"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 :"

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Accepted Answer

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$a+b=23\lambda+2$

$\lambda=0,1,2,$ ...., 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 $=12+35+42+19=108$

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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 :

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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 :

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