The number N= 12345678910111213141... till 30 is written base 23 what is the remainder when N is written in base 10 and divided by 11?
A)1 B)3 C)5 D)0 E)9

• 123456789101112131415161718192021222324252627282930 (base 23)
  
  =1*23^51+2*23^50+…….9*23^2+3*23+0*23^0(base 10)
  
  23 congruent to 1 mod 11
  
  23^2 congruent to 1 mod 11
  
  23^3 congruent to 1 mod 11
  
  And so on,23^n congruent to 1 mod 11, n being integer
  
  Answer=Sum of the digits of N mod 11
  
  =(1+2+….9+1+2+..9+1+9+2+3+….+8+9+2+8+2+9+3+0)mod 11
  
  =(55+55+55+33)mod 11
  
  =0
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• i want solution
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• plz tell me this solution
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• Apply the divisibility rule of 11. .. Answer will be zero.
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• N= 12345678910111213141... till 30 is written base 23 convert to base 10 and get N=133352791492678560604402846468248080488642226264668888206048460648040 mod 11 which is 5
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• plz can anyone give solution for this question
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• 123456789101112131415161718192021222324252627282930 (base 23)
  
  =1*23^51+2*23^50+…….9*23^2+3*23+0*23^0(base 10)
  
  23 congruent to 1 mod 11
  
  23^2 congruent to 1 mod 11
  
  23^3 congruent to 1 mod 11
  
  And so on,23^n congruent to 1 mod 11, n being integer
  
  Answer=Sum of the digits of N mod 11
  
  =(1+2+….9+1+0+1+1+1+2+.....+1+9+2+0+2+1+2+2........+2+9+3+0mod 11
  
  =(45+45+45+9+18+3)mod 11
  
  = 165 mod 11
  
  =0
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• 123456789101112131415161718192021222324252627282930 (base 23)
  after change it into base 10
  =1*23^51+2*23^50+…….9*23^2+3*23+0*23^0(base 10)
  
  23 congruent to 1 mod 11
  23^2 congruent to 1 mod 11
  23^3 congruent to 1 mod 11
  And so on,23^n congruent to 1 mod 11, n being integer
  
  Answer=Sum of the digits of N mod 11
  =(1+2+….9+1+2+..9+1+9+2+3+….+8+9+2+8+2+9+3+0)mod 11
  =(55+55+55+33)mod 11
  
  =0{Ans)->D
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• hello, are questions repeated in capgemini written test and can you share the questions pls, i have to crack the first round.
  sunilmalviya23061996@gmail.com
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