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The units digit of 7^29
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Updated on: 07 Mar 2023, 03:39
Updated on: 07 Mar 2023, 03:39
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Question Stats:
76% (01:20) correct
23% (01:19) wrong based on 71 sessions
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Quantity A |
Quantity B |
The units digit of \(7^{29}\) |
The units digit of \(3^{27}\) |
A. Quantity A is greater.
B. Quantity B is greater.
C. The two quantities are equal.
D. The relationship cannot be determined from the information given.
Drill 2
Question: 8
Page: 550
Question: 8
Page: 550
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Re: The units digit of 7^29
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29 Mar 2018, 19:05
29 Mar 2018, 19:05
1
Expert Reply
Every number, when risen to powers of itself, will produce a repeating pattern in the units digits. For example, the first 5 powers of 3 are:
3
9
27
81
243
Note that the 5th in the series ends with a 3, just as the first in the series does. If we multiply 243 by 3 again, the next in the series will end in a 9. So 3 has a 4-part cycle: 3, 9, 7, 1, 3, 9, 7, 1, etc. (Note: 13, 173, 333, and any other number ending in 3 will have the same units digit pattern as 3.) So to find the value of quantity B we should figure out where in the cycle 3^27 falls. If it were 3^28, it would go through the 4-part cycle exactly 7 times, and end with a 1. But since 3^27 is the power before that, it must end in a 7.
Let's use the same technique to find the pattern of powers of 7. To get a good score on the GRE, you should know the first 3 powers of 7: 7, 49, 343. But you don't really need to memorize past that. We can quickly figure out the rest of the pattern by just multiplying the last digit of each term by 7. So we should get a pattern as follows:
7
49
343
something ending in 1 (since 3x7 ends in 1)
something ending in 7 (since something ending in 1 times 7 must end in 7.)
Now we're back to where we started, so 7 must also have a 4-part cycle. (Not all numbers have a 4-part cycle. 5 doesn't, for example.) Anyway, using similar logic, we know that 7^28 should end with a 1, but this is the next power up, so it should end with a 7. Thus the answer is C.
3
9
27
81
243
Note that the 5th in the series ends with a 3, just as the first in the series does. If we multiply 243 by 3 again, the next in the series will end in a 9. So 3 has a 4-part cycle: 3, 9, 7, 1, 3, 9, 7, 1, etc. (Note: 13, 173, 333, and any other number ending in 3 will have the same units digit pattern as 3.) So to find the value of quantity B we should figure out where in the cycle 3^27 falls. If it were 3^28, it would go through the 4-part cycle exactly 7 times, and end with a 1. But since 3^27 is the power before that, it must end in a 7.
Let's use the same technique to find the pattern of powers of 7. To get a good score on the GRE, you should know the first 3 powers of 7: 7, 49, 343. But you don't really need to memorize past that. We can quickly figure out the rest of the pattern by just multiplying the last digit of each term by 7. So we should get a pattern as follows:
7
49
343
something ending in 1 (since 3x7 ends in 1)
something ending in 7 (since something ending in 1 times 7 must end in 7.)
Now we're back to where we started, so 7 must also have a 4-part cycle. (Not all numbers have a 4-part cycle. 5 doesn't, for example.) Anyway, using similar logic, we know that 7^28 should end with a 1, but this is the next power up, so it should end with a 7. Thus the answer is C.
Re: The units digit of 7^29
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04 Apr 2018, 16:07
04 Apr 2018, 16:07
1
Expert Reply
Explanation
To find the pattern in each sequence, write out the units digit of the first few terms in the sequence. The pattern for the units digit of powers of 7 is: 7, 9, 3, 1. The pattern for the units digit of powers of 3 is: 3, 9, 7, 1.
For both numbers, 1 repeats as the units digit every 4 powers, so the 4th power will have a units digit of 1, as will the 8th, the 12th, and so on.
Because 28 is a multiple of 4, you know that \(7^{28}\) will have a units digit of 1.
So moving forward one in the pattern, \(7^{29}\) will have a units digit of 7. Similarly, \(3^{28}\) will have a units digit of 1, so moving backward one in the pattern, \(3^{27}\) must have a units digit of 7. The quantities are equal, so the answer is choice (C).
To find the pattern in each sequence, write out the units digit of the first few terms in the sequence. The pattern for the units digit of powers of 7 is: 7, 9, 3, 1. The pattern for the units digit of powers of 3 is: 3, 9, 7, 1.
For both numbers, 1 repeats as the units digit every 4 powers, so the 4th power will have a units digit of 1, as will the 8th, the 12th, and so on.
Because 28 is a multiple of 4, you know that \(7^{28}\) will have a units digit of 1.
So moving forward one in the pattern, \(7^{29}\) will have a units digit of 7. Similarly, \(3^{28}\) will have a units digit of 1, so moving backward one in the pattern, \(3^{27}\) must have a units digit of 7. The quantities are equal, so the answer is choice (C).
