We have to consider 2 cases:
When the hundreds digit is 3, we want the sum of the tens and unit digit equal to 13. We have 4+9=5+8=6+7=13, which yield a total of 6 choices (two for each pair that adds to 13).
When the hundreds digit is 4, we want the sum of the tens and unit digit equal to 12. We have 3+9=4+8=5+7=6+6=12. The first three pairs give us 2 solutions, but the last only gives 1, so we have a total of 7 choices.
Therefore there are total of 6+7=13 integers.
There are none.
Two integers.
952 of them.
An infinite amount
Five. All nonzero integers are significant and zeros in between significant digits are significant.
There are 120 of them.
There are none.
How many
There are 20.
Two integers.
There are 125 of them.
9*9*8*7 = 4536
Since the first integer of the 3-digit numbers must be 2, the sum of the second and the third digits would be 13.Since the largest digit is 9, the 3-digit numbers could be 249, 294, 258, 285, 267, and 276.Thus, there are 6 integers between 200 and 300 whose sum of their digits is 15.
There are infinitely many. For example, consider, 45, 451, 4511, 45111, 451111 and so on.
952 of them.
An infinite amount
Five. All nonzero integers are significant and zeros in between significant digits are significant.