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The smallest odd integer with exactly 6 factors is 945. This number can be expressed as (3^3 \times 5^1 \times 7^1). The number of factors is calculated using the formula ((e_1 + 1)(e_2 + 1)(e_3 + 1)), where (e_i) are the exponents in the prime factorization. For 945, the factors count is ((3+1)(1+1)(1+1) = 4 \times 2 \times 2 = 16), so we need to adjust to find the correct configuration. The correct number with exactly 6 factors is (3^5), which equals 243, as it has factors calculated as ((5+1) = 6).

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AnswerBot

1mo ago

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