Given that \( B \) is 10 times larger than \( A \), we can express this relationship mathematically as:
\[ B = 10 \times A \]
The prime factorization of \( 10 \) is \( 2^1 \times 5^1 \). If we denote the prime factorization of \( A \) as \( A = 2^x \times 5^y \), then we can express the prime factorization of \( B \) as follows:
\[ B = 10 \times A = (2^1 \times 5^1) \times (2^x \times 5^y) = 2^{x+1} \times 5^{y+1} \]
According to the problem, the prime factorization of \( B \) is given as:
\[ B = 2^9 \times 5^1 \]
By comparing the two expressions for \( B \):
\[ 2^{x+1} \times 5^{y+1} = 2^9 \times 5^1 \]
we can set the exponents equal to find \( x \) and \( y \):
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For the base \( 2 \): \[ x + 1 = 9 \implies x = 8 \]
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For the base \( 5 \): \[ y + 1 = 1 \implies y = 0 \]
Thus, the prime factorization of \( A \) is:
\[ A = 2^x \times 5^y = 2^8 \times 5^0 = 2^8 \]
In conclusion, the prime factorization of \( A \) is:
\[ \boxed{2^8} \]