SOLUTION: If a(k) = 5+5^2+5^3+5^4+...+5^k , for which of the following values of k will a(k) be divisible by 10? F) 37 G) 51 H) 75 J) 88 K) 91

Algebra ->  Permutations -> SOLUTION: If a(k) = 5+5^2+5^3+5^4+...+5^k , for which of the following values of k will a(k) be divisible by 10? F) 37 G) 51 H) 75 J) 88 K) 91      Log On


   

Question 1084381: If a(k) = 5+5^2+5^3+5^4+...+5^k , for which of the following values of k will a(k) be divisible by 10?
F) 37 G) 51 H) 75 J) 88 K) 91
Answer by KMST(5328) About Me  (Show Source):
You can put this solution on YOUR website!
highlight%28%22J+%29+88%22%29

THE FAST WAY TO THE ANSWER:
All the terms are odd numbers, and you can never get an even sum (such as any multie of 10) by adding an odd number of terms.
That eliminates all the choices, except the one above.

THE SHOW-YOUR-WORK EXPLANATION:
That is a geometric series, the sum of consecutive terms of a geometric sequence (or geometric progression, depending on where you study math).
The sum in this problem is the sum of the first k terms of a geometric sequemce
starting with first term green%285%29 , and with a common ratio of red%285%29 .
That is green%285%29%2A%28red%285%29%5Ek-1%29%2F%28red%285%29-1%29=5%2A%28red%285%29%5Ek-1%29%2F4 .
For that to be a multiple of 10, we need red%285%29%5Ek-1%29
to be a multiple of 8.
There is a way to look at numbers grouping them according to divisibity by a fixed number of interest to us.
All numbers divisible by 8 would be said to be congruent with zero modulo 8,
because they are 0 more than a multiple of 8.
For positive numbers, we would say they are congruent with the remainder we get when we divide them by 8.
(That is modular arithmetic, where you add and multiply lime a clock,
except for modulo 8 there would only be numbers 1 through 8 on the clock)
As we go from one power of 5 to the next,
5%5E2 is congruent with 1 modulo 8;
5%5E3=5%2A5%5E2 is congruent with 5%2A1=5 ;
5%5E4=5%2A5%5E3 is congruent with 5%2A5=25,
and also congruent with 1} , because 25 is congruent with 1 .
5%5E5=5%2A5%5E4 is congruent with 5%2A1=5 ;
5%5E6 is congruent with 1,
5%5E7 is congruent with 5 ,
and so on.
So, 5%5Ek is congruent with 1 modo 8 for all even k numbers,
and it is congruent with 5 modulo 8 for all odd k numbers.
That means that 5%5Ek-1 is congruent with 1-1=0 modulo 8,
so it is a multiple of 8, making the sum a multiple of 10.