Let f(x)= (ax+1)^5 .(1+bx)^4 ; a,b ∈ N if times of x equal 62 so what are possible values of the sum a, b?


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Question Number 178347 by Acem last updated on 15/Oct/22

$L e t f (x) = {(a x + 1)}^{5} . {(1 + b x)}^{4}; a, b \in N$ $i f t i m e s o f x e q u a l 62 s o w h a t a r e p o s s i b l e v a l u e s$ $o f t h e s u m a, b ?$
	Answered by Acem last updated on 15/Oct/22

	${(1 + a x)}^{5} = 1 + 5 a x + 10 a^{2} x^{2} + . . .$ ${(1 + b x)}^{4} = 1 + 4 b x + 6 b^{2} x^{2} + . . .$ $W e t a k e t i m e s x f r o m m u l t i p l i c a t i o n$ $(5 a + 4 b) x i . e . 5 a + 4 b = 62 . . . (1)$ $W e n e e d t o (a + b) s o a n d f r o m (1)$ $4 (a + b) = - a + 62$ $S i n c e a \in N \Rightarrow 4 (a + b) < 62 \Leftrightarrow a + b < 15.5$ $S i n c e a + b \in N \Rightarrow a + b ⩽ 15 . . . (2)$ $W e r e p e a t f o r b f r o m (1)$ $5 (a + b) = b + 62, b \in N \Rightarrow 5 (a + b) > 62 \Rightarrow a + b > 12.4$ $S i n c e a + b \in N \Rightarrow a + b ⩾ 13 . . . (3)$ $(2) & (3)$ $W e f i n d t h a t t h e p o s s i b l e v a l u e s o f a + b a r e :$ $a + b = 13, a + b = 14, a + b = 15$
	Commented by Tawa11 last updated on 15/Oct/22

	$Great sir$
	Commented by Acem last updated on 15/Oct/22

	$T h a n k s S i r$
	Answered by mr W last updated on 16/Oct/22

	$c o e f . o f x t e r m i s$ $5 \times 1^{4} \times a \times 1^{4} + 4 \times 1^{3} \times b \times 1^{5} = 5 a + 4 b$ $5 a + 4 b = 62$ $\Rightarrow a = 4 k + 10 ⩾ 1 \Rightarrow k ⩾ - \frac{9}{4} \Rightarrow k ⩾ - 2$ $\Rightarrow b = - 5 k + 3 ⩾ 1 \Rightarrow k ⩽ \frac{2}{5} \Rightarrow k ⩽ 0$ $a + b = 4 k + 10 - 5 k + 3 = - k + 13$ ${(a + b)}_{m i n} = 13 a t k_{m a x} = 0$ ${(a + b)}_{m a x} = 15 a t k_{m i n} = - 2$ $\Rightarrow a + b = 13 o r 14 o r 15$
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