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Question Number 183583 by Acem last updated on 27/Dec/22

Commented by Acem last updated on 27/Dec/22

$A s t o n e o f 40 k g, i^{'} v e c u t i t i n t o 4 d i f e r e n t p a r t s,$ $n o w i c a n u s e t h e m f o r m e a s u r e w e i g h t s 1 - 40 K g$ $S u r e t h e q u s e t i o n i s w e i g h t o f e a c h p i e c e ?$ $B u t b y a l o g i c a l s o l u t i o n n o t b y e x p e r i m e n t, p l e a s e!$
	Answered by CrispyXYZ last updated on 27/Dec/22

	$1, 3, 9, 27 ?$
	Commented by Matica last updated on 27/Dec/22

	$W h a t i f y o u d i v i d e i n t o 5 p i e c e s ?$
	Commented by CrispyXYZ last updated on 27/Dec/22

	$Each piece of stone can be placed on$ $the right (+), left (-), or not (0) .$ $we can consider trinary (base 3)$ $there are 4 digits in total$ $3^{0} + 3^{1} + 3^{2} + 3^{3} = 40$ $each piece = 3^{0}, 3^{1}, 3^{2}, 3^{3}$
	Commented by Rasheed.Sindhi last updated on 27/Dec/22

	$W i t h 5 s u c h^{'} {m e a s u r e s}^{'} (1, 3, 9, 27, 81)$ $y o u c a n w e i g h u p t o 121 k g .$
	Commented by Rasheed.Sindhi last updated on 27/Dec/22

	$y o u c a n b r e a k i n o 2 p i e c e s o f a n y o n e$ $(e x c e p t 1 o f c o u r s e) t o m a k e i t$ $^{'} 5 {p e i c e s}^{'} . F o r e x a m p l e :$ $1, 3, 4, 5, 27 b u t t h i s w i l l m a k e y o u r$ $q u e s t i o n s o m e w h a t i n e f f i c i e n t .$ $B e c a u s e f o u r p e i c e s a r e$ $m i n i m u m l y r e q u i r e d a n d w i t h i t$ $t h e q u e s t i o n h a s u n i q u e s o l u t i o n .$ $(Y o u {c a n}^{'} t d o t h i s b y 3 p e i c e s o f 40 k g)$
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