Problem Solving

If n is a positive integer, what is the remainder when 3^(8n+3)+2 is divided by 5?

A.0
B.1
C.2
D.3
E.4






ANSWER IS....


(E)

Consider n=1
so, 3^11+2=(3^4)^2*3^3+2
so, last digits 1*7+2
=9 Now divide by 5, remainder=4
thats the answer E

By Eulers theorem 3^4 mod 5 = 1 .

so 3^8n mod 5 = 1

3^(8n+3) + 2 mod 5 = 3^8n mod5 * 27 mod5 + 2 mod5

= 1*2 + 2 = 4

Neo

3 taken to a crazy power is one of GMAC favorites. The beauty of 3 is that it has a repeating pattern of the units digit: 3, 9, 7, 1, 3, 9....ad infinitum.

Therefore, if you write it out:

3^1 = 3

3^2 = 9

3^3 = 27 (units digit is 7)

3^4 = n1

3^5 = n3

...
...
3^11 = n7

7 + 2 = 9 and 9/5 (as has already been shown above) = remainder of 4.

There is a temptation to do a long division here, which will give you a 1.8, so you end up without a remainder since. However, do grade 3 long division where decimal places do not yet exist. So, you end up with a remainder of 4.

Ummmm.....

While I thank AleksandrM for his super handy note about the trendy "Powers of 3", I'm curious as to why you guys decided to settle on a situation where n =1.

What if n = 2?

Oh - nevermind. I see that when n = 2, the exponent is 19 and this also results in a ones-digit value of 7.

Add a 2, which equals 9 and et voila, a remainder of 4 all over again.

Without sweating the extra possibilities, n = 3, 4, 5.. etc. I suspect that all results will bear a ones-digit value of 9 each time.

BUT! Is there any (mathematical) way to know for sure??

Answer is in question.
8n+3, every time adding 3 to any multiple. So it will always be 3rd degree
like
n3
n9
n7
n1
and repeat ..so 8n+3 will always come to n7 category....add 2 it will become 9, Hence remainder is 4.