A hard question on remainders for P5, with a step-by-step solution, diagrams and the answer. Common in the Hua Xia Cup's P3 to P6 papers; harder ones need several remainder conditions to hold at once.
The question
On a long strip of paper, Ming writes the whole numbers in order from 30 to 80 with no gaps, making one very long number:
303132…80.
What is the remainder when this long number is divided by 9?
Show the hints (3)
💡Hint 1: A number leaves the same remainder on division by 9 as the sum of its digits does.
💡Hint 2: The long number's digit sum is the digit sums of 30, 31, …, 80 added together — and each number leaves the same remainder on division by 9 as its own digit sum.
💡Hint 3: So you only need the remainder of 30 + 31 + … + 80 divided by 9.
Worked solution
Have a go first, then read the solution step by step.
1.Dividing by 9, only the digit sum matters, and the long number's digit sum is all the numbers' digit sums added together.
2.Each number leaves the same remainder on division by 9 as its digit sum (23 and 2 + 3 = 5 both leave 5), so it's enough to divide 30 + 31 + … + 80 by 9.
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