What is the remainder when we divide 5 70 + 7 70 by 74?
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The question asks for the remainder when the sum \(5^{70} + 7^{70}\) is divided by 74. This is a problem involving modular arithmetic, where we need to find the value of \(5^{70} + 7^{70} \pmod{74}\).
Let's look closely at the numbers involved: the bases are 5 and 7, and the modulus is 74.
Consider the sum of the squares of the bases:
\(5^2 + 7^2 = 25 + 49 = 74\)
Notice that the sum of the squares, \(5^2 + 7^2\), is exactly equal to the modulus, 74. This means \(5^2 + 7^2 \equiv 0 \pmod{74}\).
We need to evaluate the expression \(5^{70} + 7^{70}\). We can rewrite the exponents using the square terms we just analyzed:
\(5^{70} = (5^2)^{35}\)
\(7^{70} = (7^2)^{35}\)
So, the original expression becomes \((5^2)^{35} + (7^2)^{35}\).
This expression is in the form \(x^n + y^n\), where:
The exponent \(n = 35\) is an odd integer.
There is a powerful algebraic property related to the sum of powers that is useful in remainder problems:
Property: For any positive odd integer \(n\), the expression \(x^n + y^n\) is always divisible by \(x+y\).
In our specific problem, we have the expression \((5^2)^{35} + (7^2)^{35}\), which perfectly fits the form \(x^n + y^n\) with \(x=5^2\), \(y=7^2\), and \(n=35\) (which is indeed an odd number).
According to this property, the expression \((5^2)^{35} + (7^2)^{35}\) must be divisible by the sum \(5^2 + 7^2\).
We calculated earlier that \(5^2 + 7^2 = 25 + 49 = 74\).
Therefore, this means that \(5^{70} + 7^{70}\) is divisible by 74.
By definition, if a number is divisible by another number, the result of the division leaves no remainder. In other words, the remainder is 0.
Since we have established that \(5^{70} + 7^{70}\) is divisible by 74, the remainder when \(5^{70} + 7^{70}\) is divided by 74 is 0.
Based on the divisibility property of powers, the remainder is 0.
Understanding the divisibility rules for sums and differences of powers is crucial for efficiently solving many remainder problems. Recall these key properties:
These properties allow us to determine divisibility, and thus the remainder (which is 0 if divisible), without calculating the large powers themselves.
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