- #1

Math100

- 780

- 220

- Homework Statement
- Prove that the integer ## 53^{103}+103^{53} ## is divisible by ## 39 ##, and that ## 111^{333}+333^{111} ## is divisible by ## 7 ##.

- Relevant Equations
- None.

Proof:

First, we will prove that the integer ## 53^{103}+103^{53} ## is divisible by ## 39 ##.

Note that ## 53\equiv 14 \pmod {39}\implies 53^{2}\equiv 14^{2}\pmod {39}\equiv 196\pmod {39}\equiv 1\pmod {39} ##.

Now observe that ## 103\equiv 25\pmod {39}\equiv -14\pmod {39}\implies 103^{2}\equiv 196\pmod {39}\equiv 1\pmod {39} ##.

Thus ## 53^{103}+103^{53}\equiv (53^{2})^{51}\cdot 53+(103^{2})^{26}\cdot 103\equiv (1^{51}\cdot 53+1^{26}\cdot 103)\pmod {39}\equiv 156\pmod {39}\equiv 0\pmod {39} ##.

Therefore, the integer ## 53^{103}+103^{53} ## is divisible by ## 39 ##.

Next, we will prove that the integer ## 111^{333}+333^{111} ## is divisible by ## 7 ##.

Note that ## 111\equiv 6\pmod 7\equiv (-1)\pmod 7\implies 111^{333}\equiv (-1)^{333}\pmod 7\equiv (-1)\pmod 7 ##.

Now observe that ## 333=3\cdot 111\equiv 3\cdot (-1)\pmod 7\equiv -3\pmod 7\equiv 4\pmod 7\implies 333^{3}\equiv 4^{3}\pmod 7\equiv 1\pmod 7\implies 333^{111}\equiv (333^{3})^{37}\equiv 1^{37}\pmod 7\equiv 1\pmod 7 ##.

Thus ## 111^{333}+333^{111}\equiv (-1+1)\pmod 7\equiv 0\pmod 7 ##.

Therefore, the integer ## 111^{333}+333^{111} ## is divisible by ## 7 ##.

First, we will prove that the integer ## 53^{103}+103^{53} ## is divisible by ## 39 ##.

Note that ## 53\equiv 14 \pmod {39}\implies 53^{2}\equiv 14^{2}\pmod {39}\equiv 196\pmod {39}\equiv 1\pmod {39} ##.

Now observe that ## 103\equiv 25\pmod {39}\equiv -14\pmod {39}\implies 103^{2}\equiv 196\pmod {39}\equiv 1\pmod {39} ##.

Thus ## 53^{103}+103^{53}\equiv (53^{2})^{51}\cdot 53+(103^{2})^{26}\cdot 103\equiv (1^{51}\cdot 53+1^{26}\cdot 103)\pmod {39}\equiv 156\pmod {39}\equiv 0\pmod {39} ##.

Therefore, the integer ## 53^{103}+103^{53} ## is divisible by ## 39 ##.

Next, we will prove that the integer ## 111^{333}+333^{111} ## is divisible by ## 7 ##.

Note that ## 111\equiv 6\pmod 7\equiv (-1)\pmod 7\implies 111^{333}\equiv (-1)^{333}\pmod 7\equiv (-1)\pmod 7 ##.

Now observe that ## 333=3\cdot 111\equiv 3\cdot (-1)\pmod 7\equiv -3\pmod 7\equiv 4\pmod 7\implies 333^{3}\equiv 4^{3}\pmod 7\equiv 1\pmod 7\implies 333^{111}\equiv (333^{3})^{37}\equiv 1^{37}\pmod 7\equiv 1\pmod 7 ##.

Thus ## 111^{333}+333^{111}\equiv (-1+1)\pmod 7\equiv 0\pmod 7 ##.

Therefore, the integer ## 111^{333}+333^{111} ## is divisible by ## 7 ##.