X ^ Y denotes X raised to
power Y, find the last two digits (2361 ^ 4507) + (1741 ^ 3581)
a. 02
b. 82
c. 42
d. 62

• its 62.
  taking 2361^4507 last digit will be 1 and second last will be -> 2 (6*7=42 from 2361 and 4507) hence 21
  similarly for 1741^3581 last digit=1 and 2nd last will be ->4 (4*1=4 from 1741 and 3581) hence 41
  summing up 21+41= 62 (ANS)
  
• 11 years agoHelpfull: Yes(34) No(4)
• 62
  1^7=1 nd 6*7=42 take 2 nd for 2nd no 1^1=1 nd 4*1=4
  therefore 21+21=42
  
• 11 years agoHelpfull: Yes(3) No(5)
• 02 as 2361^4507 yield 61 last digit & 1741^3581 yield 41 last digit so 61+41=02 last two digit
• 11 years agoHelpfull: Yes(3) No(0)
• answer is 62
• 11 years agoHelpfull: Yes(1) No(3)
• d.)62 i will explain this later....very simple..try it using unit place digit and tens place digit's repitition..
  
• 11 years agoHelpfull: Yes(1) No(5)
• last two digits of(61^7 + 41 )
• 11 years agoHelpfull: Yes(0) No(5)