document.write( "Question 170679: What is the sum of the digits used to form the integers from 45 to 100, inclusive?
\n" ); document.write( "I don't know where to start with this problem. Any help is appeciated! \n" ); document.write( "

Algebra.Com's Answer #126011 by Mathtut(3670) $\"\"$ $\"About$

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well, I dont know if there is a formula or not but what this is asking is to add up the seperate digits that make up the integer numbers from 45 to 100 including 45 and 100....so for 45 we would have 4+5=9 and 46 is 4+6=10, 47 is 4+7=11 and so on up to 100 which is 1+0+0=1 and you add those result all together. in other words take the results of 45 4+5=9 and add that to the results of 46 4+6=10 on up to 100 (9+10+11+12+13+5...). I think it would be much easier to add up all the 10 digit numbers and all the ones digit numbers seperately.\r
\n" ); document.write( "\n" ); document.write( "lets take the 10's digits first..from 45-49 ( $\"4\"$ 5, $\"4\"$ 6, $\"4\"$ 7, $\"4\"$ 8, $\"4\"$ 9) you have 5(4's)=20...you then have 10(5's)in the 50's=50
\n" ); document.write( "10(6's)in the 60's=60 and so on... 10(7)=70 10(8)=80 and 10(9)=90......add those all up and that takes care of all the 10's digits.... this adds up to $\"370\"$ now lets deal with the ones digits
\n" ); document.write( "for 45-49 first you have 5+6+7+8+9= $\"35\"$ . from 50-99 you simply have a repeat of adding up (1 to 9) 5 times....in other words 1+2+3+4+5+6+7+8+9=45...45(5)= $\"225\"$ . now you have all the ones digits added up from 45-99. Lets take what we have so far and add them together : 10's digits and ones digits from 45-99 we have 370+35+225=630.....the only thing we havent covered is 100 which is 1+0+0=1........so all the digits of integers from 45-100 equals $\"630%2B1=631\"$ \n" ); document.write( "

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