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A number is divisible by 6 if it is divisible by both 2 and 3. The smallest 2-digit number is 10 and the largest is 99.

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First determine the smallest 2-digit number divisible by 6. Starting from 10 the smallest multiple of 6 is 12.

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Next find the largest 2-digit number divisible by 6. From 99 the largest multiple of 6 is 96.

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Now find the sequence of 2-digit numbers divisible by 6: 12 18 24 ... 96. This sequence is an arithmetic series where the first term \(a = 12\) and the common difference \(d = 6\).

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To find the number of terms use the formula for the \(n\)-th term of an arithmetic sequence:

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\[

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a_n = a + (n-1) \cdot d

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\]

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Set \(a_n = 96\) and solve for \(n\):

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\[

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= 12 + (n-1) \cdot 6

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\]

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