- #1

Math100

- 771

- 219

- Homework Statement
- Obtain three consecutive integers, the first of which is divisible by a square, the second by a cube, and the third by a fourth power.

- Relevant Equations
- None.

Let ## a, a+1 ## and ## a+2 ## be the three consecutive integers.

Then

\begin{align*}

&5^{2}\mid a\implies a\equiv 0\pmod {25}\\

&3^{3}\mid (a+1)\implies a+1\equiv 0\pmod {27}\implies a\equiv 26\pmod {9}\\

&2^{4}\mid (a+2)\implies a+2\equiv 0\pmod {16}\implies a\equiv 14\pmod {16}.\\

\end{align*}

Applying the Chinese Remainder Theorem produces:

## n=25\cdot 27\cdot 16=10800 ##.

This means ## N_{1}=\frac{10800}{25}=432, N_{2}=\frac{10800}{27}=400 ## and ## N_{3}=\frac{10800}{16}=675 ##.

Now we have ## 432x_{1}\equiv 1\pmod {25}, 400x_{2}\equiv 1\pmod {27} ## and ## 675x_{3}\equiv 1\pmod {16} ##.

Observe that

\begin{align*}

&432x_{1}\equiv 1\pmod {25}\implies 7x_{1}\equiv 1\pmod {25}\\

&\implies 49x_{1}\equiv 7\pmod {25}\implies -x_{1}\equiv 7\pmod {25}\\

&\implies x_{1}\equiv 18\pmod {25},\\

&400x_{2}\equiv 1\pmod {27}\implies -5x_{2}\equiv 1\pmod {27}\\

&\implies -25x_{2}\equiv 5\pmod {27}\implies 2x_{2}\equiv 5\pmod {27}\\

&\implies 28x_{2}\equiv 70\pmod {27}\implies x_{2}\equiv 16\pmod {27},\\

&675x_{3}\equiv 1\pmod {16}\implies 3x_{3}\equiv 1\pmod {16}\\

&\implies 15x_{3}\equiv 5\pmod {16}\implies -x_{3}\equiv 5\pmod {16}\\

&\implies x_{3}\equiv 11\pmod {16}.\\

\end{align*}

Since ## x_{1}=18, x_{2}=16 ## and ## x_{3}=11 ##,

it follows that ## x\equiv (0+26\cdot 400\cdot 16+14\cdot 675\cdot 11)\pmod {10800}\equiv 270350\pmod {10800}\equiv 350\pmod {10800} ##.

Thus, ## a=350, a+1=351 ## and ## a+2=352 ##.

Therefore, the three consecutive integers are ## 350, 351 ## and ## 352 ##.

Then

\begin{align*}

&5^{2}\mid a\implies a\equiv 0\pmod {25}\\

&3^{3}\mid (a+1)\implies a+1\equiv 0\pmod {27}\implies a\equiv 26\pmod {9}\\

&2^{4}\mid (a+2)\implies a+2\equiv 0\pmod {16}\implies a\equiv 14\pmod {16}.\\

\end{align*}

Applying the Chinese Remainder Theorem produces:

## n=25\cdot 27\cdot 16=10800 ##.

This means ## N_{1}=\frac{10800}{25}=432, N_{2}=\frac{10800}{27}=400 ## and ## N_{3}=\frac{10800}{16}=675 ##.

Now we have ## 432x_{1}\equiv 1\pmod {25}, 400x_{2}\equiv 1\pmod {27} ## and ## 675x_{3}\equiv 1\pmod {16} ##.

Observe that

\begin{align*}

&432x_{1}\equiv 1\pmod {25}\implies 7x_{1}\equiv 1\pmod {25}\\

&\implies 49x_{1}\equiv 7\pmod {25}\implies -x_{1}\equiv 7\pmod {25}\\

&\implies x_{1}\equiv 18\pmod {25},\\

&400x_{2}\equiv 1\pmod {27}\implies -5x_{2}\equiv 1\pmod {27}\\

&\implies -25x_{2}\equiv 5\pmod {27}\implies 2x_{2}\equiv 5\pmod {27}\\

&\implies 28x_{2}\equiv 70\pmod {27}\implies x_{2}\equiv 16\pmod {27},\\

&675x_{3}\equiv 1\pmod {16}\implies 3x_{3}\equiv 1\pmod {16}\\

&\implies 15x_{3}\equiv 5\pmod {16}\implies -x_{3}\equiv 5\pmod {16}\\

&\implies x_{3}\equiv 11\pmod {16}.\\

\end{align*}

Since ## x_{1}=18, x_{2}=16 ## and ## x_{3}=11 ##,

it follows that ## x\equiv (0+26\cdot 400\cdot 16+14\cdot 675\cdot 11)\pmod {10800}\equiv 270350\pmod {10800}\equiv 350\pmod {10800} ##.

Thus, ## a=350, a+1=351 ## and ## a+2=352 ##.

Therefore, the three consecutive integers are ## 350, 351 ## and ## 352 ##.