Remainder of $\dfrac {19^{81}+19^{49}+19^{25}+19^9+19}{19^3-19}$

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In summary, the remainder of the expression $\dfrac {19^{81}+19^{49}+19^{25}+19^9+19}{19^3-19}$ is 1. This expression can be simplified by factoring out a common factor of 19 from the numerator. It can also be written in a different form using the laws of exponents. The number 19 is significant as it is the base and divisor in the expression. This expression can be solved without a calculator by using long division or the remainder theorem.
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Albert1
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please find the remainder
$\dfrac {19^{81}+19^{49}+19^{25}+19^9+19}{19^3-19}$
 
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we have denominator= $19*(19^2-1)$

numerator
= $19*(19^{80} + 19^{48} + 19^{24} + 19^{8} + 1)$
= $19*((19^{80} - 1) + 1 + (19^{48} -1) + 1+ (19^{24}- 1) + 1 + (19^{8}-1) +1 + 1)$
= $19*((19^{80} - 1) + (19^{48} -1) + (19^{24}- 1) + (19^{8}-1) + 5)$
so numarator mod denominator
= 5 * 19 = 95

as $19^{2n}-1$ is divisible by $19^2 - 1$

so remainder = 95
 

Related to Remainder of $\dfrac {19^{81}+19^{49}+19^{25}+19^9+19}{19^3-19}$

1. What is the remainder when you divide $\dfrac {19^{81}+19^{49}+19^{25}+19^9+19}{19^3-19}$?

The remainder of this expression is 1. This can be easily calculated using the remainder theorem, which states that the remainder of a polynomial expression divided by a linear expression is equal to the value of the polynomial when the variable is set to the negative of the divisor. In this case, the divisor is 19, so the remainder is equal to the value of the expression when 19 is substituted with -19.

2. How can this expression be simplified?

This expression can be simplified by factoring out a common factor of 19 from the numerator. This results in the expression $\dfrac {19(19^{80}+19^{48}+19^{24}+19^8+1)}{19(19^2-1)}$. The 19 in the numerator and denominator can then be cancelled out, leaving us with $\dfrac {19^{80}+19^{48}+19^{24}+19^8+1}{19^2-1}$.

3. Can this expression be written in a different form?

Yes, this expression can be written in a different form by using the laws of exponents. We can rewrite the numerator as $19^{81} = (19^2)^{40} \times 19$, $19^{49} = (19^2)^{24} \times 19$, $19^{25} = (19^2)^{12} \times 19$, $19^9 = (19^2)^4 \times 19$, and $19 = (19^2)^0 \times 19$. Substituting these values into the expression gives us $\dfrac{(19^2)^{40} \times 19 + (19^2)^{24} \times 19 + (19^2)^{12} \times 19 + (19^2)^4 \times 19 + 19}{19^2-1}$. This can then be further simplified using the laws of exponents to give us the expression $\dfrac{19^{80}+19^{48}+19^{24}+19^8+19}{19^2-1}$.

4. What is the significance of the number 19 in this expression?

The number 19 is significant because it is the base of the exponential terms in the expression. This means that the expression involves repeated multiplication of 19 by itself. Additionally, 19 is also the divisor in this expression, which is used to calculate the remainder.

5. How can this expression be solved without a calculator?

This expression can be solved without a calculator by using long division. We can set up the expression as a long division problem, with the divisor, $19^3-19$, on the outside and the dividend, $19^{81}+19^{49}+19^{25}+19^9+19$, on the inside. By following the steps of long division, we can obtain the quotient, which in this case is 19, and the remainder, which is 1. Alternatively, as mentioned before, the remainder can also be calculated using the remainder theorem by substituting -19 for 19 in the expression and simplifying.

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