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Question Number 190140 by otchereabdullai last updated on 28/Mar/23

Commented by Frix last updated on 28/Mar/23

$The text is strange anyway . ‘ ‘ . . . the number$ $is increased by 8 and then divided by the$ $sum of the {digits}^{″} would mean the sum$ $of the digits a f t e r adding 8 . But then we$ $get no reasonable result .$
	Answered by Frix last updated on 28/Mar/23

	$\frac{10 x + (x - 1) + 8}{x + (x - 1)} = 8$ $\frac{11 x + 7}{2 x - 1} = 8$ $11 x + 7 = 16 x - 8$ $5 x = 15$ $x = 3$ $The number is 32$
	Commented by otchereabdullai last updated on 28/Mar/23

	$p l e a s e t h e q u e s t i o n s a y s f i n d t h e$ $t w o n u m b e r s s o w h y t h e 32 ?$
	Commented by JDamian last updated on 28/Mar/23
	what two numbers? The text talks about a "two digit number", not numbers.
	Commented by mr W last updated on 28/Mar/23

	$u s u a l l y {i t}^{'} s s p r o b l e m o f t h e$ $t r a n s l a t i o n . i t h i n k t h e q u e s t i o n i s$ $‘ ‘ f i n d t h e t w o {d i g i t s}^{″} . i n m a n y$ $l a n g u a g e s ‘ ‘ {n u m b e r}^{″} a n d ‘ ‘ {d i g i t}^{″} a r e$ $t h e s a m e w o r d .$
	Commented by otchereabdullai last updated on 28/Mar/23

	$t h a n k s a l o t$
	Answered by manxsol last updated on 29/Mar/23
	$ab^(−) b<a ((ab^(−) +8)/(a+b))=8 10a+b+8=8a+8b 2a+8=7b a+4=((7b)/2) a=((7b)/2)−4 (((7b)/2)−4)b^(−) ={32}$
	$\overset{―}{a b} b < a$ $\frac{\overset{―}{a b} + 8}{a + b} = 8$ $10 a + b + 8 = 8 a + 8 b$ $2 a + 8 = 7 b$ $a + 4 = \frac{7 b}{2}$ $a = \frac{7 b}{2} - 4$ $Missing \left or extra \right$
	Commented by otchereabdullai last updated on 29/Mar/23

	$t h a n k s s i r!$
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