There are 5000 such strings.
To determine the number of positive integers less than 1000 with distinct digits and are even, we need to consider the possible combinations of digits. Since the number must be even, the last digit must be even, giving us 5 options (0, 2, 4, 6, 8). For the hundreds digit, we have 9 options (1-9), and for the tens digit, we have 8 options (0-9 excluding the hundreds digit and the last digit). Therefore, the total number of such integers is 5 * 9 * 8 = 360.
Oh, dude, you're talking about four-digit numbers where every digit is even, right? So, we've got 2, 4, 6, and 8 to play with. Each place can have any of those four options, so it's like 4 choices for each of the 4 digits. That's 4x4x4x4, which is 256 numbers between 1000 and 9999 where all the digits are even.
To write 345 as a decimal, you place the decimal point after the last digit. So, 345 written as a decimal is 345.0. This is because the decimal point is always placed after the whole number part, even if there are no decimal digits.
Because the next greater even number is 100. That, and any subsequent even number, consists of three or more digits.
A decimal number is simply a way of representing a number in such a way that the place value of each digit is ten times that of the digit to its right. A decimal representation does not require a decimal point. So the required decimal representation is 120, exactly as in the question.
There are 500000 such numbers.
None, because you do not have three even digits. None, because you do not have three even digits. None, because you do not have three even digits. None, because you do not have three even digits.
The greatest 4-digit number with no repeated digits is... 9876
2: The number ends in 0, 2, 4, 6 or 8 4 : The last 2 digits are a multiple of 4 The 10s digit is even and the last digit is 0, 4 or 8 The 10s digit is odd and the last digit is 2 or 6 8: The last 3 digits are a multiple of 8 The 100s digit is even and the last 2 digits are a multiple of 8 The 100s digit is odd and the last 2 digits are 4 times an odd number 16: The last 4 digits are a multiple of 16 The 1,000s digit is even and the last 3 digits are a multiple of 16 The 1,000s digit is odd and the last 3 digits are 8 times an odd number 32: The last 5 digits are a multiple of 32 The 10,000s digit is even and the last 4 digits are a multiple of 32 The 10,000s digit is odd and the last 4 digits are 16 times an odd number 64: The last 6 digits are a multiple of 64 The 100,000s digit is even and the last 5 digits are a multiple of 64 The 100,000s digit is odd and the last 5 digits are 32 times an odd number 128: The last 7 digits are a multiple of 128 The 1,000,000s digit is even and the last 6 digits are a multiple of 128 The 1,000,000s digit is odd and the last 6 digits are 64 times an odd number 256: The last 8 digits are a multiple of 256 The 10,000,000s digit is even and the last 7 digits are a multiple of 256 The 10,000,000s digit is odd and the last 7 digits are 128 times an odd number 512: The last 9 digits are a multiple of 512 The 100,000,000s digit is even and the last 8 digits are a multiple of 512 The 100,000,000s digit is odd and the last 8 digits are 256 times an odd number 1,024: The last 10 digits are a multiple of 1,024 The 1,000,000,000s digit is even and the last 9 digits are a multiple of 1,024 The 1,000,000,000s digit is odd and the last 9 digits are 512 times an odd number
To find the even two-digit numbers where the sum of the digits is 5, we need to consider the possible combinations of digits. The digits that sum up to 5 are (1,4) and (2,3). For the numbers to be even, the units digit must be 4, so the possible numbers are 14 and 34. Therefore, there are 2 even two-digit numbers where the sum of the digits is 5.
12345 but if you can use a decimal, then 0.1234 would be the smallest possible number.
804
There are 5760 such numbers.
99999999
26.
I think the question is incomplete. One possibility so far is 63.197
81