find unit digit of 7^(1!+2!+.......+200!)

• Sum of the factorials=------------13(last two digits will defnitely be 13)
  so
  7^----------13 is the question
  according to cyclicity(7^1=7, power2=9(lastdigit),power 3=3,power 4=1 this will be repeated to 5,6,7 powers and so on)
  when we divide the power by 4 we get remainder as 1(divisiblity rule of 4= last 2 digits must divisible by 4)
  but here last two digits 13%4 will give 1 as remainder
  so it will be converted into
  7^1
  7 must be last digit...
  
  if it is wrong..my work will be waste
• 10 years agoHelpfull: Yes(29) No(7)
• 1!+2!+3!+4!+5!+.....+200!
  calculate manually
  (consider each and every factorials last two digit)
  1
  2
  6
  24
  23
  20
  40
  20
  80
  00(from 10! on wards we get last two digits as 00..think logically)
  by adding these we get 13 as last two digits
  Sum of the factorials=------------13(last two digits will defnitely be 13)
  so
  7^----------13 is the question
  according to cyclicity(7^1=7, power2=9(lastdigit),power 3=3,power 4=1 this will be repeated to 5,6,7 powers and so on)
  when we divide the power by 4 we get remainder as 1(divisiblity rule of 4= last 2 digits must divisible by 4)
  but here last two digits 13%4 will give 1 as remainder
  so it will be converted into
  7^1
  7 must be last digit...
  
  if it is wrong..my work will be waste
• 10 years agoHelpfull: Yes(24) No(3)
• 1!+2!+3!+...+n!=(n+1)!/n
  1!+2!+3!+...+n!=(201)!/200
  201! has so many 0's in last so 201!/200 will also have..
  and if last 2 digits are 00 then it is surely divisible by 4 (divisibilty rule)
  so 7^4n is the form of ques
  and unit digit of 7^4n is 1
  so 1 is the answer
• 10 years agoHelpfull: Yes(11) No(11)
• 1!+2!+3!=9/4=rem 1 other term will be zero..
  7^1=7
  finally ans is 7
• 10 years agoHelpfull: Yes(5) No(2)
• 1!+2!+......+200!=[200(200+1)]/2=20100
  so,7^20100
  nw 20100 is divisable by 4.
  so odd^4k unit digit must be 1.so naswer will be 1.
  
• 9 years agoHelpfull: Yes(4) No(3)
• 7^------13 will be the power
  acc to cyclicity:
  13 %4 will give 1 as remainder
  7^1=7
• 10 years agoHelpfull: Yes(3) No(3)
• I am not sure about this approach so plz correct me if its wrong
  
  1!+2!+3!.....200! can be written in reverse as 200! + 199!+198!...1!. Now taking 200! common out of this equation
  200!(1+1/200+1/199*200...)
  now this is equal to 200!*1.005 =200!(approx)
  
  200! is completely divisible by so it takes d form 4*n
  7^4*n=1 at unit place so ans is 1.
• 10 years agoHelpfull: Yes(3) No(5)
• @HARSHA LANKA.....Could you plz specify how did you find the last two digits of the factorial series.....
• 10 years agoHelpfull: Yes(2) No(0)
• simply the sum of power will be 7^(....0033)
  so, according do cyclicity the power repeats after every 4.
  here dividing power by 4 we get remainder by 1.
  So, 7^1 =7 Ans.
• 10 years agoHelpfull: Yes(2) No(0)
• please explain clearly harsha lanka
  
• 10 years agoHelpfull: Yes(0) No(1)
• please explain it clearly... i didn't get you
  
• 10 years agoHelpfull: Yes(0) No(1)
• We know reparation of last digit in 7 is 7,9,3,1 if there is a power of multiple of 4 then last digit is 1 . All the factorial greater then 4! Are divisible of 4 so 4!+5!+6!........200! Sum of these will be multiple of 4 . Now 1!+2!+3! =9 Will decide the last digit of number 9%4=1 so remainder is Ist in repeated power of 7 that is 7 so..
  
  Ans. is 7
• 9 years agoHelpfull: Yes(0) No(0)