Consider all three-digit numbers that are greater than the sum of the squares of their digits by exactly 543. What are the last three digits of the sum of these numbers?
Consider all...
There are none...
the sum of a the digits of no number is nothing, or zero.
The answer should be 626
7,2,8
To find the last three digits of the sum of these numbers, we first need to determine the range of three-digit numbers that satisfy the given condition.
Let's represent a three-digit number as ABC, where A, B, and C are the respective digits from left to right.
The condition states that the number ABC is greater than the sum of the squares of its digits (A^2 + B^2 + C^2) by 543. Mathematically, it can be expressed as:
ABC > (A^2 + B^2 + C^2) + 543
Now, we can break down this equation by considering the possible values of A, B, and C.
Since A, B, and C are digits, each can only take values from 0 to 9. Therefore, the maximum value of (A^2 + B^2 + C^2) is 9^2 + 9^2 + 9^2 = 243.
Let's solve the modified inequality:
ABC - (A^2 + B^2 + C^2) > 543
ABC - 243 > 543
ABC > 786
So, we need to find the sum of three-digit numbers greater than 786. There are several approaches to find this sum. One simple approach is to use the formula for the sum of an arithmetic series:
Sum = (n/2)(first term + last term)
In this case, the first term is 787, and the last term is 999 (the largest three-digit number). The number of terms, n, can be found by subtracting the first term from the last term and adding 1:
n = 999 - 787 + 1 = 213
Therefore, the sum of all three-digit numbers greater than 786 is:
Sum = (213/2)(787 + 999) = 213 * 893 = 190,509
Finally, we need to find the last three digits of the sum, which is 509.
Hence, the last three digits of the sum of the required numbers are 509.