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M26-02

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Senior Manager
Joined: 08 Jun 2015
Posts: 420
Location: India
GMAT 1: 640 Q48 V29
GMAT 2: 700 Q48 V38
GPA: 3.33

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26 Jul 2018, 06:12
Thanks ! This is one gem of a question
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Joined: 13 Dec 2016
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09 Sep 2018, 07:10
The thing to notice here is
17^12 is smaller than 1973^9

split it like this 1973's square root is between 40-50 for sure(appr. 45) so,
1973^9 = (45)^18

Now compare 17^12 and (45)^18
clearly second one is bigger so,
-3+1 = 2 is the unit's digit
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Joined: 14 Apr 2018
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29 Sep 2018, 05:42
I think this is a poor-quality question and I don't agree with the explanation. we can calculate as 11-03 that gives 8 as remainder, how can you say 2 as remainder
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Joined: 02 Sep 2009
Posts: 58464

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29 Sep 2018, 07:07
Biswajitsh wrote:
I think this is a poor-quality question and I don't agree with the explanation. we can calculate as 11-03 that gives 8 as remainder, how can you say 2 as remainder

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11 May 2019, 01:57
Bunuel wrote:
Official Solution:

What is the units digit of $$(17^3)^4-1973^{3^2}$$?

A. 0
B. 2
C. 4
D. 6
E. 8

Must know for the GMAT:

I. The units digit of $$(abc)^n$$ is the same as that of $$c^n$$, which means that the units digit of $$(17^3)^4$$ is the same as that of $$(7^3)^4$$ and the units digit of $$1973^{3^2}$$ is the same as that of $$3^{3^2}$$.

II. If exponentiation is indicated by stacked symbols, the rule is to work from the top down, thus: $$a^{m^n}=a^{(m^n)}$$ and not $${(a^m)}^n$$, which on the other hand equals to $$a^{mn}$$.

So:

$${(a^m)}^n=a^{mn}$$;

$$a^{m^n}=a^{(m^n)}$$.

Thus, $${(7^3)}^4=7^{(3*4)}=7^{12}$$ and $$3^{3^2}=3^{(3^2)}=3^9$$.

III. The units digit of integers in positive integer power repeats in specific pattern (cyclisity): The units digit of 7 and 3 in positive integer power repeats in patterns of 4:

1. $$7^1=7$$ (last digit is 7)

2. $$7^2=49$$ (last digit is 9)

3. $$7^3=xx3$$ (last digit is 3)

4. $$7^4=xxx1$$ (last digit is 1)

5. $$7^5=xxxxxx7$$ (last digit is 7 again!)

...

1. $$3^1=3$$ (last digit is 3)

2. $$3^2=9$$ (last digit is 9)

3. $$3^3=27$$ (last digit is 7)

4. $$3^4=81$$ (last digit is 1)

5. $$3^5=243$$ (last digit is 3 again!)

...

Thus, the units digit of $$7^{12}$$ will be 1 (4th in pattern, as 12 is a multiple of cyclisty number 4) and the units digit of $$3^9$$ will be 3 (first in pattern, as $$9=4*2+1$$).

So, we have that the units digit of $${(17^3)}^4=17^{12}$$ is 1 and the units digit of $$1973^{3^2}=1973^9$$ is 3. Also notice that the second number is much larger than the first one, thus their difference will be negative, something like $$11-13=-2$$, which gives the final answer that the units digit of $${(17^3)}^4-1973^{3^2}$$ is 2.

Hi Bunuel

Can you please explain how is 9 first in the cyclisty of number 3 ??

"the units digit of 3^9 will be 3 (first in pattern, as 9=4*2+1)"

Thanks
Math Expert
Joined: 02 Sep 2009
Posts: 58464

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11 May 2019, 03:27
JIAA wrote:
Bunuel wrote:
Official Solution:

What is the units digit of $$(17^3)^4-1973^{3^2}$$?

A. 0
B. 2
C. 4
D. 6
E. 8

Must know for the GMAT:

I. The units digit of $$(abc)^n$$ is the same as that of $$c^n$$, which means that the units digit of $$(17^3)^4$$ is the same as that of $$(7^3)^4$$ and the units digit of $$1973^{3^2}$$ is the same as that of $$3^{3^2}$$.

II. If exponentiation is indicated by stacked symbols, the rule is to work from the top down, thus: $$a^{m^n}=a^{(m^n)}$$ and not $${(a^m)}^n$$, which on the other hand equals to $$a^{mn}$$.

So:

$${(a^m)}^n=a^{mn}$$;

$$a^{m^n}=a^{(m^n)}$$.

Thus, $${(7^3)}^4=7^{(3*4)}=7^{12}$$ and $$3^{3^2}=3^{(3^2)}=3^9$$.

III. The units digit of integers in positive integer power repeats in specific pattern (cyclisity): The units digit of 7 and 3 in positive integer power repeats in patterns of 4:

1. $$7^1=7$$ (last digit is 7)

2. $$7^2=49$$ (last digit is 9)

3. $$7^3=xx3$$ (last digit is 3)

4. $$7^4=xxx1$$ (last digit is 1)

5. $$7^5=xxxxxx7$$ (last digit is 7 again!)

...

1. $$3^1=3$$ (last digit is 3)

2. $$3^2=9$$ (last digit is 9)

3. $$3^3=27$$ (last digit is 7)

4. $$3^4=81$$ (last digit is 1)

5. $$3^5=243$$ (last digit is 3 again!)

...

Thus, the units digit of $$7^{12}$$ will be 1 (4th in pattern, as 12 is a multiple of cyclisty number 4) and the units digit of $$3^9$$ will be 3 (first in pattern, as $$9=4*2+1$$).

So, we have that the units digit of $${(17^3)}^4=17^{12}$$ is 1 and the units digit of $$1973^{3^2}=1973^9$$ is 3. Also notice that the second number is much larger than the first one, thus their difference will be negative, something like $$11-13=-2$$, which gives the final answer that the units digit of $${(17^3)}^4-1973^{3^2}$$ is 2.

Hi Bunuel

Can you please explain how is 9 first in the cyclisty of number 3 ??

"the units digit of 3^9 will be 3 (first in pattern, as 9=4*2+1)"

Thanks

3^1 = 3
3^2 = 9
3^3 = 27
3^4 = 81
3^5 = ...3
3^6 = ...9

The units digits repeat in blocks of 4: {3, 9, 7, 1}. 3^9 will have the same units digit as 3^9.
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Joined: 26 Jun 2019
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GMAT 1: 730 Q49 V40
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21 Jul 2019, 07:23
I think this is a high-quality question and I agree with explanation. Deadly question
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Joined: 24 Feb 2014
Posts: 36
Location: United States (GA)
WE: Information Technology (Computer Software)

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26 Sep 2019, 22:02
I think this is a high-quality question and I agree with explanation.
Re M26-02   [#permalink] 26 Sep 2019, 22:02

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