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find the sum of all the digits used to write the digits from 0 to 9999

 Mar 23, 2019
 #1
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0

(0, 4000, 8000, 12000, 16000, 20000, 24000, 28000, 32000, 36000)>>Total Sum = 180,000
(2890, 4000, 4000, 4000, 4000, 4000, 4000, 4000, 4000, 4000) >>>Total Number = 38,890

 Mar 23, 2019
 #2
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Let's do it case by case: 1 digits, 2 digits, 3 digits, and 4 digits.

 

1 digit: The numbers are 1, 2, 3, 4, 5, 6, 7, 8, and 9. The digits sum to 45.

 

2 digits: For each group of ten numbers, there are 10 times the tens digit used plus 45. For example, 1 + 0 + 1 + 1 + 1 + 2 + 1 + 3 + 1 + 4 + 1 + 5 + 1 + 6 + 1 + 7 + 1 + 8 + 1 + 9 = (10 x 1) + 45. Repeat this to get (10 x 1 + 10 x 2 + 10 x 3 + 10 x 4 + 10 x 5 + 10 x 6 + 10 x 7 + 10 x 8 + 10 x 9) + 45 x 9 = 10 x 45 + 45 x 9 = 19 x 45.

 

3 digits: We know that each group of 100's last two digits sum to 19 x 45 (above), so we just add 100 x the hundreds digit. This is 100 x 45 for each one, and then we have 100 x 45 + 19 x 9 x 45 = 171 x 45 + 100 x 45 = 271 x 45.

 

4 digits: Do the same process we did above to get 9 x 271 x 45 + 1000 x 45 = 3439 x 45.

 

Add these up to get 20 x 45 + 271 x 45 + 3439 x 45 = 291 x 45 + 3439 x 45 = 3730 x 45 = 167850.

 

Hoping this helped, 

asdf334

 Mar 23, 2019
 #3
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asdf335: You made a mistake in your "3 digits" summation. You should get: 12,600. And also in your "4 digit" summation. You should get: 166,500.

 

0 to 9 =45
10 to 99 =855
100 to 999 =12,600
1000 to 9999 =166,500
45 + 855 + 12,600 + 166,500 =180,000 - Sum of ALL digits from 0 to 9999.

Guest Mar 23, 2019
edited by Guest  Mar 23, 2019

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