Answer:
the average of all 11 digits is 6.
Explanation:
(a1 + a2 + a3 + ... + a10) / 10 = 5.7
Multiplying both sides of the equation by 10 gives us:
a1 + a2 + a3 + ... + a10 = 57
Similarly, we are given that the average of the last 10 digits is 6.6. This can be expressed as:
(a2 + a3 + ... + a11) / 10 = 6.6
Multiplying both sides of the equation by 10 gives us:
a2 + a3 + ... + a11 = 66
Now, let's subtract the first equation from the second equation:
(a2 + a3 + ... + a11) - (a1 + a2 + a3 + ... + a10) = 66 - 57
Simplifying this equation gives us:
a11 - a1 = 9
From this equation, we can see that the difference between the last digit (a11) and the first digit (a1) is equal to 9.
Since we know that there are only 11 digits in total, we can conclude that a11 must be greater than a1 by exactly 9 units.
Now, let's consider the sum of all 11 digits:
(a1 + a2 + a3 + ... + a10) + (a2 + a3 + ... + a11) = 57 + 66
Simplifying this equation gives us:
2(a2 + a3 + ... + a10) + a11 + a1 = 123
Since we know that a11 - a1 = 9, we can substitute this into the equation:
2(a2 + a3 + ... + a10) + (a1 + 9) + a1 = 123
Simplifying further gives us:
2(a2 + a3 + ... + a10) + 2a1 = 114
Dividing both sides of the equation by 2 gives us:
(a2 + a3 + ... + a10) + a1 = 57
But we already know that (a1 + a2 + a3 + ... + a10) = 57, so we can substitute this into the equation:
57 + a1 = 57
Simplifying further gives us:
a1 = 0
Now that we know the value of a1, we can substitute it back into the equation a11 - a1 = 9:
a11 - 0 = 9
This gives us:
a11 = 9
So, the first digit (a1) is 0 and the last digit (a11) is 9.
To find the average of all 11 digits, we sum up all the digits and divide by 11:
(a1 + a2 + ... + a11) / 11 = (0 + a2 + ... + 9) / 11
Since we know that (a2 + ... + a10) = 57, we can substitute this into the equation:
(0 + 57 + 9) / 11 = (66) / 11 = 6