Question

Shakuntla Devi, Karl Gauss, Archimedes could do a lot of quick maths since their young ages.
Can you do better and find the sum of all the digits in integers from one and a billion.
Note, that's all the digits in all the numbers, not all the numbers themselves. Answer in Integer​

Answer 1

Answer:

40500000001

Step-by-step explanation:

Shakuntla Devi, Karl Gauss, Archimedes could do a lot of quick maths since their young ages.

Can you do better and find the sum of all the digits in integers from one and a billion.

Note, that's all the digits in all the numbers, not all the numbers themselves the answer is 40500000001

Answer 2

To Find :-

• Sum of all the digits in integers from one and a billion.

Solution :-

we know that,

• 1 Billion = 1,000,000,000 = 10⁹
• sum(10^d - 1) = sum{10^(d-1) - 1} * 10 + 45*{10^(d-1)}

so,

→ sum(999999999) :-

=> sum(10⁹-1) = sum(10⁸ - 1)*10 + 45*10⁸

=> sum(10⁸ - 1) = sum(10⁷ - 1)*10 + 45*10⁷

=> sum(10⁷ - 1) = sum(10⁶ - 1)*10 + 45*10⁶

=> sum(10⁶ - 1) = sum(10⁵-1)*10 + 45*10⁵

=> sum(10⁵ - 1) = sum(10⁴ - 1)*10 + 45*10⁴

=> sum(10⁴ - 1) = sum(10³ - 1)*10+ 45*10³

=> sum(10³ - 1) = sum(10² - 1)*10 + 45*10²

=> sum(10² - 1) = sum(10 - 1)*10 + 45*10 = (10*9/2)*10 + 450 = 450 + 450 = 900 .

then,

→ sum(10³ - 1) = 900*10 + 45 * 10² = 9000 + 4500 = 13500 .

→ sum(10⁴ - 1) = 13500*10 + 45*10³ = 135000 + 45*1000 = 135000 + 45000 = 180000

→ sum(10⁵ - 1) = 180000*10 + 45*10⁴ = 1800000 + 450000 = 2250000 .

→ sum(10⁶-1) = 2250000*10 + 45*10⁵ = 22500000 + 4500000 = 27000000

→ sum(10⁷ - 1) = 27000000*10 + 45*10⁶ = 315000000

→ sum(10⁸ - 1) = 315000000*10 + 45*10⁷ = 3600,000,000

therefore,

→ sum(10⁹ - 1) = 3600,000,000 * 10 + 45*10⁸ = 40,500,000,000 .

hence,

→ 1 to 1 Billion integers sum = sum(999999999) + sum(1,000,000,000) = 40,500,000,000 + 1 = 40,500,000,001 (Ans.)

Excellent Question .

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