Please help. Not sure if this is a trick question. What is the sum of the digits of all 2-digit numbers between 10 and 99. Please explain
Have you learned how to set up sums using this symbol? $\displaystyle \Sigma$?
The way to do this would be to find the sum from 1 to 99, and then subtract the sum from 1 to 9.
In sigma notation, this would be expressed as $\displaystyle \sum_{k=1}^{99}k-\sum_{k=1}^{9}k$
Note that $\displaystyle \sum_{k=1}^n k=\frac{n(n+1)}{2}$
Can you try to take it from here?
--Chris
This is an arithmetic sequence.
The sum of terms in an arithmetic sequence is given by
$\displaystyle \frac{n}{2}(a + z)$ where a is the first number of the sequence, z is the last number of the sequence, and n is the number of terms.
In this case, a = 10, z = 99, n = 90
So this sum is...
$\displaystyle \frac{90}{2}(10 + 99)=45 \times 109 = 4905$
If you're finding the sum of the numbers, all you need is the first and last numbers and the number of numbers.
There are 90 numbers in between 10 and 99, so n = 90.
The first number is 10, so a = 10
The last number is 99, so z = 99.
Then just put your a, n and z values into the formula and simplify.
But if you're finding the sum of all the digits, that's another story entirely...
Oops...misread the problem.
Note that if we add all the digits together, we can generate a pattern...
$\displaystyle 1+0+1+1+1+2+1+3+1+4+1+5+1+6+1+7+1+8+1+9$
$\displaystyle =10+(0+1+2+3+4+5+6+7+8+9)={\color{blue}10}+{\color {red}45}$
Now, if we keep going on, we see that
$\displaystyle 2+0+2+1+2+2+2+3+2+4+2+5+2+6+2+7+2+8+2+9$
$\displaystyle =20+(0+1+2+3+4+5+6+7+8+9)={\color{blue}20}+{\color {red}45}$
We start see a pattern now.
In the end, we will end up with $\displaystyle {\color{blue}10+20+30+40+50+60+70+80+90}+9({\color {red}45})=10(45)+9(45)=19(45)=\color{red}\boxed{85 5}$
I hope you can follow this!
--Chris
OK, well each group of 10 numbers will end in 0, 1, 2, 3, ..., 9.
There's 9 groups of these, so there's 9 x 1, 9 x 2, 9 x 3, ..., 9 x 9.
There's also 10 more 1's (because the first digit of the first 10 numbers is 1), 10 more 2's, 10 more 3's ... 10 more 9's.
So that makes 19 x 1 + 19 x 2 + 19 x 3 + ... + 19 x 9, or if you like,
19 x (1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9)
$\displaystyle =19 \times \frac{9}{2}(1 + 9) = 19 \times 4.5 \times 10 = 855$
Hope that helped
Hello, not happy jan!
Instead of asking "What's the formula?", we can use a primitive approach.
Imagine adding up the 90 two-digit numbers . . .What is the sum of the digits of all 2-digit numbers from 10 and 99?
$\displaystyle \text{ten 1's} \begin{Bmatrix}1&0 \\ 1&1 \\ 1&2 \\ \vdots \\ 1&9\end{Bmatrix}\;\;\text{0 to 9 = 45}$
$\displaystyle \text{ten 2's}\begin{Bmatrix}2&0 \\2&1\\2&2 \\ \vdots \\2&9\end{Bmatrix}\;\;\text{0 to 9 = 45}$
. .$\displaystyle \vdots$. . . . . $\displaystyle \vdots$ . . . . . .$\displaystyle \vdots$
$\displaystyle \text{ten 9's}\begin{Bmatrix}9&0 \\ 9&1 \\ 9& 2 \\ \vdots \\ 9&9 \end{Bmatrix}\;\;\text{0 to 9 = 45}$
In the units-column, we have the sum of the digits 0 to 9 (45) ... nine times.
Sum of the units digits: .$\displaystyle 9 \times 45 \:=\:405$
In the tens-column, we have: .$\displaystyle \text{ten 1's + ten 2's + ten 3's + . . . + ten 9's}$
. . $\displaystyle \;=\;10(1) + 10(2) + 10(3) + \cdots + 10(9) \;=\;10(1 + 2 + 3 +\cdots + 9) \;=\;10(45)$
Sum of the tens digits: .$\displaystyle 450$
Therefore, the sum of the digits is: .$\displaystyle 405 + 450 \;=\;{\color{blue}855}$
I sure did, once I figure out there were 19 sums of the first 9 digits, I used the formula for arithmetic series
$\displaystyle S_n=\frac{n}{2}(a+z)$
Where n is the number of terms, a is the first term of the sequence, and z is the last, to work out the sum and then multiplied by 19.