if x^y denotes x raised to the power y, find last two digits of (1507^3381) + (1457^3757)

• last digit=7
  7^1=07
  7^2=49
  7^3=343
  7^4=2401
  7^5=16807
  7^6=117649
  repeating cycle of last two digits of 7=4
  1507^3381=07 ; 1457^3757=07
  last two digit=07+07=14
  
• 9 years agoHelpfull: Yes(11) No(8)
• lat digit of power of 7 are as follows:
  7,9,3,1,7,9,3,1,7,9,3,1,.........(because,7^1=7,7^2=49,....)
  ((1507^4)^845)*1507 + ((1457^4)^939)*1457 = 64(last two digit)
• 9 years agoHelpfull: Yes(10) No(1)
• ((1507^4)^845)*1507 + ((1457^4)^939)*1457 =(...01^845)*1507+(...01^939)*1457=...07+...57=64
• 9 years agoHelpfull: Yes(7) No(0)
• 
  = 1507^3381 + 1457^3757 can also be written as
  
  = (1507)^(4*845)X1507 + (1457)^(4*939)X1457
  
  now if we calculate 1507^4 , and 1457^4 the last two digit for both of them is (01), and also last two digit for (01)^n is always (01) therefore,
  
  = (01)^(845)X1507 + (01)^(939)X1457
  = (01)X1507 + (01)X1457
  = 1507 + 1457
  = 2964
  
  Hence last two digit will be "64"..
• 9 years agoHelpfull: Yes(7) No(0)
• For unit place 7^1=7,,7^3757=last digit 7.so 7+7=14
  FOr 10th digit,0*1=0,,5*5=25.so 5+0=5
  so 64
• 9 years agoHelpfull: Yes(3) No(1)
• (7^3381+57^3757)/100
  3381 and 3757 can be written in the form of 4n+1
  so we have 7+57=64
  last two digits are 64.
• 9 years agoHelpfull: Yes(3) No(0)
• 4. as we know that 7 raise to the power any number will always end at 7,9,3,1. and the number 3381 can be written as 4n+1 form. hence it will end at 7. similarly 3757 can also be written as 4n+1 and also ends at 7. hence adding we will get 4 as unit digit,.
• 9 years agoHelpfull: Yes(1) No(3)
• sum is 49+93=142
• 9 years agoHelpfull: Yes(0) No(5)
• 7^1=07
  7^2=49
  7^3=343
  7^4=2401
  7^5=16807
  7^6=117649
  cycle repeats => 7^2=49(last 2 digits of 3381 since 1507=1505/5 so 2 powers remain)
  
  Similarly 81^1=81(last 2 digits of each)
  81^2=61
  81^3=41
  81^4=21
  81^5=01
  81^6=81
  this pattern repeats
  1457=1455/5 and hence 2 powers remain therefore last 2 digits= 61
  
  so,40+61=104
  so the last two digits=04
• 9 years agoHelpfull: Yes(0) No(5)
• 7 powers are 7,9,3,1 3381/4 and 3757/4
  7+7=14
• 9 years agoHelpfull: Yes(0) No(1)
• 1507^(3381)+1457^3757
  consider last digit:-
  = 7^(4*845+1)+7^(4*939+1) [cyclicity of 7 is 4)
  :7^1+7^1=14
  therefore,last two digits=14
• 6 years agoHelpfull: Yes(0) No(0)