Verified Solution • 2025–26 Session moderate

QJEEM23Ma46e2: JEE Main 2023 Mathematics - Class 11 JEEM23Ma46e2 Solution: The number of ways of selecting two numbers and and such that 2 …

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

Multiple Choice Options

  • A 186
  • B 54
  • C 108
  • D 268

Step-by-Step Verified Solution

a+b=23λ+2a+b=23\lambda+2

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

λ=1\lambda=1 (a,b)→12(a, b)\to12 pairs

λ=3\lambda=3 (a,b)→35(a,b)\to35 pairs

λ=5\lambda=5 (a,b)→42(a,b)\to42 pairs

λ=7\lambda=7 (a,b)→19(a,b)\to19 pairs

λ=9\lambda=9 (a,b)→0(a,b)\to0 pairs

⋮\vdots

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

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