1. If 3 ∙ ab^(−) + bc^(−) = 115 Find: max(a+b+c)=? 2. a,b,c∈N If (a/2) + (b/3) = (c/4) Find: min(a+b+c)=?


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Question Number 199932 by hardmath last updated on 11/Nov/23

$1 .$ $If 3 \cdot \overset{―}{ab} + \overset{―}{bc} = 115$ $Find : \max (a + b + c) = ?$ $2 .$ $a, b, c \in N$ $If \frac{a}{2} + \frac{b}{3} = \frac{c}{4}$ $Find : \min (a + b + c) = ?$
	Answered by Frix last updated on 11/Nov/23

	$1 . Only 2 possbilities$ $a = 1 b = 6 c = 7 a + b + c = 14$ $a = 2 b = 4 c = 3 a + b + c = 9$ $2 . If a, b, c \neq 0$ $c = 2 a + \frac{4 b}{3}$ $a = m \land b = 3 n$ $Minimum at$ $a = 1 b = 3 c = 6 a + b + c = 10$
	Answered by Rasheed.Sindhi last updated on 13/Nov/23
	$3 ∙ ab^(−) + bc^(−) = 115; a,b,c∈{0,1,...,9};a,c≠0 3(10a+b)+(10b+c) 30a+13b+c=115 115−13b−c=30a 115−13b−c≡0(mod 30) 13b+c≡25≡55≡85≡115(mod 30) (b,c)=(4,3),(6,7) Both pairs are successful to produce valid value of a (b,c)=(4,3): a=((115−13b−c)/(30)) a=((115−13(4)−3)/(30))=((60)/(30))=2 (b,c)=(6,7): a=((115−13(6)−7)/(30))=((30)/(30))=1 (a,b,c)=(2,4,3),(1,6,7)$
	$3 \cdot \overset{―}{ab} + \overset{―}{bc} = 115;$ $a, b, c \in {0, 1, . . ., 9}; a, c \neq 0$ $3 (10 a + b) + (10 b + c)$ $30 a + 13 b + c = 115$ $115 - 13 b - c = 30 a$ $115 - 13 b - c \equiv 0 (m o d 30)$ $13 b + c \equiv 25 \equiv 55 \equiv 85 \equiv 115 (m o d 30)$ $(b, c) = (4, 3), (6, 7)$ $B o t h p a i r s a r e s u c c e s s f u l t o$ $p r o d u c e v a l i d v a l u e o f a$ $(b, c) = (4, 3) :$ $a = \frac{115 - 13 b - c}{30}$ $a = \frac{115 - 13 (4) - 3}{30} = \frac{60}{30} = 2$ $(b, c) = (6, 7) :$ $a = \frac{115 - 13 (6) - 7}{30} = \frac{30}{30} = 1$ $(a, b, c) = (2, 4, 3), (1, 6, 7)$
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