The number of ways of selecting two numbers a | vedclass

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(C) Given $a \in \{2, 4, \ldots, 100\}$ and $b \in \{1, 3, \ldots, 99\}$.Let $a = 2m$ where $m \in \{1, 2, \ldots, 50\}$ and $b = 2n-1$ where $n \in \

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$108$

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(C) Given $a \in \{2, 4, \ldots, 100\}$ and $b \in \{1, 3, \ldots, 99\}$.Let $a = 2m$ where $m \in \{1, 2, \ldots, 50\}$ and $b = 2n-1$ where $n \in \{1, 2, \ldots, 50\}$.Then $a+b = 2m + 2n - 1 = 2(m+n) - 1$.We want $a+b \equiv 2 \pmod{23}$,so $2(m+n) - 1 = 23k + 2$,which implies $2(m+n) = 23k + 3$.Since $2(m+n)$ is even,$23k+3$ must be even,so $k$ must be odd. Let $k = 2j-1$.Then $2(m+n) = 23(2j-1) + 3 = 46j - 23 + 3 = 46j - 20$.So $m+n = 23j - 10$.Since $1 \le m, n \le 50$,we have $2 \le m+n \le 100$.Possible values for $j$ are $1, 2, 3, 4, 5$:If $j=1$,$m+n = 13$. Number of pairs $(m, n)$ is $12$.If $j=2$,$m+n = 36$. Number of pairs $(m, n)$ is $35$.If $j=3$,$m+n = 59$. Number of pairs $(m, n)$ is $42$.If $j=4$,$m+n = 82$. Number of pairs $(m, n)$ is $19$.If $j=5$,$m+n = 105$ (not possible as max $m+n=100$).Total ways $= 12 + 35 + 42 + 19 = 108$.

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

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