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17^27 has a units digit of:

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by Jay@ManhattanReview » Wed Jan 09, 2019 10:04 pm
BTGmoderatorDC wrote:17^27 has a units digit of:

(a) 1
(b) 2
(c) 3
(d) 7
(e) 9

OA C

Source: Manhattan Prep
Let's understand the power cycle of 7.

"¢ 7^1 = 7; unit digit = 7;
"¢ 7^2 = 49; unit digit = 9;
"¢ 7^3 = 343; unit digit = 3;
"¢ 7^4 = ...1; unit digit = 1;

"¢ 7^5 = ...7; unit digit = 7;

We see that units digit of the power of 7 repeats after every 4 cycle.

Let's see 17^27 now.

17^27 = 17^(24 + 3) = 17^(6*4 + 3)

Note that units digit of 17^27 will be the same that of 17^3 since 6*4 has a cycle of 4.

Again, the units digit of 17^3 will have the same units digit as has 7^3.

Now 7^3 = 343; thus, the units digit of 17^27 = 3.

The correct answer: C

Hope this helps!

-Jay
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by Brent@GMATPrepNow » Thu Jan 10, 2019 6:38 am
BTGmoderatorDC wrote:17^27 has a units digit of:

(a) 1
(b) 2
(c) 3
(d) 7
(e) 9
Look for a pattern

17^1 = 17
17^2 = (17)(17) = ---9 [aside: we need not determine the other digits. All we care about is the units digit]
17^3 = (17)(17^2) = (17)(---9) = ----3
17^4 = (17)(17^3) = (17)(---3) = ----1
17^5 = (17)(17^4) = (17)(---1) = ----7

NOTICE that we're back to where we started.
17^5 has units digit 7, and 17^1 has units digit 7
So, at this point, our pattern of units digits keep repeating 7, 9, 3, 1, 7, 9, 3, 1, . . .
We say that we have a "cycle" of 4, which means the digits repeat every 4 powers.

So, we get:
17^1 = --7
17^2 = ---9
17^3 = ----3
17^4 = ----1
17^5 = ----7
17^6 = ---9
17^7 = ----3
17^8 = ----1
17^9 = ----7
17^10 = ----9
etc.

Notice that when the exponent is a MULTIPLE of 4 (4, 8, 12, 16, ...), the units digit will be 1
Since 24 is a MULTIPLE of 4, we know that the units digit of 17^24 will be 1
Continuing on, we get:
17^25 = ----7
17^26 = ---9
17^27 = ----3

Answer: C

Here's an article I wrote on this topic (with additional practice questions): https://www.gmatprepnow.com/articles/un ... big-powers

Cheers,
Brent
Brent Hanneson - Creator of GMATPrepNow.com
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by fskilnik@GMATH » Thu Jan 10, 2019 8:59 am
BTGmoderatorDC wrote:17^27 has a units digit of:

(a) 1
(b) 2
(c) 3
(d) 7
(e) 9
Source: Manhattan Prep
$$? = \left\langle {{{17}^{27}}} \right\rangle = \left\langle {{7^{27}}} \right\rangle $$
$$\left\langle {{7^4}} \right\rangle = \left\langle {{7^2} \cdot {7^2}} \right\rangle = \left\langle {\left\langle {{7^2}} \right\rangle \cdot \left\langle {{7^2}} \right\rangle } \right\rangle = 1$$
$$\left\langle {{7^{24}}} \right\rangle = \left\langle {{7^4} \cdot {7^4} \cdot \ldots \cdot {7^4}} \right\rangle = {\left\langle {{7^4}} \right\rangle ^6} = 1$$
$$? = \left\langle {{7^{24}} \cdot {7^3}} \right\rangle = \left\langle {{7^{24}}} \right\rangle \cdot \left\langle {{7^3}} \right\rangle = 1 \cdot 3 = 3$$

This solution follows the notations and rationale taught in the GMATH method.

Regards,
Fabio.
Fabio Skilnik :: GMATH method creator ( Math for the GMAT)
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Portuguese-speakers :: https://www.gmath.com.br
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by Scott@TargetTestPrep » Mon Jan 21, 2019 5:42 pm
BTGmoderatorDC wrote:17^27 has a units digit of:

(a) 1
(b) 2
(c) 3
(d) 7
(e) 9

Since we only care about units digits, we can rewrite the expression as:

7^27

Now we can evaluate the pattern of the units digits of 7^n for positive integer values of n. That is, let's look at the pattern of the units digits of powers of 7. When writing out the pattern, notice that we are ONLY concerned with the units digit of 7 raised to each power.

7^1 = 7

7^2 = 9

7^3 = 3

7^4 = 1

7^5 = 7

The pattern of the units digit of powers of 7 repeats every 4 exponents. The pattern is 7-9-3-1. In this pattern, all positive exponents that are multiples of 4 will produce a 1 as its units digit. Thus:

7^28 has a units digit of 1, and 7^27 has a units digit of 3.

Answer: C

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