\Rightarrow p = m + 1$ $\Rightarrow s (x) + 2 s (y) + 1 = y - 4 ⩽ 3 s (y) + 1$ $\Rightarrow 10^{m - 1} - 4 ⩽ 3 m + 1 \Rightarrow 10^{m - 1} ⩽ 3 m + 5 \Rightarrow m \in {1, 2} m = 1$ $s (x) ⩽ s (y) \Rightarrow s (x) = 1$ $m = 1 \Rightarrow s (x) + 3 = y - 4$ $\Rightarrow y = 8$ $x = 2$ $z = 2 + 8 + 2 = 12$ $(x, y, z) = {(2, 8, 12)}$ $m = 2$ $s (x) ⩽ 2$ $\Rightarrow s (x) + 5 = y - 4 \Rightarrow s (x) = y - 9 ⩽ 2 \Rightarrow y ⩽ 11$ $y \in {10, 11}$ $s (z) = 3 \Rightarrow$ $z = 2 y - 4 ⩾ 100 absurd$ $\Rightarrow {(x, y, z)} = {(2, 8, 12)}$
	Commented by aliesam last updated on 30/Oct/19

	$e l e g a n t s o l u t i o n s i r . g o d b l e s s y o u$
	Commented by mind is power last updated on 30/Oct/19

	$Thank you sir$
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