Find the last digit of A if A = 1+4+4^2+4^3+...4^2017+4^2018
Answers
Answer:
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Step-by-step explanation:
Last digit of 14 is 1.
Last digit of 24 is 6.
Last digit of 3 4 is 1.
Last digit of 4 4 is 6.
Last digit of 5 4 is 5 .
Last digit of 6 4 is 6.
Last digit of 7 4 is 1.
Last digit of 8 4 is 6.
Last digit of 9 4 is 1.
Last digit of 10 4 is 0.
Sum of all the last digits upto 10 is 1+6+1+6+5+6+1+6+1+0= 33.
Now, last digit of 114 or 214 will be the same as the last digit of 14 (you'll understand why if you binomially expand 214 as (20+1)4 ).
Similarly, last digit of 124 or 32 4 will be the same as the last digit of 24 (you'll understand why if you binomially expand 32 4 as (30+2)4 ).
So, the sum of last digits for numbers 1-10, 11-20, 21-30 etc. will be 33.
So, sum of last digits upto 2010 is 201∗33=6633
Now, sum of last digits from 2011 to 2017 is 1+6+1+6+5+6+1=26.
So, total sum of last digits is= 6633+26=6659.
Last digit of the sum is 9.Last digit of 14 is 1.
Last digit of 24 is 6.
Last digit of 3 4 is 1.
Last digit of 4 4 is 6.
Last digit of 5 4 is 5 .
Last digit of 6 4 is 6.
Last digit of 7 4 is 1.
Last digit of 8 4 is 6.
Last digit of 9 4 is 1.
Last digit of 10 4 is 0.
Sum of all the last digits upto 10 is 1+6+1+6+5+6+1+6+1+0= 33.
Now, last digit of 114 or 214 will be the same as the last digit of 14 (you'll understand why if you binomially expand 214 as (20+1)4 ).
Similarly, last digit of 124 or 32 4 will be the same as the last digit of 24 (you'll understand why if you binomially expand 32 4 as (30+2)4 ).
So, the sum of last digits for numbers 1-10, 11-20, 21-30 etc. will be 33.
So, sum of last digits upto 2010 is 201∗33=6633
Now, sum of last digits from 2011 to 2017 is 1+6+1+6+5+6+1=26.
So, total sum of last digits is= 6633+26=6659.
Last digit of the sum is 9.