I have no idea what this problem means:

For how many two-digit numbers if the ones digit larger than the tens-digit? Can you find a systematic way to arrive at the number?

Ok, every 2-digit number looks like xy, where x is the 10's digit and y is the one's digit. It should be apparent that x runs from 1 to 9.
If x is 1 then y runs from 2 to 9, or 8 digits
If x is 2 then y runs from 3 to 9, or 7 digits
.
.
.
If x is 8 then y = 9, or 1 digit.
For x=9, 0 digits.
If we count up the numbers possible for y we add the integers 1+2+3+4+5+6+7+8
Put that into google if you want or just add it up.

The numbers you seek look like the following, the one's digit being preceded by each of the integers to the left of the right hand number

12
123
1234
12345
...
...
...
123456789

So you have
2 + 3 + 4+ 5 + 6 + 7 + 8 + 9 = ?

The sum of the first n integers starting with 1 is S == n(n += 1)/2

Therefore, the sum of your numbers is S = n(n + 1) - 1

Please HElPPP!!!!!!!!!!

Is it possible to arrange numbers 1 through 16 in a 4 by 4 grid so that consecutive numbers are not touching? If so, how did you do it and also can you find more than one solution?

Please Help I don't understand THIS !!!!!!!!!!!

THANKS!!!!!!!!

Does this work?
1,

To solve the first problem of finding how many two-digit numbers have the ones digit larger than the tens digit, we can analyze the possible values for both digits.

Let's consider the two-digit number in the form of "xy", where x represents the tens digit and y represents the ones digit.

It should be apparent that the tens digit, x, can range from 1 to 9.

If x is 1, then y can range from 2 to 9, giving us 8 possible digits for y.

If x is 2, then y can range from 3 to 9, giving us 7 possible digits for y.

Continuing this pattern, we see that if x is 8, then y can only be 9, resulting in 1 possible digit for y.

Finally, if x is 9, there are no possible digits for y since there are no numbers greater than 9.

To calculate the total number of two-digit numbers where the ones digit is larger than the tens digit, we can sum up the number of possible digits for y.

Adding the numbers 8 + 7 + 6 + 5 + 4 + 3 + 2 + 1 gives us 36.

Therefore, there are 36 two-digit numbers where the ones digit is larger than the tens digit.

To verify this calculation, we can use the formula for the sum of the first n integers starting from 1: S = n(n+1)/2.

Plugging in n = 8, we get S = 8(8+1)/2 = 36, which matches our calculation.

For the second problem of arranging numbers 1 through 16 in a 4 by 4 grid so that consecutive numbers are not touching, it is not possible to do so.

To understand why, let's visualize the numbers arranged in a 4 by 4 grid:

1 2 3 4
5 6 7 8
9 10 11 12
13 14 15 16

If we observe, there are many pairs of consecutive numbers that are adjacent to each other horizontally, vertically, or diagonally. For example, 2 and 3 are adjacent horizontally, 6 and 7 are adjacent vertically, and 10 and 11 are adjacent diagonally.

Since the problem explicitly asks for consecutive numbers to not touch, it is not possible to arrange numbers 1 through 16 in a 4 by 4 grid satisfying this condition.

Therefore, there is no solution to this problem.