Latest self Aptitude Question SOLUTION: Q. Find last two digits of 44^70.

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Q. Find last two digits of 44^70.

jayakishore (11 years ago)

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K.ARun  (11 years ago)

44^70=> (2*22)^70
2^70 * 22^70;
(2^10)^7 * (2*11)^70;
24*(2^70)*(11^70);
24*(2^10)^7 * 01;
24*24*01;
576;
so last two digits are 76;

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K.ARun  (11 years ago)

Cyclicity for 2:
2^1=2
2^2=4
2^3=8
2^4=6
2^5=2
so 2,4,8,6,2,4,8,6...
Cyclicity for 3:
3^1=3
3^2=9
3^3=7
3^4=1
3^5=3
so 3,9,7,1,3.....
like these
for 4: 4,6,4,....
for 5: 5,5,5,...
for 6: 6,6,6
for 7: 7,9,3,1,7,9,...
for 8: 8,4,2,6
for 9: 9,1,9,1....

there are four types present
1)ends in one,
2)ends in 3,7,9
3)ends in 4,6,8
4)ends in 2

@jayakishore lets see ends in one:
21^786
let say ab^cde
so what you have to do is a*e and b^e.
in order to find the last two digits
so 2*6=12(take only the last digit i.e. 2)
and 1^6.(Ans:1)
so last two digit of 21^786=21

just take another example
as we know that 11^2=121
so last two digits 21.
jst identify this using our method.
1*2 and 1^2
so 21.

So, is this understandable for you?
if you are ok with this then we can see the remaining three types.

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jayakishore  (11 years ago)

i cant understand how we get 24*(2^70)*(11^70)

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K.ARun  (11 years ago)

@Jayakishore: we have to convert the numbers in terms of cyclicity..
Last two digis of (2^10) is 24 so it is raised to odd power.i.e.24.
and if it is raised to even power means then 76.
but here odd power i.e.raised to 7... so 24*24*01=576;
so 76.
if the given explanation is not enough, let me know.

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