An expert question on cycles for P5–P6, with a step-by-step solution, diagrams and the answer. Cycle questions appear in the Hua Xia Cup from the P1 first round to the P5 papers: which comes in a given place, which day of the week, which last digit.
The question
The maths club plays a number chain: the first number is 3, the second 2, and each after that is the sum of the two before it, so it begins 3, 2, 5, 7, 12, 19, … They keep going to the 1973rd number.
How many of the 1973 numbers leave 1 when divided by 3?
Show the hints (3)
💡Hint 1: The numbers soon get huge; writing them all is hopeless. Look only at each number's remainder on division by 3: how does the next remainder come from the two before?
💡Hint 2: Write the remainders on until two neighbours are 0, 2 again. What happens after that?
💡Hint 3: Count how many remainders make one cycle, how many of them fit, and where they come in it. How many whole cycles are in 1973 numbers, and how many of the ones left over fit?
Worked solution
Have a go first, then read the solution step by step.
1.Look only at remainders on division by 3: the remainder of a sum is the sum of the remainders, reduced again, so each remainder comes from the two before without writing the big numbers.
2.The remainders run 0, 2, 2, 1, 0, 1, 1, 2, 0, 2, … The 9th and 10th are 0, 2 again, as at the start, so from then on they repeat every 8.
3.In each cycle places 4, 6, 7 leave 1 when divided by 3: 3.
4.1973 ÷ 8 = 246 remainder 5: 246 cycles give 246 × 3 = 738; the 5 left are the first 5 of a cycle, with 1. In all 738 + 1 = 739.
⚠️Common mistake: starting the cycle as soon as one remainder comes back (0 again, say); only when two neighbouring remainders match the start does everything repeat.
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