# MATH 101 problem solving

• Mar 9th 2009, 06:21 AM
phillyfan09
MATH 101 problem solving
Just wanted to see if someone could help me out and show me the answer you got as well as your process for solving so as I can match it with my work...

PROBLEM:

"In the first problem, we found that Jen needed to write 1080 digits to number the 396 pages in her biology notebook. What do the 1080 digits add up to?"

Any help appreciated!
• Mar 9th 2009, 06:44 AM
u2_wa
Quote:

Originally Posted by phillyfan09
Just wanted to see if someone could help me out and show me the answer you got as well as your process for solving so as I can match it with my work...

PROBLEM:

"In the first problem, we found that Jen needed to write 1080 digits to number the 396 pages in her biology notebook. What do the 1080 digits add up to?"

Any help appreciated!

Lets check what is the sum of digits in between 1-99, leaving zeros
digit $1$: 10+10=20 (1,10,11,...19,21,31,...91) (total value=20)
digit $2$: same 20 (total value=20*2)
hence same with digits $3,4,5,6,7,8,9$
over all total in this above range = $20(1+2+3...9)=900$

Lets check what is the sum of digits in between 100-199, leaving zeros

Note: units and tens are the same as in 1-99 but only thing 1 digit is repeating from 100-199 (so there are 100 1's extra in this step)

over all total in this above range = $100 +900 = 1000$

Lets check what is the sum of digits in between 200-299, leaving zeros
Note: units and tens are the same as in 1-99 but only thing 2 digit is repeating from 200-299 (so there are 100 2's extra in this step)
over all total in this above range = $200 + 900 = 1100$

Lets check what is the sum of digits in between 300-399, leaving zeros
Note: units and tens are the same as in 1-99 but only thing 3 digit is repeating from 300-399 (so there are 100 3's extra in this step)
over all total in this above range = $300 + 900 = 1200$

but range given is upto 396 so we have to deduct $397(3+9+7),398(3+9+8),399(3+9+9)$
$= -60$
hence total value of digits is $= 900 + 1000 + 1100 + 1200 - 60 = 4140$
• Mar 9th 2009, 09:29 AM
phillyfan09
why did you divide by 396 by 2 and multiply that by 396 + 1?

where do the 2 and 1 come from?
• Mar 9th 2009, 10:13 AM
u2_wa
Just ignore the previous post. I have edit my last post.
• Mar 11th 2009, 11:08 AM
laurmarissa
I do not understand the first part of the problem.
"Lets check what is the sum of digits in between 1-99, leaving zeros
digit http://www.mathhelpforum.com/math-he...6f75849b-1.gif: 9+10=19 (1,10,11,...19,21,31,...91) (total value=19)
digit http://www.mathhelpforum.com/math-he...cc14862c-1.gif: same 19 (total value=19*2)
hence same with digitshttp://www.mathhelpforum.com/math-he...a5fdc224-1.gif
over all total in this above range = http://www.mathhelpforum.com/math-he...c5eca352-1.gif"
Can you please clarify this? Thank you!
• Mar 12th 2009, 12:06 AM
u2_wa
Hello Laurmarissa:
There were some calculation mistakes which i've now corrected.
Let me clarify you the first part in detail,

First check how many times these digits appear in the range $1-99$
Digit 1: (1,10,11,12,13,14,15,16,17,18,19,21,31,41,51,61,71, 81,91)
=20 times, and if we add 1's we'll get $1*20=20$

Digit 2: (2,12,20,21,22,23,24,25,26,27,28,29,32,42,52,62,72, 82,92)
=20 times, and if we 2's we'll get $2*20=40$
Now we can deduce that all the remaining digits from 3-9 will occur 20 times in the above range.
Now lets calculate the total sum of these digits: $20(1+2+3+..+9)=900$
Hope this helps!