If a + b = 1001 & HCF(a, b) = 13 then how many set of a & b.


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Question Number 193180 by BaliramKumar last updated on 06/Jun/23

$If a + b = 1001 & HCF (a, b) = 13$ $then how many set of a & b .$
	Answered by Subhi last updated on 06/Jun/23

	$a = 13 k$ $b = 13 m$ $a + b = 1001 = 13 (k + m)$ $k + m = 77$ $t h e r e a r e 78 p a i r s$ $(77, 0), (76, 1) . . . . . . . . . . . . . (1, 76), (0, 77)$ $t h e r e a r e r e s t r i c r i o n s w h e n m a n d k a r e d i v i s i b l e b y 11 o r 7$ $m = 7 l$ $k = 7 n$ $k + m = 77 = 7 (l + n)$ $l + n = 11$ $(l, n) = (10, 1), (9, 2), (8, 3), (7, 4), (6, 5) . . . . . . . . (1, 10)$ $10 p a i r s a r e e x c l u d e d$ $(i i) m = 11 g$ $k = 11 h$ $k + m = 77 = 11 (g + h)$ $g + h = 7$ $(g, h) = (6, 1), (5, 2), (4, 3) . . . . . (1, 6)$ $6 p a i r s a r e e x c l u d e d$ $78 - 6 - 10 = 62$
	Commented by MM42 last updated on 06/Jun/23

	$t h e r e i s n o d i f f r e n c e b e t w e e n t h e t w o n u m b e r s$ $a = 1, b = 76 a n d a = 76, b = 1 a n d . . .$
	Commented by Subhi last updated on 06/Jun/23

	$(1, 76) i s d i f f e r e n t t o (76, 1)$ $v a l u e s o f a a n d b d i f f e r$ $, b u t y o u a r e r i g h t t h a t t h e r e r e s t r i c t i o n s f o r s o m e v a l u e s f o r a, b w h e n f a c t o r s k, m a r e d i v i s i b l e b y 11 o r 7$
	Commented by Subhi last updated on 06/Jun/23

	$T h a n k y o u f o r t h a t p o i n t$
	Commented by BaliramKumar last updated on 07/Jun/23

	$Sir$ $0, 77 \Rightarrow 13 \times (0, 77) = (0, 1001)$ $HCF (0, 1001) = ?$
	Commented by MM42 last updated on 07/Jun/23

	$y o u a r e r i g h t$
	Answered by MM42 last updated on 07/Jun/23

	$a = 13 a^{'} & b = 13 b^{'}; (a^{'}, b^{'}) = 1$ $\Rightarrow 13 a^{'} + 13 b^{'} = 1001 \Rightarrow a^{'} + b^{'} = 77$ $W = {{a^{'}, b^{'}} : a^{'} + b^{'} = 77, (a^{'}, b^{'}) = 1}$ $S = {{m, n} : m + n = 77} = {{1, 76}, {2, 75}, . . ., {38, 39}} \Rightarrow∣ S ∣= 38$ $A = {{m, n} : m + n = 77, (m, n) = 7}$ $\Rightarrow A = {{7, 70}, {14, 63}, {21, 56}, {28, 49}, {35, 42}} \Rightarrow∣ A ∣= 5$ $B = {{m, n} : m + n = 77, (m, n) = 11}$ $\Rightarrow A = {{11, 66}, {22, 55}, {33, 44}} \Rightarrow∣ B ∣= 3$ $\Rightarrow∣ W ∣= 38 - 8 = 30$
	Answered by BaliramKumar last updated on 18/Nov/23

	$\frac{ϕ (\frac{1001}{13})}{2} = \frac{ϕ (\frac{7 \times 11 \times 13}{13})}{2} = \frac{ϕ (7 \times 11)}{2}$ $\Rightarrow \frac{(7 - 1) (11 - 1)}{2} = \frac{6 \times 10}{2} = 30 unordered pairs$ $\Rightarrow 60 ordered pairs$
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