7^1+7^2+7^3+.......+7^205. Find out how many numbers present which unit place contain 3?

• Ans:51
  SIMPLE,
  7^1=7
  7^2=49
  7^3=343
  7^4=2401
  7^5=16807
  7^6=117649,
  obsereve unit place 7,9,3,1 are repeating so,
  total=205%4=51
• 9 years agoHelpfull: Yes(18) No(0)
• 51.
  Follow power cycle method.
• 9 years agoHelpfull: Yes(11) No(4)
• answer: 51.
  7^3, 7^7, 7^11... has 3 in its unit place.
  thus for every power increment by 4 places, we get 3 in the unit place.
  Hence, 205%4=51.
• 9 years agoHelpfull: Yes(4) No(2)
• 7^1=7(in the base),7^2=9,7^3=3,7^4=1.............and this cycle repeats like this onward 7,9,3,1
  (7^1=7 + 7^2=9 + 7^3=3 + 7^4=1)....................+ 7 ^205=(7+9+3+1+7+9+3+1................. and this cycle continue until 204th term)+7^1(205th term)=7+9+3+1=20 base is 0 so
  (7+9+3+1)*54 + 7^1= 0*54 + 7 = 7 (ans)
  7(1)
  9(2)
  3(3)
  1(4)
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  .(series reapts until 54 term=7^204) =20*54(20+20+20+20+20.......54 times will will give unit base as 0).......so 0+7=7(ans)
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• 9 years agoHelpfull: Yes(3) No(6)
• we are just considering the unit place....so
  7^1=7
  7^2=9
  7^3=3
  7^4=1
  -----------
  7^5=7
  7^6=9
  7^7=3
  7^8=1
  --------
  so we see that in a cycle of 4 there is one 3 on the unit place
  so divide 205/4=51 + reminder=1 it doesn't matter so answer will be 51........
• 9 years agoHelpfull: Yes(2) No(1)
• follow power of 7 cycle is 4.
  ans.51
• 9 years agoHelpfull: Yes(1) No(1)
• 7*9 gives 3 at its unit place
  so series will be
  7*9,7*19,7*29........,7*199,
  no of terms in this series are can be calculated by tn=a+(n-1)d
  7*199=7*9+(n-1)*70
  =>199=9+(n-1)10
  =>n=20 the 20 is the correct answer
• 9 years agoHelpfull: Yes(0) No(4)
• Please post the question clearly, this ain't a series its a sum...The question isn't clear
• 9 years agoHelpfull: Yes(0) No(0)
• Most simple way is we have units place 7^1=7,7^2=9,7^3=3,7^4=1
  Thus cycles goes on repeating 3,7,11,15------ we can see common difference is 4... Thus to find upto how much time 3 is contained as unit dgt. One way is keep costing manually with difference of 4 like 3,7,11,15,19------ or one way is find a factor of 4 as last dgt is 7^205, where only 204 is lst dgt dvsble by 4 nd contains 3 in unit plce.. So ap forms 3,7,11,15,--------204 so n=51
• 9 years agoHelpfull: Yes(0) No(0)
• + sign should not be there
• 9 years agoHelpfull: Yes(0) No(0)
• Frst power like 7^3 = its units digit 3. SO, similarly we will get 7^3+7^7+7^11+.........+7^204. (beacuse 7 has a 4 cycle).totaly we will get 3 to 204 means 68 numbers.
  ANS:68
• 9 years agoHelpfull: Yes(0) No(0)
• unit place will be 3 only on no whose power will give remainder as 3 when divided by 4.
  so a series is formed in such case....
  series goes like this................. 3,7,11,15......
  last term will be 203 which lie in this series
  so,
  203= 3 + (n-1)*4
  => n=51.
  hope it will help.
• 9 years agoHelpfull: Yes(0) No(0)
• No option please follow power cycle method:)
• 8 years agoHelpfull: Yes(0) No(0)
• Unit digit of first 6 terms are 7, 9, 3, 1, 7, 9 hence it forms a pattern. For every 4 terms we get one term with 3 in its unit digit. So there are total of 205/4 = 51 sets and each set contains one term with 3 as it's unit digit.
  So the ans is 51.
• 6 years agoHelpfull: Yes(0) No(0)