\nFirst, count the total occurrences of the digit 5 from 1 to 99. The digit 5 appears in the units place in 5, 15, 25, 35, 45, 55, 65, 75, 85, 95 (10 times). It appears in the tens place in 50, 51, 52, 53, 54, 55, 56, 57, 58, 59 (10 times). The number 55 contains the digit 5 twice. So, the digit 5 appears a total of 10 + 10 = 20 times.
\nNext, identify the numbers between 1 and 99 that contain the digit 5 and are divisible by 3. These numbers are:
\n– 15 (1+5=6, divisible by 3) – contains one 5
\n– 45 (4+5=9, divisible by 3) – contains one 5
\n– 51 (5+1=6, divisible by 3) – contains one 5
\n– 54 (5+4=9, divisible by 3) – contains one 5
\n– 57 (5+7=12, divisible by 3) – contains one 5
\n– 75 (7+5=12, divisible by 3) – contains one 5
\nThese are 6 numbers. Let’s count the occurrences of the digit 5 *within* these numbers: 15 (one 5), 45 (one 5), 51 (one 5), 54 (one 5), 57 (one 5), 75 (one 5). Total occurrences of 5 in these numbers are 6.
\nThe numbers containing 5 are: 5, 15, 25, 35, 45, 50, 51, 52, 53, 54, 55, 56, 57, 58, 59, 65, 75, 85, 95.
\nThe numbers from this list that are divisible by 3 are: 15, 45, 51, 54, 57, 75.
\nWe need to count the 5s in the numbers *not* divisible by 3: 5, 25, 35, 50, 52, 53, 55, 56, 58, 59, 65, 85, 95.
\nCounting the 5s in this remaining list:
\n5 (one 5), 25 (one 5), 35 (one 5), 50 (one 5), 52 (one 5), 53 (one 5), 55 (two 5s), 56 (one 5), 58 (one 5), 59 (one 5), 65 (one 5), 85 (one 5), 95 (one 5).
\nTotal count = 1 + 1 + 1 + 1 + 1 + 1 + 2 + 1 + 1 + 1 + 1 + 1 + 1 = 16.
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