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To determine how many times the digit 3 appears from 1 to 500, we need to consider each place value separately. In the units place, the digit 3 appears 50 times (3, 13, 23, ..., 493). In the tens place, the digit 3 appears 5 times in each set of 10 numbers (30, 31, 32, ..., 39), totaling 50 times. In the hundreds place, the digit 3 appears 100 times (300, 301, 302, ..., 399). Therefore, the digit 3 appears 200 times from 1 to 500.

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ProfBot

βˆ™ 1w ago
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Wiki User

βˆ™ 15y ago

Instead of going from 1 to 500, let's just subtract one from the set and come up with "How many 3's appear from 1 to 499?"

Total possible combinations in this, including leading zeros, is 5*10*10, i.e. 5 possible in the first set(0,1,2,3,4), and 10 in the second two(0,1,2,,4,5,6,7,8,9).

Instead of figuring out how many have a 3, lets figure out how many don't. Well there is 4 possibilities in the first (0,1,2,4), and nine in the second two (0,1,2,4,5,6,7,8,9).

So, total = 5 * 10 * 10 = 500

Without 3 = 4 * 9 * 9 = 324

500-324 = 176

There are 176 numbers from 1 to 500 that have the digit 3 in them.

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Q: How many 3's appear from 1 to 500?
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