Prove that [x] + [2x] + [4x] + [8x] + [16x] + [32x] = 12345 has no solution.


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Question Number 13389 by Tinkutara last updated on 19/May/17

$Prove that$ $[x] + [2 x] + [4 x] + [8 x] + [16 x] + [32 x] = 12345$ $has no solution .$
Commented by prakash jain last updated on 20/May/17

$x = n + y, n \in Z^{+}, 0 ⩽ y < 1$ $[x] + [2 x] + [4 x] + [8 x] + [16 x] + [32 x]$ $= n + [y] + 2 n + [2 y] + . . + 32 n + [32 y]$ $= 63 n + [y] + [2 y] + [4 y] + [8 y] + [16 y] + [32 y]$ $= 12345$ $\Rightarrow n ⩽ 195$ $x = 195 + y$ $0 ⩽ y < 1$ $[y] + [2 y] + [4 y] + [8 y] + [16 y] + [32 y] = 60$ $m a x ([y]) = 0$ $m a x ([2 y]) = 1$ $m a x ([4 y]) = 3$ $m a x ([8 y]) = 7$ $m a x ([16 y) = 15$ $m a x ([32 y]) = 31$ $m a x ([y] + [2 y] + [4 y] + [8 y] + [16 y] + [32 y]) = 57$ $n o v a l u e o f y i s p o s s i b l e w h i c h$ $w i l l g i v e$ $[y] + [2 y] + [4 y] + [8 y] + [16 y] + [32 y] = 60$ $h e n c e n o s o l u t i o n$
Commented by mrW1 last updated on 20/May/17

$v e r y n i c e!$
Commented by mrW1 last updated on 20/May/17

$b u t, c a n w e f i n d a s o l u t i o n f o r$ $[y] + [2 y] + [4 y] + [8 y] + [16 y] + [32 y] ⩽ 57 ?$ $o r h o w c a n w e f i n d ?$
Commented by prakash jain last updated on 20/May/17

$y = . 99 or . 999$ $w i l l g i v e$ $[y] + [2 y] + [4 y] + [8 y] + [16 y] + [32 y] = 57$ $For other value, s a y 50$ $we can try various option$ $0 + 1 + 3 + 6 + 13 + 27 = 50$ $so we can pick y = . 85$ $[y] = 0$ $] 2 y] = 1$ $[4 y] = 3$ $[8 y] = 6$ $[16 y] = 13$ $[32 y] = 27$
Commented by mrW1 last updated on 20/May/17

$T h a n k s!$
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