Question

Qn: what is the unit digit in {((264)^102)+((264)^103)} ?

ans: ((264)^102)+((264)^103)=((264)^102){1+264}=((264)^102)+265

required unit digit = unit digit in [(4)^102 * 5]
= unit digit in [((4)^4)^25 + 4^2 * 5]
= unit digit in (6*6*5)=0


I really cant get through this logic...how did this happen??

Answer 1

HINT: Unit digit means taking the modulus 10

Answer 2

Also, \( 264^{102}+264^{103}=264^{102}{(1+264)}= 264^{102} \times 265 \)

Answer 3

It's written a little wrong
\[264^{102}+264^{103}\]
\[264^{102}(1+264)\]
\[264^{102} \times 265\]
Now if we have 14 , and we need to find the unit digit in 14^2=196, we know it's 4 but it can also be found out by finding the unit digit of 4^2=>16, it's also 6
so in place of 264, we'll use 4, and for 265 we'll use 5. Th unit digit will be the unit digit of the product
\[4^{102} \times 5\]
4^{102} is an even no. Any even no. multiplied by 5, will have 0 as the unit digit.
So the unit digit is 0

Answer 4

\[ 264^{102} \equiv 4^{102} \equiv 6^{51} \equiv 6 \mod 10 \]

Answer 5

we know it's "6"
Sorry made a typographical error

Answer 6

and \( 265 \equiv 5 \pmod {10} \) combining we get \( 264^{102} \times 265 \equiv 0 \pmod{10} \)

Hence the answer.

Answer 7

all this is right, but i maybe it is easier to see

Answer 8

if we start with
\[264^{102}(1+264)\] and then
\[264^{102}\times 265\] then all we need to know is that
\[264^{102}\] is even

Answer 9

Lolz :D Indeed sat!

Answer 10

not sure why all the other stuff is necessary

Answer 11

Others stuffs are necessary because it's the formal way of doing things but this approach is a neat one :D