The number of ways of selecting two numbers a and b, a 2,4,6, ., 100 and b 1,3,5, .

Question

The number of ways of selecting two numbers $a$ and $b, a \in\{2,4,6, \ldots ., 100\}$ and $b \in\{1,3,5, \ldots . ., 99\}$ such that 2 is the remainder when $a+b$ is divided by 23 is :

Accepted Answer

$$a+b=23\lambda+2$$

$$\lambda=0,1,2,$$ ...., but $$\lambda$$ cannot be even as $a+b$ is odd

$$\lambda=1$$ $$(a, b)\to12$$ pairs

$$\lambda=3$$ $$(a,b)\to35$$ pairs

$$\lambda=5$$ $$(a,b)\to42$$ pairs

$$\lambda=7$$ $$(a,b)\to19$$ pairs

$$\lambda=9$$ $$(a,b)\to0$$ pairs

$$ \vdots $$

Total $=12+35+42+19=108$

Solutions · 1