if x^y denotes X raised to the power y. find the last two digits of (1941^3781) * (1961^4181)
a) 12 b)22 c)42 d)44

• i guess none of these... as (1941)^3781 will give 1 at the unit's plc coz lst digit is 1... n same in the case of 1961^4181.... n since lst terms are 1 in both..thn 1*1=1 wl be the lst digit.... but no optn with lst digit as 1...
• 12 years agoHelpfull: Yes(10) No(3)
• it is coming 01
• 12 years agoHelpfull: Yes(5) No(2)
• ans is 01
  41*61= 2501
  last 2 digit is 01
• 12 years agoHelpfull: Yes(3) No(0)
• periodicity of both 1941 and 1961 is 6
  so ans would be 41^1+61^5=41+01=42
  power 1 and 5 obtained resp. by finding remainder of power divided by 6
• 12 years agoHelpfull: Yes(2) No(4)
• 41*61 = 01
• 12 years agoHelpfull: Yes(2) No(2)
• last two digits are 22
  bcz 1941 repeats its lat term after 6 times so 3781/6=rem3 so last two digits are 21
  bcz 1961 repeats its lat term after 6 times so 3781/6=rem5 so last two digits are 01
  so last two digit of that questn =21+01=22
  
• 12 years agoHelpfull: Yes(1) No(2)
• baral bhai question is wrong...i mean solution
• 12 years agoHelpfull: Yes(0) No(2)
• @nirvan gupta:::hey can u please explain how to find periodicity of 1941 and 1961??
• 12 years agoHelpfull: Yes(0) No(1)