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Find the sum of all 3-digit numbers which leave a remainder of 5 when divided by 8.

A. 36,555
B. 27,777
C. 25,555
D. 24,444

Answer :

Final answer:

The sum of all 3-digit numbers which leave a remainder of 5 upon division by 8 is 62037, which is not an available option.

Explanation:

The 3-digit numbers which leave a remainder of 5 after being divided by 8 are actually a series starting with the number 101 (the first 3-digit number to satisfy this condition) and ending at 997 (the last 3-digit number to satisfy the condition). These numbers form an arithmetic sequence, and so we can use the formula for the sum of an arithmetic sequence to calculate their sum:

sum = n/2 * (a + l)

where n is the number of terms, a is the first term, and l is the last term.

In this case, the difference between each consecutive term in the sequence is 8 (since each number is 8 higher than the previous number that left a remainder of 5 after division). The number of terms can be calculated as:

n = (997 - 101) / 8 + 1 = 113)

The sum of the sequence can be found to be:

sum = 113/2 * (101 + 997) = 62037

Hence, none of the options a, b, c, or d is correct.

Learn more about Arithmetic Sequence here:

https://brainly.com/question/33807844

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