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.
Number of terms: ( 180 \div 18 = 10 ) multiples of 9 with even multipliers (2,4,6,…,20) → yes, 10 terms. My Pals Are Here Maths Pdf 5a
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.) Number of terms: ( 180 \div 6 =
Ravi added, "And now we can reassemble the exam papers correctly." Number of terms: ( 180 \div 18 =