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Question Number 163763 by Rasheed.Sindhi last updated on 10/Jan/22


Commented by Rasheed.Sindhi last updated on 10/Jan/22

$You can't use 'macro parameter character #' in math mode$
	Answered by Rasheed.Sindhi last updated on 10/Jan/22

	$II .) D e t e r m i n e t h e s e t c o u p l e$ $a, b \in N^{*} s u c h a s$ $\underline{LCM (a, b) + GCD (a, b) = b + 9}$ $GCD (a, b) = d \Rightarrow LCM (a, b) = \frac{a b}{d}$ $\frac{a b}{d} + d = b + 9$ $b (\frac{a}{d} - 1) = 9 - d$ $d = 1, 2, 3, . . ., 8$ $C a s e : d = 1 : a b + 1 = b + 9$ $a b - b = 8$ $b (a - 1) = 8 = 1 \times 8 = 8 \times 1 = 2 \times 4 = 4 \times 2$ $b = 1, a = 9 ✓$ $b = 4, a = 3 ✓$ $b = 2, a = 5 ✓$ $C a s e : d = 2 \Rightarrow a, b ⩾ 2$ $b (\frac{a}{2} - 1) = 9 - 2 = 7 = 1 \times 7 = 7 \times 1$ $b = 7, \frac{a}{2} - 1 = 1 \Rightarrow a = 4 [GCD (4, 7) \neq 2]]$ $C a s e : d = 3 \Rightarrow a, b ⩾ 3$ $b (\frac{a}{3} - 1) = 9 - 3 = 6 = 3 \times 2$ $b = 3, \frac{a}{3} - 1 = 2 \Rightarrow a = 9 ✓$ $C a s e : d = 4 \Rightarrow a, b ⩾ 4$ $b (\frac{a}{4} - 1) = 9 - 4 = 5 = 5 \times 1$ $b = 5, \frac{a}{4} - 1 = 1 \Rightarrow a = 8 [GCD (5, 8) \neq 4]$ $C a s e : d = 5 \Rightarrow a, b ⩾ 5$ $b (\frac{a}{5} - 1) = 9 - 5 = 4 [N o v a l u e o f a, b ⩾ 5]$ $C a s e s : d ⩾ 5 \Rightarrow a, b ⩾ d [N o v a l u e o f a, b ⩾ d]$ $(a, b) = (9, 1), (3, 4), (5, 2), (9, 3)$
	Answered by Rasheed.Sindhi last updated on 10/Jan/22

