My Pals Are Here Maths: Pdf 5a

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.

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

Ravi added, "And now we can reassemble the exam papers correctly." Better: A: 6×(odd) = 18k

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. For this to be multiple of 18: 12m+6

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