Find the sum of all the natural numbers from 1 to 400, inclusive, that are not divisible by 11. The answer is not 35840.

To find the sum of all the natural numbers from 1 to 400 that are not divisible by 11, we can first find the sum of all the natural numbers from 1 to 400, and then subtract the sum of all the natural numbers from 1 to 400 that are divisible by 11.

The sum of the natural numbers from 1 to 400 is equal to $\frac{400 \cdot 401}{2} = 80150$.

There are 400/11 = 36 numbers that are divisible by 11 between 1 and 400, inclusive. The sum of these numbers is equal to $11 + 22 + 33 + \dots + 396 = 11(1 + 2 + 3 + \dots + 36)$. The sum of the first 36 natural numbers is equal to $\frac{36 \cdot 37}{2} = 666$. Therefore, the sum of the numbers from 1 to 400 that are divisible by 11 is equal to $11 \cdot 666 = 7326$.

Subtracting the sum of the numbers that are divisible by 11 from the sum of all the numbers from 1 to 400 gives us a final answer of $80150 - 7326 = 72824$. This is not equal to 35840, as stated in the problem.

mhanifa mhanifa · Answer 2 · 2023-01-07T11:09:14-05:00

Answer:

The sum is 72874

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Use the sum of the first n terms formula for an AP:

[tex]S_n=n(t_1+t_n)/2[/tex]

The sum of all natural numbers from 1 to 400:

[tex]S_{400}=400(1+400)/2=200*401=80200[/tex]

The greatest natural number less than 400 but divisible by 11:

400/11 ≈ 36

Sum of the numbers divisible by 11:

1*11 + 2*11 + ... + 36*11 =
11(1 + 2 + ... + 36) =
11*36*(1 + 36)/2 =
11*18*37 =
7326

Sum of the numbers not divisible by 11:

80200 - 7326 = 72874