it is divisble by...................

(2power51+2power52+2power53+2power54+2power55)............. options:- (a)23 (b)58 (c)124 (d)127

• Nandani
• 12 May
• 12 Answers

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• • Albin
  • 03 May

  23

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• • Ranjithkumar
  • 28 Apr

  a) 23

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• • Rohan
  • 28 Apr

  2^51+ 2^52+2^53+2^54+2^55= 2^51(1+2+2^2+2^3+2^4) = 2^51 * 31 = 2^49 * 124
  Therefore answer is 124

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• • Ayush
  • 28 Apr

  option c. 124

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• • Anamika
  • 19 Apr

  ANS= 124

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• • Rishab
  • 19 Apr

  Ans. 124 Explanation: (2pow51 + 2pow52+2pow53+2pow54+2pow55) = 2pow51*(1 + 2 + 4 + 8 + 16) // take 2pow51 common = 31*2pow51 =124*2pow49

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• • Abhishek
  • 19 Apr

  127

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• • Mitali
  • 12 Apr

  127

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• • Sunil
  • 29 Jul

  In the above given series,
  the units digit of 2^51 is 8 &
  the units digit of 2^52 is 6 &
  the units digit of 2^53 is 2 &
  the units digit of 2^54 is 4 &
  the units digit of 2^55 is 8.
  The sum of all the units digits is 8+6+2+4+8=28.
  The sum of digits of 28 is2+8= 10 & the sum of digits of 10 is 1+0=1.
  There from options,
  (a)  sum of the digits of 23 is 2+3=5 &
  (b) sum of the digits of 58 is 5+8=13 & again 1+3=4. &
  (c) sum of the digits of 124 is 1+2+4= 7 &
  (d) sum of digits of 127 is 1+2+7=10 & again 1+0 = 1.
  Therefore, the sum of digits of option d is equal to the sum of digits of units place of the given series i.e, 1=1.
   Therefore,
  the given series is divisible by 127.
   

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