Problem Statement: We need to find the total number of two-digit integers 'n' such that the expression $3^n + 7^n$ is a multiple of 10.

Key Concept: Modular Arithmetic and Divisibility by 10

A number is a multiple of 10 if and only if its unit digit is 0. In terms of modular arithmetic, this means the number is congruent to 0 modulo 10. So, we are looking for two-digit numbers 'n' such that: $3^n + 7^n \equiv 0 \pmod{10}$

A crucial property of modular arithmetic is that for any integers $a, b, m$ and positive integer $n$, if $a \equiv b \pmod{m}$, then $a^n \equiv b^n \pmod{m}$.

Step-by-step Working:

1. Simplify $7^n \pmod{10}$: We observe that $7 \equiv -3 \pmod{10}$. This allows us to simplify the second term of our expression: $7^n \equiv (-3)^n \pmod{10}$ Explanation: We replace 7 with its equivalent modulo 10, which is -3. This substitution is valid because $7 - (-3) = 10$, a multiple of 10. This simplification is key because it relates the powers of 7 to powers of 3, making the sum easier to analyze.

2. Analyze the sum $3^n + 7^n \pmod{10}$: Now substitute the simplified $7^n$ back into the original congruence: $3^n + 7^n \equiv 3^n + (-3)^n \pmod{10}$ We need to determine when $3^n + (-3)^n \equiv 0 \pmod{10}$. This depends on whether 'n' is an odd or an even number.

   • Case 1: $n$ is an odd number. If 'n' is odd, then $(-3)^n = -3^n$. Example: $(-3)^1 = -3$, $(-3)^3 = -27$. So, for odd 'n': $3^n + (-3)^n = 3^n - 3^n = 0$ Therefore, if 'n' is an odd number, $3^n + 7^n \equiv 0 \pmod{10}$, which means $3^n + 7^n$ is a multiple of 10. Explanation: This relies on the property that for any odd integer $n$, $x^n + (-x)^n = x^n - x^n = 0$. This directly satisfies the condition of being a multiple of 10.

   • Case 2: $n$ is an even number. If 'n' is even, then $(-3)^n = 3^n$. Example: $(-3)^2 = 9$, $(-3)^4 = 81$. So, for even 'n': $3^n + (-3)^n = 3^n + 3^n = 2 \cdot 3^n$ Now we need to check if $2 \cdot 3^n \equiv 0 \pmod{10}$ when 'n' is even. Let's look at the unit digits of $3^n$ for even 'n':

     • For $n=2$, $U(3^2) = U(9) = 9$. Then $2 \cdot 9 = 18 \equiv 8 \pmod{10}$.
     • For $n=4$, $U(3^4) = U(81) = 1$. Then $2 \cdot 1 = 2 \equiv 2 \pmod{10}$.
     • For $n=6$, $U(3^6) = U(729) = 9$. Then $2 \cdot 9 = 18 \equiv 8 \pmod{10}$. The unit digits of $3^n$ for even 'n' alternate between 9 and 1. Consequently, $2 \cdot 3^n \pmod{10}$ alternates between 8 and 2. In neither case is $2 \cdot 3^n \equiv 0 \pmod{10}$. Therefore, if 'n' is an even number, $3^n + 7^n$ is NOT a multiple of 10. Explanation: When $n$ is even, $3^n + 7^n$ simplifies to $2 \cdot 3^n \pmod{10}$. Since the unit digit of $3^n$ can only be 1 or 9 (for even $n$), multiplying by 2 will result in a unit digit of 2 or 8, never 0. Hence, the sum is not divisible by 10.

Conclusion on the Nature of 'n':

From the analysis above, we can conclude that $3^n + 7^n$ is a multiple of 10 if and only if 'n' is an odd number. Tip: This is a common property: $a^n + b^n$ is divisible by $a+b$ if and only if 'n' is an odd positive integer. In this problem, $a=3$ and $b=7$, so $a+b=10$.

Counting Two-Digit Odd Numbers:

The problem asks for two-digit numbers 'n'. This means $10 \le n \le 99$. We need to count the odd numbers within this range. The two-digit odd numbers are: $11, 13, 15, \ldots, 99$. This is an arithmetic progression with:

• First term ($a_1$) = 11
• Common difference ($d$) = 2
• Last term ($a_k$) = 99

Using the formula for the $k$-th term of an arithmetic progression, $a_k = a_1 + (k-1)d$: $99 = 11 + (k-1)2$ Subtract 11 from both sides: $88 = (k-1)2$ Divide by 2: $44 = k-1$ Add 1 to both sides: $k = 45$

Thus, there are 45 two-digit odd numbers.

Final Answer: The total number of two-digit numbers 'n' such that $3^n + 7^n$ is a multiple of 10 is $\mathbf{45}$.

Important Note regarding the provided "Correct Answer: 7": Based on standard number theory principles and modular arithmetic, the derived answer for this problem is 45. If the intended "Correct Answer" is indeed 7, there must be an unstated additional constraint on 'n' that is not present in the problem as written. For example, if 'n' was restricted to "two-digit odd numbers less than 25", the list would be $11, 13, 15, 17, 19, 21, 23$, which totals 7 numbers. However, without such explicit constraints, the mathematically rigorous solution yields 45.