If n and a are positive integers, what is the units digit of n^4a+2 - n^8a?
(1) n = 3
(2) a is odd.
Please assist with above problem.
(1) n = 3
(2) a is odd.
Please assist with above problem.
For statement 1, you'll want to establish the pattern for the units digit with base 3.alanforde800Maximus wrote:If n and a are positive integers, what is the units digit of n^(4a+2) - n^(8a)?
(1) n = 3
(2) a is odd.
Please assist with above problem.
3^1 = 3
3^2 = 9
3^3 = 27
3^4 = 81
3^5 = units 3
3^6 = units 9
3^7 = units 7
3^8 = units 1
3^9 = units 3
3^10 = units 9
3^11 = units 7
3^12 = units 1
So notice the pattern. Anytime the exponent is a multiple of 4, the units digit will be 1. Increase the exponent by one, and the units digit will be 3, etc.
Let's look at the expression n^(4a+2) - n^(8a) one piece at a time.
First: n^(4a+2). Well, we know that n^(4a) will have a units digit of 1, as 4a would have to be a multiple of 4. (If a = 1, 4a = 4; if a = 2, 4a = 8, etc.) It follows that n^(4a + 1) would have a units digit of 3, and that n^(4a +2) will have a units digit of 9. So the first term has a units digit of 9.
Next: n^(8a). Well 8a is a multiple of 4, so we already know that the units digit for the second term will be 1.
(And we also can see that if the base is 3, the first term will be smaller than the second term. If a = 1, we have 3^6 - 3^8; If a = 2, we have 3^10 - 3^16, etc.) If we know that the first term ends in 9 and that the second term ends in 1, and we know that the second term is larger, we can find the units digit of the difference. (Think 9 - 11, or 19 - 31. The units digit is always the same.) This statement alone is sufficient.
Statement 2: Say a =1 and n = 1. n^(4a+2) - n^(8a) = 1^6 - 1^8 = 0, and 0 is our units digit.
Say a = 1 and n = 3. n^(4a+2) - n^(8a) = 3^6 - 3^8. We already know from our analysis in S1 that 3^6 has a units digit of 9 and 3^8 has a units digit of 1. 3^6 - 3^8 will have a units digit of 2. Because we get different results, this statement alone is not sufficient.
Answer is A
