is there a special rick or is it just trial and error to solve this porblem

it would take a really long time to find he asnwer by trial and answer

Best way to do this is to break things down. Start with 100. Clearly 1+0 = 1 which is not 0, so that doesn't work. But 101 does since 1+0 = 1 which is the last digit. Now ask yourself: are there any more numbers in the range of 100 to 109 that work? The answer is no since 1+0 = 1 is unique to this set of numbers (ie there are no more numbers in this range that have a 1 as the last digit) So again, in the list of numbers from 100 to 109, the only upright number is 101 Similarly, for numbers in the range 110 to 119, the only upright number is 112 Keep going to find the third upright number to be 123 Etc, etc When you come upon numbers that start with 19, you won't find any upright numbers since 1+9 = 10, which is not a single digit. So , in the range of 100 to 199, there are 8 numbers that are upright numbers So from 200 to 299, there are 7 numbers that work. Why 7 and not 8? notice that for numbers that start with 29 add to 2+9 = 11, but that's not a single digit. Likewise, numbers starting with 28 add to 2+8 = 10, again not a single digit. However, the rest work. Keep going and you'll find that only 6 numbers work from 300 to 399 etc etc After all that, you'll have the following number counts: 8, 7, 6, 5, 4, 3, 2, 1, 0 Add them all up to get 36 So the answer is choice C

Interesting question. Here's how I tackled it:Look at the first few 3 digit numbers.100101102103104105106107108109They all have the same first two digits, but different last digits. You'll notice that exactly one of these numbers is upright. You'll find the same thing for 110 through 119. I can keep going unless the sum of the first two digits is greater than 9. For example, any number with 67 as the first two digits can never be upright because 6+7 is a two digits number.So to start counting upright numbers, add up all the possible 2 digit combinations you can get. you can use 10 through 99, so that's 90 possible two digit pairs. Now you need to subtract off the pairs that don't work, like 67. 10, 11, 12, 13, ... , 18 all work, but 19 doesn't. That's 9 upright numbers. 20, 21, ... , 27 all work, but 28 and 29 don't. That's 8 more upright numbers. 90 works, but 91, 92, ... , 99 don't. That's just one more upright number. See the pattern? So find 1+2+3+4+5+6+7+8, and you have your answer.

you have the right idea pmilano, but you fell into the trap that there's a number in the range of 190 to 199 when there isn't one

By the way, the fast way to add the numbers 1 through 8 (or, more generally, 1 through n) is by using the formula (n)(n+1)/2 So 8(8+1)/2 = 36

Thanks jim.

my answer is D)45. There is actually a pattern if you try to list down the first numbers. Starting at 1__, there are 9 possible answers. As you move up to the next hundreds digit, you decrease that number by one, because you can't have a sum of more than 9. 9+8+7+...+2+1 = 45

I actually realized that before I posted but didn't proof read my post well enough (hence my correct answer and incorrect counting).

My bad, I was ignoring zero Start with the range 100-199 and moving to the next 200-299, etc, the following upright numbers are found in the tens digits listed below 0, 1, 2, 3, 4, 5, 6, 7, 8 0, 1, 2, 3, 4, 5, 6, 7 0, 1, 2, 3, 4, 5, 6 0, 1, 2, 3, 4, 5 0, 1, 2, 3, 4 0, 1, 2, 3 0, 1, 2 0, 1 0 So there are 45 here

i am getting 36?

Here is another method lol (incase you didnt have enough) When the last digit is 'n', you want to ask yourself how many solutions are there to: x + y = n where x and y are non-negative integers? For example, if n was 3, then these are the possible solutions for x and y: x = 0, y = 3 x = 1, y = 2 x = 2, y = 1 x = 3, y = 0 These correspond to the three digits numbers: 033 123 213 303 Of course we would discard the '033' case. Anywhos, the equation that gives you the number of solutions to "x + y = n" is: \[\left(\begin{matrix}n+2-1 \\ 2-1\end{matrix}\right) = \left(\begin{matrix}n+1 \\ 1\end{matrix}\right)\] We need to do this for n = 1, 2, 3,....,9, while remembering to not count the case of the first digit being 0. So we end up with: \[\left(\begin{matrix}2 \\ 1\end{matrix}\right)+\left(\begin{matrix}3 \\ 1\end{matrix}\right)+\left(\begin{matrix}4 \\ 1\end{matrix}\right)+\ldots +\left(\begin{matrix}10 \\ 1\end{matrix}\right)-9\]\[ = 2+3+4+\ldots+10-9 = 1+2+3+\ldots+9 = 45\]

First set of numbers: 101,112,123,134,145,156,167,178,189 ---> 9 numbers. For 2__, that is one less than the 1 hundreds because u can only use 0-7 for the 2nd digit. The last possible number is 909, the only number in the 9 hundreds.

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