Sure! Let's solve the problem step-by-step.
First, represent the three consecutive multiples of 8 as:
[tex]\[ 8n, \; 8n+8, \; \text{and} \; 8n+16. \][/tex]
The sum of these three multiples is:
[tex]\[ 8n + 8n + 8 + 8n + 16 = 24n + 24. \][/tex]
Next, represent the first two consecutive multiples of 6 as:
[tex]\[ 6m \; \text{and} \; 6m + 6. \][/tex]
The sum of these two multiples is:
[tex]\[ 6m + 6m + 6 = 12m + 6. \][/tex]
According to the problem, the sum of the three multiples of 8 is equal to eight times the sum of the first two multiples of 6:
[tex]\[ 24n + 24 = 8 \times (12m + 6). \][/tex]
Simplify the equation:
[tex]\[ 24n + 24 = 8 \times 12m + 8 \times 6, \][/tex]
[tex]\[ 24n + 24 = 96m + 48. \][/tex]
Subtract 24 from both sides to further simplify:
[tex]\[ 24n = 96m + 24. \][/tex]
Divide every term by 24:
[tex]\[ n = 4m + 1. \][/tex]
Now we need to find integers [tex]\( m \)[/tex] and [tex]\( n \)[/tex] such that the above equation holds.
Let's start with the smallest positive integer [tex]\( m = 1 \)[/tex]:
[tex]\[ n = 4 \times 1 + 1 = 5. \][/tex]
Thus, the three consecutive multiples of 8 are:
[tex]\[ 8n = 8 \times 5 = 40, \][/tex]
[tex]\[ 8(n+1) = 8 \times 6 = 48, \][/tex]
[tex]\[ 8(n+2) = 8 \times 7 = 56. \][/tex]
So, the three consecutive multiples of 8 are:
[tex]\[ \boxed{40, 48, \; \text{and} \; 56}. \][/tex]
These values satisfy the given condition in the problem.
