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The sum of three prime numbers is 100. If one of them exceeds another by 36, then one of the numbers is:

  • A.7
  • B.29
  • C.41
  • D.67

Answer: D

\(x+(x+36)+y=100 \\ \Rightarrow 2x+y=64\)
Therefore \(y\) must be even prime, which is 2.
Therefore,
\(2x+2=64 \\ \Rightarrow x=31\)
Third prime number
\(= (x+36)\\= (31+36)\\= 67\)

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