Prove that the integer ## 53^{103}+103^{53} ## is divisible by....

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In summary, an integer is divisible by another integer if the result of dividing the first integer by the second integer is a whole number. To prove this, one can use the division algorithm or divisibility rules, or show that the remainder is 0. The expression "53^{103}+103^{53}" serves as the integer being proven to be divisible by another integer. An integer is a divisor if it evenly divides into another integer. To prove that the expression is divisible by a specific integer, one can use algebraic manipulation, mathematical properties, modular arithmetic, or induction.
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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 ##.
 
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Math100 said:
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 ##.
Correct. If you want to avoid too long lines, then you can use the following structure:

\begin{align*}
53^{103}+103^{53}&\equiv (53^{2})^{51}\cdot 53+(103^{2})^{26}\cdot 103\pmod {39}\\
&\equiv (1^{51}\cdot 53+1^{26}\cdot 103)\pmod {39}\\
&\equiv 156\pmod {39}\\
&\equiv 0\pmod {39}
\end{align*}

which results in

\begin{align*}
53^{103}+103^{53}&\equiv (53^{2})^{51}\cdot 53+(103^{2})^{26}\cdot 103\pmod {39} \\
&\equiv (1^{51}\cdot 53+1^{26}\cdot 103)\pmod {39}\\
&\equiv 156\pmod {39}\\
&\equiv 0\pmod {39}
\end{align*}
 
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1. How do you prove that an integer is divisible by a specific number?

To prove that an integer is divisible by a specific number, you can use the division algorithm or perform long division to show that there is no remainder. You can also use the fact that if a number is divisible by another number, their greatest common divisor will be the smaller number.

2. What is the significance of the number 53 in the given integer?

The number 53 is significant because it is the base of the first term in the given expression. This means that 53 is being raised to a power, which will result in a large number that can potentially be divided by other numbers.

3. How does the exponent of the second term affect the divisibility of the given integer?

The exponent of the second term, 103, does not directly affect the divisibility of the given integer. However, since 103 is an odd number, it will result in an odd number being added to the first term. This can potentially affect the divisibility by certain numbers, such as 2, but not by others, such as 3.

4. Can you use modular arithmetic to prove the divisibility of the given integer?

Yes, you can use modular arithmetic to prove the divisibility of the given integer. By using the properties of modular arithmetic, you can show that the given integer is congruent to 0 (mod n), where n is the number you are trying to prove it is divisible by.

5. What is the significance of the number 103 in the given integer?

The number 103 is significant because it is the base of the second term in the given expression. This means that 103 is being raised to a power, which will result in a large number that can potentially be divided by other numbers.

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