What is the range of h(x)? –∞ < h(x) < ∞ h(x) ≤ 5 h(x) ≥ 5 h(x) ≥ 3

Seven tenths Forty seven and nine tenths Eighteen and eight one hundredths One hundred twenty five and twenty three thousandths One hundred and twenty six thousandths One hundred fifty and seventy five thousandths

7 0

3 years ago

Find the smallest 4 digit number such that when divided by 35, 42 or 63 remainder is always 5

alex41 [277]

The smallest such number is 1055.

We want to find $x$ such that

$\begin{cases}x\equiv5\pmod{35}\\x\equiv5\pmod{42}\\x\equiv5\pmod{63}\end{cases}$

The moduli are not coprime, so we expand the system as follows in preparation for using the Chinese remainder theorem.

$x\equiv5\pmod{35}\implies\begin{cases}x\equiv5\equiv0\pmod5\\x\equiv5\pmod7\end{cases}$

$x\equiv5\pmod{42}\implies\begin{cases}x\equiv5\equiv1\pmod2\\x\equiv5\equiv2\pmod3\\x\equiv5\pmod7\end{cases}$

$x\equiv5\pmod{63}\implies\begin{cases}x\equiv5\equiv2\pmod 3\\x\equiv5\pmod7\end{cases}$

Taking everything together, we end up with the system

$\begin{cases}x\equiv1\pmod2\\x\equiv2\pmod3\\x\equiv0\pmod5\\x\equiv5\pmod7\end{cases}$

Now the moduli are coprime and we can apply the CRT.

We start with

$x=3\cdot5\cdot7+2\cdot5\cdot7+2\cdot3\cdot7+2\cdot3\cdot5$

Then taken modulo 2, 3, 5, and 7, all but the first, second, third, or last (respectively) terms will vanish.

Taken modulo 2, we end up with

$x\equiv3\cdot5\cdot7\equiv105\equiv1\pmod2$

which means the first term is fine and doesn't require adjustment.

Taken modulo 3, we have

$x\equiv2\cdot5\cdot7\equiv70\equiv1\pmod3$

We want a remainder of 2, so we just need to multiply the second term by 2.

Taken modulo 5, we have

$x\equiv2\cdot3\cdot7\equiv42\equiv2\pmod5$

We want a remainder of 0, so we can just multiply this term by 0.

Taken modulo 7, we have

$x\equiv2\cdot3\cdot5\equiv30\equiv2\pmod7$

We want a remainder of 5, so we multiply by the inverse of 2 modulo 7, then by 5. Since $2\cdot4\equiv8\equiv1\pmod7$ , the inverse of 2 is 4.

So, we have to adjust $x$ to

$x=3\cdot5\cdot7+2^2\cdot5\cdot7+0+2^3\cdot3\cdot5^2=845$

and from the CRT we find

$x\equiv845\pmod2\cdot3\cdot5\cdot7\implies x\equiv5\pmod{210}$

so that the general solution $x=210n+5$ for all integers $n$ .

We want a 4 digit solution, so we want

$210n+5\ge1000\implies210n\ge995\implies n\ge\dfrac{995}{210}\approx4.7\implies n=5$

which gives $x=210\cdot5+5=1055$ .

5 0

3 years ago