If XY denotes X is raised to the power Y Find the last two digits of 19413843 19614181 a 12

ans is 82. if you will find the remainder of powers divided by 4 then u will get 3 and 1 respectively for 3843 & 4181...3 power of 1941 is 7312...21 and 1 power of 1961 is 1961 after this add last two digit of these values it will be..21+61=82.
9 years agoHelpfull: Yes(5) No(0)
82
1941^3843=unit digit^(unit digit of power)=1^3=1
1961^4181=unit digit^(unit digit of power)=1^1=1 thus, unit digit would be 1+1=2
for tens place =tens place^(unit digit of power)=4^3=12 but we will take only tens place=2
similarly, 6^1=6 thus adding 6+2=8 (value at tens place)
final ans 82
9 years agoHelpfull: Yes(3) No(2)
ans is 42
9 years agoHelpfull: Yes(0) No(2)
the question actually is
1941^3843 + 1961^4181
correct ans is 82.. can anyone explain?
9 years agoHelpfull: Yes(0) No(1)
Ans is 82 but best approach is to use Binomial Theorem as the units digit is 1 in both cases..
Like (1941^3843)/100 =(41^3843)?100
=(1+40)^3843 is divided by 100
=nC0 x1^3843x40^0+nC1x1^3842x40^1+....(Other terms will be having atleast 2 Zero so neglect them)
=(1+3843x40) divided by 100 as nC1=n and nC0=1
=..21/100 (interested in only last two)
=21 remainder similarly find other terms remainder..
9 years agoHelpfull: Yes(0) No(0)

If XY
denotes X is raised to the
power Y, Find the last two digits
of 19413843+ 19614181
a)12 b)22
c)42 d)82