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17^27 has a units digit of:

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17^27 has a units digit of:

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17^27 has a units digit of:

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

OA C

Source: Manhattan Prep

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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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GMAT/MBA Expert

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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/units-digits-big-powers

Cheers,
Brent

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Brent Hanneson – Creator of GMATPrepNow.com
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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.

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Fabio Skilnik :: GMATH method creator ( Math for the GMAT)
English-speakers :: https://www.gmath.net
Portuguese-speakers :: https://www.gmath.com.br

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