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."

Number of terms: ( 180 \div 6 = 30 ) multiples of 6, but only odd multipliers → half of them? Let’s check: Multiples of 6 up to 180 = 6×1 to 6×30 (30 numbers). Odd multipliers: 1,3,5,…,29 → that’s 15 terms.

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.

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.

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.

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.

Ravi added, "And now we can reassemble the exam papers correctly."

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.)

[ \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.

Sum of intersection: 18+54+90+126+162 = (18+162)=180, (54+126)=180, plus 90 → 180+180+90=450. Stack C = Total − (Sum A + Sum B − Intersection) = 16,290 − (1,395 + 990 − 450) = 16,290 − (2,385 − 450) = 16,290 − 1,935 = 14,355 . Step 7: The twist Lin announced, "Miss Lee, Stack C's total is 14,355."

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

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