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

Printable View

- Aug 15th 2008, 12:35 AMnot happy janProblem Solving
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

- Aug 15th 2008, 12:39 AMChris L T521
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 - Aug 15th 2008, 12:44 AMProve It
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...

$\displaystyle \frac{90}{2}(10 + 99)=45 \times 109 = 4905$ - Aug 15th 2008, 12:46 AMProve It
- Aug 15th 2008, 12:46 AMnot happy jan
You lost me. Thats way beyond my ability

- Aug 15th 2008, 12:49 AMProve It
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... - Aug 15th 2008, 12:50 AMnot happy jan
I need the sum of the digits

- Aug 15th 2008, 12:55 AMChris L T521
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 - Aug 15th 2008, 12:58 AMProve It

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 - Aug 15th 2008, 01:04 AMChris L T521
- Aug 15th 2008, 09:00 AMSoroban
Hello, not happy jan!

Instead of asking "What's the formula?", we can use a primitive approach.

Quote:

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: .$\displaystyle 405 + 450 \;=\;{\color{blue}855}$

- Aug 15th 2008, 06:42 PMProve It
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.