If n is a positive integer such that \frac{6^{n+3} \cdot 15^{2n-1}}{10^{n+1} \cdot 9^{n+1}… — GMAT

If $n$ is a positive integer such that $\frac{6^{n+3} \cdot 15^{2n-1}}{10^{n+1} \cdot 9^{n+1}} = 8{}100$ , what is the value of $n$ ?

Answer: 4

Rewriting all composite bases (

6, 15, 10, 9

) into prime bases (

2, 3, 5

) simplifies the equation to

4 \cdot 3^n \cdot 5^{n-2} = 8,100

. Dividing by 4 yields

3^n \cdot 5^{n-2} = 2,025 = 3^4 \cdot 5^2

, which gives

n = 4

Express each composite base in the fraction using its prime factor decomposition.

Numerator:

(2 \cdot 3)^{n+3} \cdot (3 \cdot 5)^{2n-1} = 2^{n+3} \cdot 3^{n+3} \cdot 3^{2n-1} \cdot 5^{2n-1} = 2^{n+3} \cdot 3^{3n+2} \cdot 5^{2n-1}

.
Denominator:

(2 \cdot 5)^{n+1} \cdot (3^2)^{n+1} = 2^{n+1} \cdot 5^{n+1} \cdot 3^{2n+2}

Breaking composite numbers down into prime factors allows terms with identical bases to be combined using exponent rules.

Simplify the fraction by subtracting the denominator exponents from the numerator exponents for each prime base.

\frac{2^{n+3} \cdot 3^{3n+2} \cdot 5^{2n-1}}{2^{n+1} \cdot 3^{2n+2} \cdot 5^{n+1}} = 2^{(n+3)-(n+1)} \cdot 3^{(3n+2)-(2n+2)} \cdot 5^{(2n-1)-(n+1)} = 2^2 \cdot 3^n \cdot 5^{n-2} = 4 \cdot 3^n \cdot 5^{n-2}

Applying the quotient rule for exponents:

\frac{a^x}{a^y} = a^{x-y}

Set the simplified expression equal to

8,100

and solve for

n

4 \cdot 3^n \cdot 5^{n-2} = 8,100 \implies 3^n \cdot 5^{n-2} = 2,025

. Prime factorization of

2,025 = 81 \cdot 25 = 3^4 \cdot 5^2

. Equating exponents gives

n = 4

(and

n-2 = 2

, which confirms

n=4

Because 3 and 5 are distinct prime bases, the unique prime factorization theorem guarantees that

3^n \cdot 5^{n-2} = 3^4 \cdot 5^2

implies

n = 4

Simplifying exponential expressions using prime factorization and quotient laws

Estimated Time:2m 0s

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