What is the units digit of 6^15 - 7^4 - 9^3?
A:8
B:7
C:6
D:5
E:4
This can be easily solve but if the numbers are big, we can take a approach using the cyclicity of a number.
cyclicity of 6 is
6*6 = 36
36 * 6 --->6*6 = 36 ---> take unit digit of 36 i.e 6 and again multiply by 6
So we can see that 6 repeats itself in unit place in every multiplication
therefore cyclicity of 6 is 1, this means 6^n will always give 6 in unit digit
6^15 will have 6 in unit place
similary, cyclicity of 7 can be found as follows
7 * 7 = 49
49 * 7 --->9 *7 = 63
63 * 7 ----> 3*7 =21
21 * 7-----> 1* 7 = 7
7 * 7 this is repetition again
cyclicity of 7 is 4 (7,9,3,1)
7^4 unit digit can be found by dividing power by cyclicity = 4/ 4 will give remainder 0
0 wil point to 1 in (7,9,3,1) as 0 is above 1 . (7--->1, 9--->2, 3--->3, 1--->4,0)
therefore , unit digit is 1
9^3
cyclicity is 2 i.e (9,1)
if I divide 3/2 remainder is 1 which gives 9 as (9--->1, 1--->2,0)
6^15 - 7^4 - 9^3
Now we can have 6 - 1-9 = 6-10
as we are dealing with unit digit 10 will become 0 therefore
6 -10 ---> 6 - 0 = 6