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For all positive integers w and y , where w>y , let the operation be defined by
 wy=2w+y2wy . For how many positive integers w is w1 equal to 4 ?
A. None
B. One
C. Two
D. Four
E. More than four

Answer
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Hint: We solve this question by considering each option. Since w>y so we check w1 equal to 4 or not for all values of w such that w>1 by using wy=2w+y2wy .

Complete step by step solution:
Option (A): Suppose w=2 . Then we get
 21=22+1221=4 .
  w has at least one positive integer such that w1 equal to 4.
Option (A) is incorrect.
Option (B):
Suppose w=2 . Then we get
 21=22+1221=4 .
Similarly suppose w=3 . Then we get
 31=23+1231=4 .
  w has at least two positive integers such that w1 equal to 4.
  Option (B) is incorrect.
Option (C):
Suppose w=2 . Then we get
 21=22+1221=4 .
Similarly suppose w=3 . Then we get
 31=23+1231=4 .

Similarly suppose w=4 . Then we get
 41=24+1241=4 .
  w has at least three positive integers such that w1 equal to 4.
Option (C) is incorrect.
Option (D): Suppose w=k , where k is any positive integer such that k>1 then, we get
 k1=2k+12k1=4 .
  w has more than four positive integers such that w1 equal to 4.
 Option (D) is incorrect.
Option (E): Suppose w=k , where k is any positive integer such that k>1 then, we get
 k1=2k+12k1=4 .
  w has more than four positive integers such that w1 equal to 4.
Option (E) is correct.
So, the correct answer is “Option E”.

Note: Note that the given condition is important. Sometimes many people neglect and suffer to get the exact solutions. The number we are assuming has to satisfy a given condition else it cannot be considered.

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