Let's determine the three consecutive multiples of 11 whose sum is 1023.
1. Let's denote these three multiples of 11 as:
- The first multiple: [tex]\(11n\)[/tex]
- The second multiple: [tex]\(11(n+1)\)[/tex]
- The third multiple: [tex]\(11(n+2)\)[/tex]
2. To find their sum, we can write the equation:
[tex]\[
11n + 11(n+1) + 11(n+2) = 1023
\][/tex]
3. Simplify the equation by factoring out 11:
[tex]\[
11n + 11n + 11 + 11n + 22 = 1023
\][/tex]
[tex]\[
33n + 33 = 1023
\][/tex]
4. Solve for [tex]\(n\)[/tex] by isolating it on one side of the equation:
[tex]\[
33n + 33 = 1023
\][/tex]
[tex]\[
33n = 1023 - 33
\][/tex]
[tex]\[
33n = 990
\][/tex]
[tex]\[
n = \frac{990}{33}
\][/tex]
[tex]\[
n = 30
\][/tex]
5. Now that we have [tex]\(n = 30\)[/tex], we can find the actual multiples:
- The first multiple: [tex]\(11n = 11 \times 30 = 330\)[/tex]
- The second multiple: [tex]\(11(n+1) = 11 \times 31 = 341\)[/tex]
- The third multiple: [tex]\(11(n+2) = 11 \times 32 = 352\)[/tex]
Thus, the three consecutive multiples of 11 whose sum is 1023 are [tex]\(330\)[/tex], [tex]\(341\)[/tex], and [tex]\(352\)[/tex].
