Question

Find the sum of the positive integers less than 200 which are not multiples of 4 and 7​

Answer 1

Answer:

$12942$ is the sum of positive integers between $1$ (inclusive) and $199$ (inclusive) that are not multiples of $4$ and not multiples $7$.

Step-by-step explanation:

For an arithmetic series with:

• $a_1$ as the first term,
• $a_n$ as the last term, and
• $d$ as the common difference,

there would be $\displaystyle \left(\frac{a_n - a_1}{d} + 1\right)$ terms, where as the sum would be $\displaystyle \frac{1}{2}\, \displaystyle \underbrace{\left(\frac{a_n - a_1}{d} + 1\right)}_\text{number of terms}\, (a_1 + a_n)$.

Positive integers between $1$ (inclusive) and $199$ (inclusive) include:

$1,\, 2,\, \dots,\, 199$.

The common difference of this arithmetic series is $1$. There would be $(199 - 1) + 1 = 199$ terms. The sum of these integers would thus be:

$\begin{aligned}\frac{1}{2}\times ((199 - 1) + 1) \times (1 + 199) = 19900 \end{aligned}$.

Similarly, positive integers between $1$ (inclusive) and $199$ (inclusive) that are multiples of $4$ include:

$4,\, 8,\, \dots,\, 196$.

The common difference of this arithmetic series is $4$. There would be $(196 - 4) / 4 + 1 = 49$ terms. The sum of these integers would thus be:

$\begin{aligned}\frac{1}{2}\times 49 \times (4 + 196) = 4900 \end{aligned}$

Positive integers between $1$ (inclusive) and $199$ (inclusive) that are multiples of $7$ include:

$7,\, 14,\, \dots,\, 196$.

The common difference of this arithmetic series is $7$. There would be $(196 - 7) / 7 + 1 = 28$ terms. The sum of these integers would thus be:

$\begin{aligned}\frac{1}{2}\times 28 \times (7 + 196) = 2842 \end{aligned}$

Positive integers between $1$ (inclusive) and $199$ (inclusive) that are multiples of $28$ (integers that are both multiples of $4$ and multiples of $7$) include:

$28,\, 56,\, \dots,\, 196$.

The common difference of this arithmetic series is $28$. There would be $(196 - 28) / 28 + 1 = 7$ terms. The sum of these integers would thus be:

$\begin{aligned}\frac{1}{2}\times 7 \times (28 + 196) = 784 \end{aligned}$.

The requested sum will be equal to:

• the sum of all integers from $1$ to $199$,
• minus the sum of all integer multiples of $4$ between $1\!$ and $199\!$, and the sum integer multiples of $7$ between $1$ and $199$,
• plus the sum of all integer multiples of $28$ between $1$ and $199$- these numbers were subtracted twice in the previous step and should be added back to the sum once.

That is:

$19900 - 4900 - 2842 + 784 = 12942$.