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How many non negative integral values of $x$ satisfy the equation :$$\biggl\lfloor{\dfrac{x}{5}}\bigg\rfloor=\bigg\lfloor{\dfrac{x}{7}}\bigg\rfloor$$.

My try:

Writing few numbers and putting in the required equation.

$\underbrace{0,1,2,3,4}_{\implies 0=0},\underbrace{5,6}_{\displaystyle\times},\underbrace{7,8,9}_{\implies 1=1},\underbrace{10,11,12,13}_{\displaystyle\times} ,\underbrace{14}_{\implies 2=2},\underbrace{15,16,17,18,19,20}_{\text{LHS gives 3 but RHS gives 2}},\underbrace{21,22,23,24}_{\text{LHS gives 4 but RHS gives 3}}$.

The point which I wish to make here is that after $14$ we'll get $5's$ multiple before than $7's$, hence $\forall x \geq 15 \implies \biggl\lfloor{\dfrac{x}{5}}\bigg\rfloor>\bigg\lfloor{\dfrac{x}{7}}\bigg\rfloor$

Hence the only solution are: $\{0,1,2,3,4,7,8,9,14\}$. Does these make sense?

I need to know other solutions that don't incorporate such analyzing, means using some properties and framing it to solve above.

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5 Answers 5

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Rewrite the given condition as $$5n\le x<5n+5,\quad 7n\le x<7n+7$$ where $n$ is an integer. Since $x\ge0$, $n\ge0$ as well, so the above is equivalent to $7n\le x<5n+5$. The only nonnegative integers $n$ satisfying $7n<5n+5$ are $0$, $1$, and $2$. And so just a little cleanup is needed.

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By the division rule,

$$x=5q+r=7q+s$$ with $0\le r<5,0\le s<7$.

From this we draw $$2q=r-s\in[-6,4],$$

thus the only possibilities are with $q=0,1,2$.

  • $q=0$: $r=s$, possible for $x\in[0,4]\cap[0,6]=[0,4]$;

  • $q=1$: $5+r=7+s$, possible for $x\in[5,9]\cap[7,13]=[7,9]$;

  • $q=2$: $10+r=14+s$, possible for $x\in[10,14]\cap[14,20]=[14]$.

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  • $\begingroup$ why intersection? $\endgroup$
    – mnulb
    Commented Jan 25, 2017 at 16:52
  • $\begingroup$ Because the two bracketings must hold. What else would you do ? $\endgroup$
    – user65203
    Commented Jan 25, 2017 at 16:53
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    $\begingroup$ did you came up with those intersections by hit and trials or something different, and if something new, please don't mind adding there. $\endgroup$
    – mnulb
    Commented Jan 25, 2017 at 17:08
  • $\begingroup$ @Ayushakj: the intervals are drawn from the equations and inequations. $\endgroup$
    – user65203
    Commented Jan 25, 2017 at 17:11
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For a solution the LHS and RHS are a positive integer, let's call this $c$: $$c = \left \lfloor \frac{x}{5} \right \rfloor = \left \lfloor \frac{x}{7} \right \rfloor$$

Then, there are integers $0 \leq a < 5, 0 \leq b < 7$, s.t.: $$x = 5c + a = 7c + b$$ Thus: $$a - b = 2c$$

With the restrictions to $a,b$, this is enough to find all solutions.

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Analysis:

Let us take the general form where $p>q$ :

$$ \left\lfloor{\dfrac{x}{q}}\right\rfloor=\left\lfloor{\dfrac{x}{p}}\right\rfloor $$

Solution for above equation exists when $\dfrac{x}{q}-\dfrac{x}{p}<1\Rightarrow x<\dfrac{p\cdot q}{p-q}$

So there is no need to check for integers greater than$\left\lfloor\dfrac{p\cdot q}{p-q}\right\rfloor$

Also $\left\lfloor{\dfrac{x}{q}}\right\rfloor=\left\lfloor{\dfrac{x}{p}}\right\rfloor=k$ implies $$ \begin{align} &x\in [k\cdot p,(k+1)p) \cap\ [k\cdot q,(k+1)\cdot q)\\ \Rightarrow\ \ &x\in [k\cdot p,(k+1)\cdot q) \end{align} $$

The integers which satisfy the equation are in $$[0p,1q)\ \cup\ [1p,2q)\ \cup\ [2p,3q) \cup\ ...\ \cup\ [(n-1)p,nq)$$ where $nq<\left\lfloor\dfrac{p\cdot q}{p-q}\right\rfloor$ .

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  • $\begingroup$ Well done but you are missing the case $k=0$, giving the interval $[0,q)$. $\endgroup$
    – user65203
    Commented Jan 26, 2017 at 7:41
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$$\left[\frac{{x}}{\mathrm{5}}\right]=\left[\frac{{x}}{\mathrm{7}}\right]={n}=\left\{\mathrm{0},\mathrm{1},\mathrm{2}\right\} \\ $$ $${x}=\left[\mathrm{5}{n};\mathrm{5}{n}+\mathrm{5}\right)\cap\left[\mathrm{7}{n};\mathrm{7}{n}+\mathrm{7}\right) \\ $$ $${x}=\left[\mathrm{7}{n};\mathrm{5}{n}+\mathrm{5}\right) \\ $$ $${x}=\left\{\left[\mathrm{0};\mathrm{5}\right),\left[\mathrm{7};\mathrm{10}\right),\left[\mathrm{14};\mathrm{15}\right)\right\} \\ $$ $${x}=\left\{\mathrm{0},\mathrm{1},\mathrm{2},\mathrm{3},\mathrm{4},\mathrm{7},\mathrm{8},\mathrm{9},\mathrm{14}\right\} \\ $$

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