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What is the sum of all individual digits from 1 to 100. In other words, 1+2+3+4+5+6+7+8+9+(1+0)+(1+1)+(1+2) and so on until 100.

 

Any help would be greatly appreciated!

 Oct 23, 2021
 #1
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Actually putting your mind to it would help.

 Oct 23, 2021
 #2
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is it 901?

 Oct 23, 2021
 #3
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Hello Guest,

 

from the sum beetween 1 and 9 is (\(1+9*(\frac{9}{2})=\)) 45, the sum beetween 10 and 19 is (45+10 =) 55, the sum beetween 20 and 29 is (45+20 =) 65, do this trick every time and you'll get the answer. May you response and tell me the answer?

 

Straight

 Oct 23, 2021
 #4
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You know the sum beetween 1 and 9 is 45, multiply this by 10 (because there are 11 and 19, 21 and 29, ... 91 and 99) and that equals 450. The first digit could be all digits (except 0) and the second digit too, but it could be 0, but x + 0 = is still x. So we double the 450 and that is (450 * 2 =) 900 and now + 1 because of 100 (1 + 0 + 0 = 1). So (900 + 1 =) 901 is the solution.

 

Straight

Straight  Oct 23, 2021

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