	$III .)$ $D i v i d e n d = a, D i v i s o r = b, R e m a i n d e r = r < b$ $Q u o t i e n t = q$ $T h e s e q u a n t i t i e s a r e r e l a t e d a s u n d e r :$ $a = b q + r; 0 ⩽ r < b$ $r = 25 \land b > r \Rightarrow b > 25$ $∴ M i n i m u m v a l u e o f b = 26$ $a = b q + r \Rightarrow a = 26 \times 17 + 25$ $a = 467$ $(a, b) = (467, 26)$
	Answered by kdaramaths last updated on 10/Jan/22
	$I) N=2975 ; gcd(a.b)=d lcm(a.b)=m (s): a^2 + b^2 = m^2 − N a.) N=2975 = 5^2 ×119=5^2 ×7×17 the divisors of n ={1. 5 . 25....2975} the divisors of asked are {1 . 5 } b.) ensemble of divisors of 120 we have: 120 =2^3 ×3×5 D_(120) ={1.2.3.4.6.8.12.20.24.15.30.60.120} c.)solve of the equation s (S): a^2 +b^2 = m^2 − N let a=a′d . b= b′d with d(a′.b′)=1 a×b =md⇒ a′b′d^2 =md⇒m=a′b′d in (s) (a′d)^2 + (b′d)^2 =(a′b′d)^2 −N a′^2 + b′^2 −(a′b′)^2 = −(N/d^2 ) a′^2 b′^2 −a′^2 −b′^2 =(N/d^2 ) b′^2 ( a′^2 −1) − b′^2 +1−1 =... (a′^(2 ) −1)(b′^2 −1) =(N/d^2 ) +1 d is a divisors of N and with the answers of a.) the possible valus of d is {1 . 5} for d=1 ⇒(a′^2 −1)(b′^2 −1) =2976 (a rejete car ∄(a.b) such that (a^2 −1)(b^2 −1)=2976. for d=5⇒(a′^2 −1)(b′^2 −1)=120 (a′^2 −1)(b′^2 −1)=1×120 =4×30 =8×15 =2×60 =5×24 the valable possibilite is : (a′^2 −1)(b′^2 −1)= 8×15 {a⟨b} ⇒ a′^2 −1=8 ⇒a′=3 b′^2 −1=15⇒b′=4 a=a′d⇒ a=3×5 ⇒ a=15 b=b′d ⇒b=4×5 ⇒ b=20 S_N^(2 ) ={(15 . 20)}$
	$I) N = 2975; g c d (a . b) = d l c m (a . b) = m$ $(s) : a^{2} + b^{2} = m^{2} - N$ $a .) N = 2975 = 5^{2} \times 119 = 5^{2} \times 7 \times 17$ $t h e d i v i s o r s o f n = {1 . 5 . 25 . . . . 2975}$ $t h e d i v i s o r s o f a s k e d a r e {1 . 5}$ $b .) e n s e m b l e o f d i v i s o r s o f 120$ $w e h a v e : 120 = 2^{3} \times 3 \times 5$ $D_{120} = {1.2.3.4.6.8.12.20.24.15.30.60.120}$ $c .) s o l v e o f t h e e q u a t i o n s$ $(S) : a^{2} + b^{2} = m^{2} - N$ $l e t a = a^{'} d . b = b^{'} d w i t h d (a^{'} . b^{'}) = 1$ $a \times b = m d \Rightarrow a^{'} b^{'} d^{2} = m d \Rightarrow m = a^{'} b^{'} d$ $i n (s) {(a^{'} d)}^{2} + {(b^{'} d)}^{2} = {(a^{'} b^{'} d)}^{2} - N$ $a^{' 2} + b^{' 2} - {(a^{'} b^{'})}^{2} = - \frac{N}{d^{2}}$ $a^{' 2} b^{' 2} - a^{' 2} - b^{' 2} = \frac{N}{d^{2}}$ $b^{' 2} (a^{' 2} - 1) - b^{' 2} + 1 - 1 = . . .$ $(a^{' 2} - 1) (b^{' 2} - 1) = \frac{N}{d^{2}} + 1$ $d i s a d i v i s o r s o f N a n d w i t h t h e$ $a n s w e r s o f a .) t h e p o s s i b l e v a l u s$ $o f d i s {1 . 5}$ $f o r d = 1 \Rightarrow (a^{' 2} - 1) (b^{' 2} - 1) = 2976$ $(a r e j e t e c a r ∄ (a . b) s u c h t h a t$ $(a^{2} - 1) (b^{2} - 1) = 2976 .$ $f o r d = 5 \Rightarrow (a^{' 2} - 1) (b^{' 2} - 1) = 120$ $(a^{' 2} - 1) (b^{' 2} - 1) = 1 \times 120$ $= 4 \times 30$ $= 8 \times 15$ $= 2 \times 60$ $= 5 \times 24$ $t h e v a l a b l e p o s s i b i l i t e i s :$ $(a^{' 2} - 1) (b^{' 2} - 1) = 8 \times 15$ ${a ⟨ b} \Rightarrow a^{' 2} - 1 = 8 \Rightarrow a^{'} = 3$ $b^{' 2} - 1 = 15 \Rightarrow b^{'} = 4$ $a = a^{'} d \Rightarrow a = 3 \times 5 \Rightarrow a = 15$ $b = b^{'} d \Rightarrow b = 4 \times 5 \Rightarrow b = 20$ $S_{N^{2}} = {(15 . 20)}$
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