If x^y denotes x raised to the power of y find last two digits of (1507^3381)+(1437^3757)

• soln is 44
  bcoz power cycle of 7 is 4.when 3381 is divided by 4 it gives remainder 1, so (07)^1=07
  similarly in 3757 it is (37)^1=37
  07+37 =44
  but tcs question was(1457^3757)
• 8 years agoHelpfull: Yes(6) No(3)
• solution is 44
  because we know that (07)^4 gives unit digit is 1 . so we can write power 3381 in the multiplication of power 4. now 3381 divide by 4 than remainder is 1 , so (07)^1 =07.
  
  similarly (37)^3757 gives the remainder 1 when 3757 is divided by 4 . so (37)^1=37.
  
  now 07 + 37 = 44
• 8 years agoHelpfull: Yes(4) No(0)
• 07+61^939*37=07+17=24
• 8 years agoHelpfull: Yes(3) No(1)
• 7^1=7 (last digits )
  7^2=9
  7^3=3
  7^4=1 further same series will repeats. convert the eqn into powers of 7^4 then
  (7^4)^845*7 + (7^4)^939*7
  last digits = 1*7 + 1*7 =14
  .". last digit =4
• 8 years agoHelpfull: Yes(1) No(2)
• We have ,7^(multiple of 4)=1
  So let us consider last two digits of 3381 i.e. 81 and divide 81 by 4 so that we will get 40
  Quotient and remainer as 1,so 7^4*40=1 and as remainder is 1 so 7^1=7
  Now consider second no.=3757 take its last two digits=57
  Divide by 4 wll get quotient=14,so 7^4*14=1
  And remainder is =1,so 7^1=7
  So, in both cases we have (7^1)+(7^1)=14
  Hence and.=14
• 8 years agoHelpfull: Yes(1) No(1)
• solution is 24
  
• 8 years agoHelpfull: Yes(1) No(2)
• 4
  Since last digit is 7 in both the nos. ,we can observe easily that 7 raised till power 4 repeat itself .
• 8 years agoHelpfull: Yes(0) No(1)
• 78 (49+29)
• 8 years agoHelpfull: Yes(0) No(2)
• (1507^3381)+(1437^3757)=76
• 8 years agoHelpfull: Yes(0) No(1)
• is it asked again?
• 8 years agoHelpfull: Yes(0) No(2)
• 07+17 = 24
• 8 years agoHelpfull: Yes(0) No(1)
• First for cyclicity of 7 is 4 ,for a unit place odd number the whatever the power it should give 01,for another one we should go for calculation so 1507^3381 can be written as 1507^(4*845)+1so for this the last digit for 1507^4*835 is always 01 and for 1507^1 it is 07 so last 2 digit is 01*07=7..similarly for 1437^(4*939)+1 is 01*37 so 37 and the sum of last 2 digit will be 07+37=44...
• 8 years agoHelpfull: Yes(0) No(1)