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- Thread starter jacks
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For (2), 2012 since the floor of 2012/2011 = 1 and 2012/2012 = 1.(1) find real value of $k$ for which $k\lfloor x \rfloor = \lfloor kx \rfloor$

(2) find value of $x$ for which $\displaystyle \lfloor \frac{x}{2011}\rfloor = \lfloor \frac{x}{2012}\rfloor$

where $\lfloor x \rfloor = $ floor function

If you want values though, it would be $x\in [2012,4021]$

in general [tex]\lfloor x \rfloor = a \Rightarrow a \leq x < a+1 [/tex]

[tex]\lfloor x k \rfloor = k \lfloor x \rfloor [/tex]

[tex]k \lfloor x \rfloor \leq x k < k \lfloor x \rfloor +1 [/tex]

[tex]0 \leq x k - k \lfloor x \rfloor <1 [/tex]

[tex]0 \leq k(x - \lfloor x \rfloor) <1 [/tex]

[tex]x - \lfloor x \rfloor = {0,1} [/tex] if x integer 0 if x is not integer 1

[tex] 0 \leq k < 1 [/tex]

But

[tex]\lfloor kx \rfloor , \lfloor x \rfloor [/tex] are both integers which leave k to be integer so k=0 and k=1

[tex]\lfloor x k \rfloor = k \lfloor x \rfloor [/tex]

[tex]k \lfloor x \rfloor \leq x k < k \lfloor x \rfloor +1 [/tex]

[tex]0 \leq x k - k \lfloor x \rfloor <1 [/tex]

[tex]0 \leq k(x - \lfloor x \rfloor) <1 [/tex]

[tex]x - \lfloor x \rfloor = {0,1} [/tex] if x integer 0 if x is not integer 1

[tex] 0 \leq k < 1 [/tex]

But

[tex]\lfloor kx \rfloor , \lfloor x \rfloor [/tex] are both integers which leave k to be integer so k=0 and k=1

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- Feb 7, 2012

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Those are the numbers for which $\displaystyle \left\lfloor \frac{x}{2011}\right\rfloor = \left\lfloor \frac{x}{2012}\right\rfloor = 1.$ There are also the numbers $0\leqslant x\leqslant 2010$ for which $\displaystyle \left\lfloor \frac{x}{2011}\right\rfloor = \left\lfloor \frac{x}{2012}\right\rfloor = 0.$ Then there are the numbers $4024 \leqslant x\leqslant 6032$ for which $\displaystyle \left\lfloor \frac{x}{2011}\right\rfloor = \left\lfloor \frac{x}{2012}\right\rfloor = 2,$ and a whole lot of other intervals going right up to the number $X = 4044120 = 2012\times 2010.$ This has the property that $\displaystyle \left\lfloor \frac{X}{2011}\right\rfloor = \left\lfloor \frac{X}{2012}\right\rfloor = 2010,$ and it is the largest number for which $\displaystyle \left\lfloor \frac{X}{2011}\right\rfloor$ and $\left\lfloor \dfrac{X}{2012}\right\rfloor$ are equal.For (2), 2012 since the floor of 2012/2011 = 1 and 2012/2012 = 1.(1) find real value of $k$ for which $k\lfloor x \rfloor = \lfloor kx \rfloor$

(2) find value of $x$ for which $\displaystyle \lfloor \frac{x}{2011}\rfloor = \lfloor \frac{x}{2012}\rfloor$

where $\lfloor x \rfloor =$ floor function

If you want values though, it would be $x\in [2012,4021]$

in general [tex]\lfloor x \rfloor = a \Rightarrow a \leq x < a+1 [/tex]

[tex]\lfloor x k \rfloor = k \lfloor x \rfloor [/tex]

[tex]k \lfloor x \rfloor \leq x k < k \lfloor x \rfloor +1 [/tex]

[tex]0 \leq x k - k \lfloor x \rfloor <1 [/tex]

[tex]0 \leq k(x - \lfloor x \rfloor) <1 [/tex]

[tex]x - \lfloor x \rfloor = {0,1} [/tex] if x integer 0 if x is not integer 1

[tex] 0 \leq k < 1 [/tex]

But

[tex]\lfloor kx \rfloor , \lfloor x \rfloor [/tex] are both integers which leave k to be integer so k=0 and k=1

Or you can solve it in other way if x=1

[tex] k = \floor k \rfloor [/tex] so k should be integer

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the trivial solution is k={0,1}, suppose k is not 0 or 1

choose x = 1/k, if k>0 [tex]\lfloor \frac{1}{k} \rfloor = 0 [/tex] if k <0 [tex]\lfloor \frac{1}{k} \rfloor = -1 [/tex]

[tex] k \lfloor \frac{1}{k} \rfloor = {0,-1} [/tex]

[tex] \floor k \left(\frac{1}{k}\right) \rfloor = 1 [/tex]