GMAT Question of the Day - Daily to your Mailbox; hard ones only

 It is currently 23 Oct 2019, 05:51

### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

# M26-02

Author Message
TAGS:

### Hide Tags

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

### Show Tags

26 Jul 2018, 06:12
Thanks ! This is one gem of a question
_________________
" The few , the fearless "
Intern
Joined: 13 Dec 2016
Posts: 4

### Show Tags

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
Intern
Joined: 14 Apr 2018
Posts: 4

### Show Tags

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
Math Expert
Joined: 02 Sep 2009
Posts: 58464

### Show Tags

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

_________________
Manager
Joined: 18 Jul 2018
Posts: 52
Location: United Arab Emirates

### Show Tags

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

### Show Tags

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.
_________________
Intern
Joined: 26 Jun 2019
Posts: 22
Location: United States (CA)
GMAT 1: 730 Q49 V40
GPA: 3.95

### Show Tags

21 Jul 2019, 07:23
I think this is a high-quality question and I agree with explanation. Deadly question
Intern
Joined: 24 Feb 2014
Posts: 36
Location: United States (GA)
WE: Information Technology (Computer Software)

### Show Tags

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

Go to page   Previous    1   2   [ 28 posts ]

Display posts from previous: Sort by

# M26-02

Moderators: chetan2u, Bunuel