MA wrote:
If x is a positive integer, what is the units digit of (24)^(2x + 1)*(33)^(x + 1)*(17)^(x + 2)*(9)^(2x)?
A. 4
B. 6
C. 7
D. 8
E. 9
Since we are only concerned with the units digit, we can simplify the expression as:
(4)^(2x + 1)*(3)^(x + 1)*(7)^(x + 2)*(9)^(2x)
This simplified expression will have the same units digit as the given expression. Next, let’s look at the units digit patterns of powers of 4, 3, 7, and 9, respectively:
Units digits of powers of 4: 4-6 (the patterns repeats in a cycle of 2 with 4^odd = 4 and 4^even = 6)
Units digits of powers of 3: 3-9-7-1 (the patterns repeats in a cycle of 4 with 3^(a multiple of 4) = 1)
Units digits of powers of 7: 7-9-3-1 (the patterns repeats in a cycle of 4 with 7^(a multiple of 4) = 1)
Units digits of powers of 9: 9-1 (the patterns repeats in a cycle of 2 with 9^odd = 9 and 9^even = 1)
Since 2x + 1 is always odd regardless of what integer x is, 4^(2x + 1) = 4^odd = 4. Similarly, since 2x is always even regardless of what integer x is, 9^(2x) = 9^even = 1. However, since x + 1 (the exponent of 3) and x + 2 (the exponent of 7) are sometimes odd and sometimes even depending on what integer x is, we are going to change tactics in analyzing the units digit of (3)^(x + 1)*(7)^(x + 2). Notice that:
(3)^(x + 1)*(7)^(x + 2) = (3)^(x + 1)*(7)^(x + 1)*7 = (3*7)^(x + 1)*(7) = (21)^(x + 1)*(7)
Since we are only concerned with the units digit, we can simplify (21)^(x + 1)*(7) as (1)^(x + 1)*(7). Since 1 raised to any power is 1, the units digit of (1)^(x + 1)*(7) or (21)^(x + 1)*(7) is 1*7 = 7. With this, we can see that the units digit of (4)^(2x + 1)*[(3)^(x + 1)*(7)^(x + 2)]*(9)^(2x) is 4*[7]*1 = 28, i.e., 8.
Answer: D
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