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If $[x+0.19]+[x+0.20]+[x+0.21]+\cdots[x+0.91]=546$ Find the value of $[100 x].$
Option: 1
$743$

Option: 2
$243$

Option: 3
$512$

Option: 4
$125$

Answers (1)

First note that the number of terms is $73 .$ Also, if we look at $[x+a]$ where $a$ ranges from $0.19\: to\: 0.91 ,$ then we see it can only reach two values; those are $n=[x+0.19]$ and possibly, but not necessarily, $n+1.$ We know though that

$73 n \leq[x+0.19]+[x+0.20]+\cdots+[x+0.91]<73(n+1)$

and so we deduce $n=7.$ We can also find exactly where the value of $[x+a]$ changes from $n$ to $n+1$ since, $546-73 \cdot 7=35,$ so there are $35$ terms $[x+a]=n+1,$ so the last $a$ such that $[x+a]=n$ is $a=0.56.$ So, $[x+0.56]=7\: but\: [x+0.57]=8.$ This means that

$7.43 \leq x<7.44 \text {. So, } 743 \leq 100 x<744 \text {, so }[100 x]=743 \text {. }$

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