Aptitude - Numbers - Discussion

Remember, all digits at the power of 5 gives the value having unit place containing the original digit. Special digits (0,1,5,6) raised to any power always give the value having unit place containing the original digit.

In this, 1793/5= ### and remainder is 3,
Unit value of 6374 is to be raised to power of 3 = 4^3= Unit value is 4.

Similarly 317/5= ### and remainder is 2,
Unit value of 625 is to be raised to power of 2 = 5^2= Unit value is 5.

Last factor comes in special digit i.e. 1 and will always give 1 at unit place,if raised to any power.

Hence, the unit place of the result will be 4x5x1= 20; 0 at unit place.

To find unit digit in a large powered number, simply consider its last digit only. for suppose in (6374)^1793 take 4^1793 only. now note that any odd power of 4 gives a number with its unit digit as 4.so unit digit in (6374)^1793 is '4'.and then any power of 5 gives a number with 5 as its unit digit.so u.d in (625)^317 is '5'.then any power of 1 gives only 1 itself. so u.d in 341^491 is '1'. finally multiply all numbers i.e., 4*5*1 which gives 20 with u.d as 0.hence that result.

The unit digit is simply another term for the ones place..
the unit digit in 2343546 is 6
the unit digit in 234 is 4.. etc :)

What is unit digit ?

How can you divide 1793 simply by 4 and write that remainder? is it applicable for any number like 6^, 7^.

The process generally is like this

Say 4^0=1 5^0=1 1^0=1 (1 time)
4^1=4 5^1=5 1^1=1 (2nd time)
4^2=16 5^2=25 1^2=1 (3rd time)
4^3=64 5^3=625 1^4=1 (4th time)

So as per problem units number must be the result of
(4^1793)*(5^317)*(1^491)

From the above calculations done it is evident that
1^anything =1 and 5^any positive number gives 5 as units digit
Similarly with 4 powers

Now take 4^1793.......First divide the power 1793/4 gives remainder 1.

So Units digit is (4^1)*5*1=20 whose
units digit is 0

Find the unit digit in (264)^102 + (264)^103

Find the unit digit in (264)^102 + (264)^102.

Thank you seema. Got it.

What is Unit digit ?