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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!!!!!!!!

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