Among 200 integers 2^3, 2^6, 2^9, 2^{12},…., 2^{600}, how

[Max@Math Revolution]

[GMAT math practice question]

Among 200 integers 2^3, 2^6, 2^9, 2^{12},...., 2^{600}, how many numbers are there whose unit digit is 4?

A. 20
B. 25
C. 40
D. 50
E. 100

[Max@Math Revolution]

=>

2^3=8
2^6=(2^3)^2 = 8^2 = 64 ~ 4
2^9 = (2^3)^3 = 8^3 = 32 ~ 2
2^{12}=(2^3)^4 = 8^4 =16 ~ 6
............

When we repeat multiplications by 8, the units digits repeats 8, 4, 2, and 6.
2^3, 2^6, 2^9, 2^{12},...., 2^{600} are 8^1, 8^2, 8^3, 8^4,...., 8^{200}.
Among 8^2, 8^6, ..., 8^{198} have the units digit 4.
The number of those terms is (198-2)/4 + 1 = 196/4 + 1 = 49 + 1 = 50.

Therefore, D is the answer.
Answer: D

[Scott@TargetTestPrep]

[GMAT math practice question]

Among 200 integers 2^3, 2^6, 2^9, 2^{12},...., 2^{600}, how many numbers are there whose unit digit is 4?

A. 20
B. 25
C. 40
D. 50
E. 100

Since 2^1 = 2, 2^2 = 4, 2^3 = 8, 2^4 = 16, 2^5 = 32, 2^6 = 64, etc., we see that the powers of 2 will have a units digit of 4 if the exponent is 2 more than a multiple of 4. In the sequence of powers of 2 given, the exponents are multiples of 3. We see that the first power of 2 that will have a units digit of 4 is 2^6, and the subsequent ones will be, exponent-wise, 12 (i.e., the LCM of 3 and 4) more, i.e., 2^18, 2^30, etc. Therefore, the last one is 2^594 and the number of such integers in the sequence is:

(594 - 6)/12 + 1 = 49 + 1 = 50

Answer: D