Mathematical thinking turns a printing disaster into a solvable puzzle—one page at a time. If you have the My Pals Are Here Maths 5A PDF, you’ll find these topics in Chapters 1–4 (Whole Numbers, Factors & Multiples, Four Operations). You can use this story as a word problem for practice or to help students see the real-life application of those chapters.

Number of terms: ( 180 \div 18 = 10 ) multiples of 9 with even multipliers (2,4,6,…,20) → yes, 10 terms.

Miss Lee smiled. "Correct. But here's the useful part: In real life, problems aren't always in order. You used to sort, LCM to avoid double-counting, and sum formulas to check totals without re-adding thousands of pages. That's why we learn these skills—not just for exams, but to organize real-world chaos."

Sum of Stack A = (\frac{15}{2} \times (6 + 180) = 7.5 \times 186 = 1,395). Stack B = 18, 36, 54, …, 180. First term 18, last term 180, common difference 18.

Sum of Stack B = (\frac{10}{2} \times (18 + 180) = 5 \times 198 = 990). Numbers in both A and B are multiples of both 6 and 9 → multiples of LCM(6,9)=18. From Stack A: multiples of 18 with odd multiplier (18×1=18, 18×3=54, 18×5=90, 18×7=126, 18×9=162) → 5 numbers. From Stack B: multiples of 18 with even multiplier (18×2=36, 18×4=72, 18×6=108, 18×8=144, 18×10=180) → different set! Wait — this means no number is in both A and B , because A requires odd ×6, B requires even ×9. Let’s check 18: A: 6×3 (3 odd, yes), B: 9×2 (2 even, yes) — oh! 18 is in both! So my earlier assumption wrong — 18 satisfies both. But 36? A: 6×6 (6 even → not in A). So intersection is numbers divisible by 18 with multiplier odd for A (×3,×9,×15… no, that's wrong — let's methodically solve.)

She called two students, Lin and Ravi, from the My Pals Are Here Maths 5A class for help.

[ \text{Total} = \frac{n \times (n + 1)}{2} = \frac{180 \times 181}{2} = 90 \times 181 = 16,290 ] Stack A = 6, 18, 30, …, 180. This is an arithmetic sequence: first term 6, last term 180, common difference 12.

Miss Lee, the head of the mathematics department, had a problem. The printer in the office had jammed while printing the end-of-year exam papers for Primary 5. When the technician fixed it, the papers printed in a scattered, messy pile—completely out of order. The problem? The pages were numbered from 1 to 180 , but they were stacked in reverse and in chunks.

Better: A: 6×(odd) = 18k? Let odd=2m+1. Then 6(2m+1)=12m+6. For this to be multiple of 18: 12m+6 divisible by 18 → 12m+6=18p → divide 6: 2m+1=3p → 2m+1 odd multiple of 3. B: 9×(even)=9×2n=18n. So A∩B = numbers that are 18×k where k is both an odd integer (from A) and any integer (from B) → Wait B's even multiplier: 9×2n=18n, so B includes all multiples of 18. A's odd multiplier: 6×(odd) = 6,18,30,42,54,66,78,90,102,114,126,138,150,162,174. Multiples of 18 in that list: 18,54,90,126,162 → yes 5 numbers. Those are in A∩B. So intersection size = 5.