How many pairs of positive integers x, y exist such that HCF (x, y) + LCM(x, y) = 91?


Question and Answers Forum

All Questions Topic List
Arithmetic Questions
Previous in All Question Next in All Question
Previous in Arithmetic Next in Arithmetic
Question Number 189357 by BaliramKumar last updated on 15/Mar/23

How many pairs of positive integers x, y exist such that HCF (x, y) + LCM(x, y) = 91?
Commented by mr W last updated on 15/Mar/23

$5 p a i r s :$ $(1, 90)$ $(2, 45)$ $(5, 18)$ $(7, 84)$ $(9, 10)$
Commented by BaliramKumar last updated on 15/Mar/23

$s i r, a n s w e r i s 8$
Commented by som(math1967) last updated on 15/Mar/23

$(13, 78), (21, 28), (26, 39)$ $91 = 1 \times 7 \times 13 ∴ H . C . F o f x, y$ $e i t h e r 1 o r 7 o r 13$
Commented by mr W last updated on 15/Mar/23

$t h a n k s s i r s!$
Commented by BaliramKumar last updated on 15/Mar/23

$n i c e s i r$
	Answered by BaliramKumar last updated on 15/Mar/23

	$L e t H C F (x, y) = k$ $x = k \cdot a, y = k \cdot b, L C M (x, y) = k \cdot a \cdot b$ $H C F (x, y) + L C M (x, y) = 91$ $k + k \cdot a \cdot b = 91$ $k (1 + a \cdot b) = 91 . . . . . . . . . . (i)$ $k = 1, 7, 13, 91$ $p u t k = 1 i n e q u . (i)$ $1 (1 + a b) = 91$ $a b = 90 = 2^{1} \times 3^{2} \times 5^{1} [3 d i s t i n c t p r i m e f a c t o r o f 90]$ $N o . o f c o - p r i m e p a i r (a, b) = 2^{3 - 1} = 4$ $\begin{array}{cc} (a, b) & (1, 90) & (2, 45) & (5, 18) & (9, 10) \\ (x, y) = (k a, k b) & (1, 90) & (2, 45) & (5, 18) & (9, 10) \end{array}$ $p u t k = 7 i n e q u . (i)$ $7 (1 + a b) = 91$ $a b = 12 = 2^{2} \times 3^{1} [2 d i s t i n c t p r i m e f a c t o r o f 12]$ $N o . o f c o - p r i m e p a i r (a, b) = 2^{2 - 1} = 2$ $\begin{array}{cc} (a, b) & (1, 12) & (3, 4) \\ (x, y) = (k a, k b) & (7, 84) & (21, 28) \end{array}$ $p u t k = 13 i n e q u . (i)$ $13 (1 + a b) = 91$ $a b = 6 = 2^{1} \times 3^{1} [2 d i s t i n c t p r i m e f a c t o r o f 6]$ $N o . o f c o - p r i m e p a i r (a, b) = 2^{2 - 1} = 2$ $\begin{array}{cc} (a, b) & (1, 6) & (2, 3) \\ (x, y) = (k a, k b) & (13, 78) & (26, 39) \end{array}$ $T o t a l p a i r = 4 + 2 + 2 = 8 A n s w e r$
	Answered by Rasheed.Sindhi last updated on 15/Mar/23

	$hcf (x, y) + lcm (x, y) = 91$ $lcm (x, y) = 91 - hcf (x, y)$ $\frac{lcm (x, y)}{hcf (x, y)} = \frac{91 - hcf (x, y)}{hcf (x, y)} \in Z^{+}$ $[∵ hcf (x, y) ∣ lcm (x, y)]$ $\Rightarrow hcf (x, y) = 1, 7, 13$ $\Rightarrow lcm (x, y) = 90, 84, 78$ $Case 1 : hcf (x, y) = 1 \land lcm (x, y) = 90$ $hcf (x, y) \times lcm (x, y) = x \times y$ $(1) \times lcm (x, y) = x \times y$ $lcm (x, y) = x \times y = 90$ $(x, y) = (1, 90), (2, 45), (5, 18), (9, 10)$ $(4 p a i r s)$ $Case 2 : hcf (x, y) = 7 \land lcm (x, y) = 84$ $hcf (x, y) \times lcm (x, y) = x \times y$ $(7) \times lcm (x, y) = x \times y$ $lcm (x, y) = \frac{x \times y}{7} = 84$ $x \times y = 588 = 2^{2} \times 3 \times 7^{2}$ $(3 \times 7, 2^{2} \times 7) = (21, 28)$ $(7, 4 \times 3 \times 7) = (7, 84)$ $(2 p a i r s)$ $Case 3 : hcf (x, y) = 13 \land lcm (x, y) = 78$ $hcf (x, y) \times lcm (x, y) = x \times y$ $(13) \times lcm (x, y) = x \times y$ $lcm (x, y) = \frac{x \times y}{13} = 78$ $x \times y = 13 \times 78 = 1014 = 2 \times 3 \times 13^{2}$ $(x, y)$ $= (13, 2 \times 3 \times 13) = (13, 78)$ $= (2 \times 13, 3 \times 13) = (26, 39)$ $(2 p a i r s)$ $4 + 2 + 2 = 8 p a i r s$ $B u t a s i f (x, y) i s s o l u t i o n s o a s$ $(y, x) i s a l s o s o l u t i o n .$ $S o T o t a l p a i r s = 16$
	Commented by BaliramKumar last updated on 15/Mar/23

	$N i c e$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum