There are 9 choices for the first digit (1,2,3,4,5,6,7,8,9) No zero because numbers don't start with zeroes. 10 choices for the second digit (0,1,2,3,4,5,6,7,8,9). 10 choices for the third digit (0,1,2,3,4,5,6,7,8,9) Okay now 9x10x10=900.
Answer:900
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102 = 100 which is the first possible three digit number that is a perfect square. 312 = 961 which is the last possible three digit number that is a perfect square. So there are 22 three digit positive numbers that are perfect squares.
There are 200 positive four digit integers that have 1 as their first digit and 2 or 5 as their last digit. There are 9000 positive four digit numbers, 1000 through 9999. 1000 of them have 1 as the first digit, 1000 through 1999. 200 of them have 2 or 5 as their last digit, 1002, 1005, 1012, 1015, ... 1992, and 1995.
Assuming positive integers, and no leading zeros, the range of five digit numbers is 10000 to 99999. The ones that end in zero can be found by taking the four digit numbers: 1000 to 9999 and multiplying each by ten. {1000,1001, 1002, ...9999}, multiplied by ten is {10000,10010,10020,....99990}. There are 9000 of them.
There are only 999 three digit whole numbers.
24 three digit numbers if repetition of digits is not allowed. 4P3 = 24.If repetition of digits is allowed then we have:For 3 repetitions, 4 three digit numbers.For 2 repetitions, 36 three digit numbers.So we have a total of 64 three digit numbers if repetition of digits is allowed.