Powers of Two Divisibility

CAT 2019 Slot 1 · QA · Hard · Number Theory

The number of integers xx such that 0.25<2x<2000.25 < 2^x < 200, and 2x+22^x + 2 is perfectly divisible by either 3 or 4, is

Answer

5

Explanation

0.25=22<2x<2001x70.25 = 2^{-2} < 2^x < 200 \Rightarrow -1 \le x \le 7 for integer xx. Possible integer values of xx: {1,0,1,2,3,4,5,6,7}\{-1, 0, 1, 2, 3, 4, 5, 6, 7\}. Test values:

  • x=1x = -1: 21+2=2.52^{-1} + 2 = 2.5 (not integer)
  • x=0x = 0: 20+2=32^0 + 2 = 3 (divisible by 3) - Yes
  • x=1x = 1: 21+2=42^1 + 2 = 4 (divisible by 4) - Yes
  • x=2x = 2: 22+2=62^2 + 2 = 6 (divisible by 3) - Yes
  • x=3x = 3: 23+2=102^3 + 2 = 10 (not div by 3 or 4)
  • x=4x = 4: 24+2=182^4 + 2 = 18 (divisible by 3) - Yes
  • x=5x = 5: 25+2=342^5 + 2 = 34 (not div by 3 or 4)
  • x=6x = 6: 26+2=662^6 + 2 = 66 (divisible by 3) - Yes
  • x=7x = 7: 27+2=1302^7 + 2 = 130 (not div by 3 or 4) Satisfying values: x{0,1,2,4,6}x \in \{0, 1, 2, 4, 6\}. Total = 5.

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