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Question Number 69569 by Ajao yinka last updated on 25/Sep/19

	Answered by naka3546 last updated on 25/Sep/19

	Commented by Rasheed.Sindhi last updated on 26/Sep/19

	$I t h i n k^{'} p o s i t i v e {i n t e g e r}^{'} a p p l i e s t o$ $x & y o n l y . (S o c m a y b e z e r o)$ $F o r a = 2, b = 1 & c = 0 :$ $x = 20, y = 5$ $F o r (a, b, c) = (3, 1, 0) :$ $x = 50, y = 3$
	Commented by Ajao yinka last updated on 26/Sep/19

	$c o o l$
	Answered by Rasheed.Sindhi last updated on 27/Sep/19
	$10^a ≤3636⇒a<4 a>b>c: possibilities for (a,b,c) (3,2,1),(3,2,0),(3,1,0),(2,1,0) (a,b,c)=(3,2,1) _(−) 10^3 +10^2 +10^1 +x^2 +5^y =3636 x^2 +5^y =3636−1110=2526 x=(√(2526−5^y )) Testing for y=1,2,3,... x∈Z^+ only for y=3 ∴ for (a,b,c)=(3,2,1): y=3,x=49 (a,b,c)=(3,2,0) _(−) 10^3 +10^2 +10^0 +x^2 +5^y =3636 x^2 +5^y =3636−1101=2535 x=(√(2535−5^y )) x is positive integer for no value of y. ∴ No solution in (a,b,c)=(3,2,0). (a,b,c)=(3,1,0) _(−) 10^3 +10^1 +10^0 +x^2 +5^y =3636 x=(√(2625−5^y )) y=3⇒x=50 (only positive integers) ∴ for (a,b,c)=(3,1,0):y=3,x=50 (a,b,c)=(2,1,0) _(−) 10^2 +10^1 +10^1 +x^2 +5^y =3636 x=(√(3525−5^y )) y=5⇒x=20 (only positive integers) ∴for (a,b,c)=(2,1,0): y=5,x=20 { (((a,b,c)=(3,2,1):(x,y)=(49,3))),(((a,b,c)=(3,2,0):(x,y)=φ)),(((a,b,c)=(3,1,0):(x,y)=(50,3))),(((a,b,c)=(2,1,0):(x,y)=(20,5))) :}$
	$10^{a} ⩽ 3636 \Rightarrow a < 4$ $a > b > c :$ $p o s s i b i l i t i e s f o r (a, b, c)$ $(3, 2, 1), (3, 2, 0), (3, 1, 0), (2, 1, 0)$ $\underline{(a, b, c) = (3, 2, 1)}$ $10^{3} + 10^{2} + 10^{1} + x^{2} + 5^{y} = 3636$ $x^{2} + 5^{y} = 3636 - 1110 = 2526$ $x = \sqrt{2526 - 5^{y}}$ $T e s t i n g f o r y = 1, 2, 3, . . .$ $x \in Z^{+} o n l y f o r y = 3$ $∴ f o r (a, b, c) = (3, 2, 1) : y = 3, x = 49$ $\underline{(a, b, c) = (3, 2, 0)}$ $10^{3} + 10^{2} + 10^{0} + x^{2} + 5^{y} = 3636$ $x^{2} + 5^{y} = 3636 - 1101 = 2535$ $x = \sqrt{2535 - 5^{y}}$ $x i s p o s i t i v e i n t e g e r f o r n o v a l u e o f y .$ $∴ N o s o l u t i o n i n (a, b, c) = (3, 2, 0) .$ $\underline{(a, b, c) = (3, 1, 0)}$ $10^{3} + 10^{1} + 10^{0} + x^{2} + 5^{y} = 3636$ $x = \sqrt{2625 - 5^{y}}$ $y = 3 \Rightarrow x = 50 (o n l y p o s i t i v e i n t e g e r s)$ $∴ f o r (a, b, c) = (3, 1, 0) : y = 3, x = 50$ $\underline{(a, b, c) = (2, 1, 0)}$ $10^{2} + 10^{1} + 10^{1} + x^{2} + 5^{y} = 3636$ $x = \sqrt{3525 - 5^{y}}$ $y = 5 \Rightarrow x = 20 (o n l y p o s i t i v e i n t e g e r s)$ $∴ f o r (a, b, c) = (2, 1, 0) : y = 5, x = 20$ ${\begin{cases} (a, b, c) = (3, 2, 1) : (x, y) = (49, 3) \\ (a, b, c) = (3, 2, 0) : (x, y) = ϕ \\ (a, b, c) = (3, 1, 0) : (x, y) = (50, 3) \\ (a, b, c) = (2, 1, 0) : (x, y) = (20, 5) \end{cases}$
	Commented by Ajao yinka last updated on 27/Sep/19

	$i s a i d p o s i t i v e i n t e g e r s o l u t i o n, z e r o i s n o t a p o s i t i v e i n t e g e r$
	Commented by Rasheed.Sindhi last updated on 27/Sep/19

	$a, b & c a r e a r b i t r a r y c o n s t a n t s a n d x & y a r e$ $v a r i a b l e s . T h e c o n d i t i o n ‘ ‘ p o s i t i v e {i n t e g e r s i p}^{″}$ $a r e a p l i c a b l e t o v a r i a b l e s h e r e .$
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