Question

If \(k\) is a positive integer and \(10^k\) is a divisor of the number \(9^{11} + 11^9\), then the greatest value of \(k\) is:

• A. \(1\)
• B. \(2\)
• C. \(3\)
• D. \(4\)

Hint:

To find the highest power of \(10^k\) dividing a number, factor 10 as \(2 \times 5\) and compute the highest powers of 2 and 5 that divide the number separately.

Answer 1

The Correct Option is B

Step 1: We are given that \(10^k\) divides \(9^{11} + 11^9\).
So, we must find the highest power of 10 that divides \(9^{11} + 11^9\).
Step 2: Note that:
\[ 10^k = 2^k \cdot 5^k \] So, we need to find the highest power of both 2 and 5 that divide \(9^{11} + 11^9\).
Step 3: Check divisibility by 2 and 5:
Let’s calculate \(9^{11} + 11^9 \mod 2\):
\[ 9^{11} \equiv 1^{11} \equiv 1 \mod 2 \\ 11^9 \equiv 1^9 \equiv 1 \mod 2 \\ \Rightarrow 9^{11} + 11^9 \equiv 1 + 1 = 0 \mod 2 \] So, divisible by 2.
Step 4: Check modulo 4:
\[ 9 \equiv 1 \mod 4 \Rightarrow 9^{11} \equiv 1^{11} = 1 \mod 4 \\ 11 \equiv 3 \mod 4 \Rightarrow 11^2 \equiv 1 \Rightarrow 11^8 \equiv 1 \Rightarrow 11^9 \equiv 3 \mod 4 \\ \Rightarrow 9^{11} + 11^9 \equiv 1 + 3 = 0 \mod 4 \] So, divisible by \(4 = 2^2\).
Step 5: Check divisibility by 5:
Since \(9 \equiv -1 \mod 5\), \(11 \equiv 1 \mod 5\):
\[ 9^{11} \equiv (-1)^{11} \equiv -1 \mod 5 \\ 11^9 \equiv 1^9 \equiv 1 \mod 5 \\ \Rightarrow 9^{11} + 11^9 \equiv -1 + 1 = 0 \mod 5 \] So divisible by 5.
Step 6: Check divisibility by 25:
Now try modulo 25:
\[ 9^2 = 81, 9^4 = 6561, 9^8 = 43046721, \text{but let's look for patterns:} \] Euler’s theorem: Since \(\phi(25) = 20\), \(a^{20} \equiv 1 \mod 25\) if \(\gcd(a,25)=1\)
We can use successive powers or verify:
\[ 9^5 = 59049 \Rightarrow 59049 \mod 25 = -1 \Rightarrow 9^{10} \equiv 1 \mod 25 \Rightarrow 9^{11} \equiv 9 \mod 25 \\ 11^2 = 121, 11^4 = 14641, 11^8 = 214358881, \text{find } 11^9 \mod 25: \\ \text{Since } 11^2 = 121 \equiv 121 \mod 25 = 121 - 100 = 21 \\ \Rightarrow 11^2 \equiv 21 \mod 25 \\ 11^4 = (11^2)^2 \equiv 21^2 = 441 \Rightarrow 441 \mod 25 = 441 - 425 = 16 \\ 11^8 \equiv 16 \Rightarrow 11^9 = 11^8 \cdot 11 \equiv 16 \cdot 11 = 176 \Rightarrow 176 \mod 25 = 1 \] So:
\[ 9^{11} + 11^9 \equiv 9 + 1 = 10 \mod 25 \Rightarrow \text{Not divisible by }25 \] Hence, the maximum power of 5 is \(5^1\), and of 2 is \(2^2\).
\[ \Rightarrow \text{Greatest }k \text{ such that } 10^k \mid 9^{11} + 11^9 \text{ is } \boxed{2} \]