+1443 The sum of all the digits used to write the whole numbers 10 through 13 is 1+0+1+1+1+2+1+3 = 10. What is the sum of all the digits used to write the whole numbers 1 through 110, inclusive?
i think you would just count up the 1's then the 2's and so on, then multiply and add.
so there is 1, eleven 1's in 10-19, eight 1's in 20-99, thirteen 1's in 100-110,
1+11+8+13 = 33
so there are 2, 12, and 102, then eleven 2's in 20-29, seven 2's in 30-101.
(3*2) + (11*2) + (7*2) = 42
i see some patterns, but test it out by doing the 3's and 4's?
(4*3) + (11*3) + (6*3) =63 based on patterns
would the 4's = 84? test and see
0, 1, 2, 3..............106, 107, 108, 109, Will add first and last=0 +109 =10
Will add second to second last =1 + 108 =10........and so on.
Since we have: 110/2 =55 pairs, each equaling 10, we therefore have:
55 x 10 =550
The only number left is the 110 itself, and since it has a value of 2, we therefore have:
550 + 2 = 552. And that is what I think !!.
+1443 i only have two chances if i get it wrong 2 times ill have to start all over again with new problems... tough work
The method used by Guest #2 works ONLY for numbers all ending in 9s. Therefore, the following is the more accurate method:
0, 1, 2, 3..............96, 97, 98, 99. Will add first and last=0 +99 =18
Will add second to second last =1 + 98 =18........and so on.
Since we have: 100/2 =50 pairs, each equaling 18, we therefore have:
50 x 18 =900
The only numbers left 100 to 110 itself, and since they have:1+2+3+4+5+6+7+8+9+10+2=57
900 + 57 =957 - Sum of all digits from 1 to 110.
+2446 Here is an extremrly basic mockup of a number table from 0-99 for you to reference as I make the adding process easier.
I notice a few patterns here. The tens digit remains the same. Let's attempt to add the numbers 0-9.
\(\underbrace{0+\underbrace{1+\underbrace{2+\underbrace{3+\underbrace{4+5}+6}+7}+8}+9}\\ \)
The numbers enclosed by the underbraces sum to nine (9+0=9, 1+8=9, 2+7=9, 3+6=9. 4+5=9). There are 5 lots of this occurring with the numbers one to nine, so the sum of the tens digit in the first columns equals \(9*5\text{ or }45\). This phenomenon occurs ten times or once per column, so let's multiply this by 10. \(45*10=450\).
Now, let's do the ones column. The ones column is different because each column increments the amount by 1. The sum of the ones column can be represented by the following sequence. \(0*10+1*10+2*10+...+8*10+9*10\). We can use algebra to group the common factor, so it simplifies to \(10(1+2+3+...+8+9)\). Of course, we already know that the sum of the numbers 1-9 sums to 45 by our first calculation. 10 times that amount yields \(450\).
Of course, we have only dealt with the sum of the digits from 0-99. However, if you think about it, the sum of 100-109 should be the same as adding the ones digit and the hundreds digits. Of course, this is a special case since the tens digits are all 0. We already know that \(1+2+3+...+8+9=45\). We know that the hundreds digit of the numbers 100-109 sums to \(1*10=10\). However, we have forgotten 110, which has a value of 2. \(45+10+2=57\) for the sum of the digits from 100-110. Now, let's add everything together.
\(450+450+57=957\) or the sum of the digits from 0-110
Summed up on Wolfram/Alpha and it gives:
57 + sum_(b=0)^9( sum_(a=0)^9(a + b)) =957
https://www.wolframalpha.com/input/?i=57+%2B+%E2%88%91%5Ba%2Bb%5D,+a%3D0+to+9,+b%3D0+to+9
+129852 11 "1s" in the hundreds digits = 11
10 * 45 + 1 = 451 = sum of tens digits
11 * 45 = 495 = sum of units digits
So
11 + 10*45 + 1 + 11*45 = 957
Here is a very simple method to add them all up.
Write each number as two digits as follows:
01, 02, 03..........to 97, 98, 98
Now, since we have 100 numbers each of which consists of 2 digits, we, therefore, have:
2 x 100 = 200 individual digits. The average value of each digit is: 1+2+3+4+5+6+7+8+9 =45/10 =4.5.
200 x 4.5 = 900 - this is the sum of all numbers from 1 to 100. And to this add the sum of all numbers from 100 to 110 inclusive =57.
So, the total is 900 + 57 = 957 - the sum of all numbers from 1 to 110.
