Let A = nN : HCF(n,45)=1 and B = 2k : k1,2,,100.

Question

Let $A = \{n\in\mathbb{N} : 	ext{HCF}(n,45)=1\}$ and $B = \{2k : k\in\{1,2,\ldots,100\}\}$. Then the sum of all the elements of $A\cap B$ is \_\_\_\_.

Answer: 5264

Accepted Answer

$A$ = numbers not divisible by 3 or 5 (since $45=3^2	imes5$). $B=\{2,4,\ldots,200\}$.

$A\cap B = \{2k : k=1,\ldots,100,\ k 	ext{ not divisible by 3 or 5}\}$ (since 2 is coprime to 3 and 5, "$2k$ divisible by 3" $\iff$ "$k$ divisible by 3", similarly for 5).

So the sum is $2	imes$ (sum of $k=1,\ldots,100$ not divisible by 3 or 5):
- Sum of all $k$: $5050$
- Sum divisible by 3: $3(1+\cdots+33) = 3	imes561 = 1683$
- Sum divisible by 5: $5(1+\cdots+20) = 5	imes210 = 1050$
- Sum divisible by 15: $15(1+\cdots+6) = 15	imes21 = 315$
- Sum divisible by 3 or 5: $1683+1050-315 = 2418$
- Sum not divisible by 3 or 5: $5050-2418 = 2632$

Total: $2	imes2632 = 5264$

Solutions · 1