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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.

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โˆ™ 2012-01-15 22:25:14
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Q: How many three digit positive integers are square numbers?
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Continue Learning about Statistics

How may two digit-numbers contain the digit 4?

18 positive integers and 36 integers (negative and positive)


How many positive four digit integers have 1 as their first digit and 2 or 5 as their last digit?

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.


How many three digit positive integers are there?

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


How many 5 digit numbers that end with 0 are there?

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.


How many 3 digit numbers can be formed?

It depends upon whether the numbers can be used more than once. If the numbers can be used more than once, then there are 1,000 possible combinations; if not, then there are 720 possible combinations. ========== Assuming you are talking about integers you can calculate it this way: you can have any one of 9 digits for the first digit (zero is excluded because that would make it only a 2 digit number) You can have any one of 10 digits for the second and any one of 10 digits for the third digit. That gives you 9x10x10 = 900 different combinations for 3 digit numbers (not 1000). If you can include both negative and positive numbers as different numbers you get twice as many (2x900=18000). If you can count decimal numbers as 3 digit numbers (i.e. not restricted to integers) you can have 900 integers, 900 with the form xx.x, 1000 with the form x.xx (if zero is permitted to be the first digit and count as one of the 3 digits) or 900 (if a leading zero is NOT counted as one of the 3 digits). If a leading zero is NOT counted as one of the 3 digits, you could also have 1000 numbers of the form 0.xxx or just .xxx

Related questions

How may two digit-numbers contain the digit 4?

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How many triple digit numbers are there?

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