Maths Question 18 – AP-EAMCET 2026
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To count the total occurrences of a specific digit (say, '5') in a range of numbers, we analyze each place value (units, tens, hundreds) separately. This method correctly accounts for numbers that contain the digit multiple times (e.g., 551 has two '5's). We consider the range from 1 to 999 and then check 1000 separately.
Remember to count each instance of the digit, not just the number containing the digit. For example, 55 counts as two '5's.
The digit '5' appears in the units place for numbers like 5, 15, 25, ..., 95, 105, ..., 995. In every block of 100 numbers (e.g., 1-100, 101-200), '5' appears 10 times in the units place. Since there are 10 such blocks from 1 to 1000 (i.e., numbers ending in '5' in 1-100, 101-200, ..., 901-1000), the total occurrences in the units place is:
The digit '5' appears in the tens place for numbers like 50-59, 150-159, 250-259, ..., 950-959. In every block of 100 numbers, '5' appears 10 times in the tens place (e.g., 50, 51, ..., 59). Again, there are 10 such blocks from 1 to 1000. The total occurrences in the tens place is:
The digit '5' appears in the hundreds place for numbers like 500-599. This range consists of 100 numbers (500, 501, ..., 599), where '5' is the hundreds digit. The digit '5' does not appear in the hundreds place for any other block of 100 numbers within 1 to 1000. The number 1000 also does not contain the digit '5'.
Sum the occurrences from each place value:
This systematic approach avoids missing any occurrences, especially in numbers like 550 or 505.
