7^1+7^2+7^3+...........+7^2015 find out how many numbers present which unit place contain 3 ?

• 7^3,7^7,7^11,7^15,7^19.........7^2015 contains last digit=3.
  Now,we see
  3,7,11,15....2015 are in AP
  tn=a+(n-1)d
  2015=3+(n-1)4
  2015=3+4n-4
  4n=2016
  n=504
  Ans=504.
• 9 years agoHelpfull: Yes(20) No(5)
• ans 54
  explanation:
  cyclic pattern of 7 is 7, 9, 3, 1 i.e., 4 is cyclic pattern number of 7
  7^1 = 7
  7^2 = 4(9)
  7^3 = 34(3)
  7^4 = 240(1)
  pattern repeats
  quotnt(2015/4)= 53 and remainder is 3 ie 7^2015 contains 3 in units digit
  so 54 numbers contain 3 in their units place
• 9 years agoHelpfull: Yes(8) No(7)
• The number of 11 letter words formed by using English alphabets if all vowels are used such that no two vowels are adjacent to each other is:
  a. 5! * 20! / 15!
  b. 5! * 21! / 15!
  c. 21! / 10!
  d. 21! / 15!
  
  
  
  20 feb tcs
• 9 years agoHelpfull: Yes(6) No(3)
• Vinaya, Please learn how to divide...:D
  and Pawan this is not the way of solving it..
  Unit digit theorem is enough to explain the answer..
  2015/4 =503 (according to cyclic rule of power 7 has a cycle of 4)
  2015%4=3 so the last 3 will be 7,9,3...
  so we can conclude, (503+1)= 504...:)
• 9 years agoHelpfull: Yes(4) No(1)
• 2015=4k+3; k=503
  3=4k+3; k=0
  Therefore , total nos with last digit 3 = 504
• 9 years agoHelpfull: Yes(2) No(1)
• 7^1=7
  7^2=49
  7^3=343
  7^4=2401
  so for every 4 terms we get one term with 3 in its unit digit
  so there are total of
  2015/4 we will get 503.6 as answer so answer is 504
• 9 years agoHelpfull: Yes(1) No(1)