Find all School-related info fast with the new School-Specific MBA Forum

 It is currently 02 Sep 2015, 22:55

### 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

# Events & Promotions

###### Events & Promotions in June
Open Detailed Calendar

# Algebra :: M26-02

Author Message
Director
Joined: 03 Sep 2006
Posts: 884
Followers: 6

Kudos [?]: 340 [1] , given: 33

Algebra :: M26-02 [#permalink]  02 Jan 2013, 05:53
1
KUDOS
6
This post was
BOOKMARKED
What is the units digit of $$(173)^4-1973^{3^2}$$

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

I think the answer should be 8 (1-3 or 11-3=8). Unable to figure out why the official answer is not 8.
Status: Tutor
Joined: 05 Apr 2011
Posts: 536
Location: India
Concentration: Finance, Marketing
Schools: XLRI (A)
GMAT 1: 570 Q49 V19
GMAT 2: 700 Q51 V31
GPA: 3
WE: Information Technology (Computer Software)
Followers: 72

Kudos [?]: 388 [2] , given: 47

Re: Algebra :: M26-02 [#permalink]  02 Jan 2013, 06:07
2
KUDOS
LM wrote:
What is the units digit of $$(173)^4-1973^{3^2}$$

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

I think the answer should be 8 (1-3 or 11-3=8). Unable to figure out why the official answer is not 8.

Because 1973^9 is greater than 173^4 so the difference will be negative... Final answer will be a negative number with units digit (3-1)=2... So answer will be B

Hope it helps!
_________________
Senior Manager
Joined: 15 Sep 2011
Posts: 306
Location: United States
WE: Corporate Finance (Manufacturing)
Followers: 4

Kudos [?]: 169 [0], given: 40

Re: Algebra :: M26-02 [#permalink]  22 Jun 2013, 18:48
LM wrote:
What is the units digit of $$(173)^4-1973^{3^2}$$

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

I think the answer should be 8 (1-3 or 11-3=8). Unable to figure out why the official answer is not 8.

I made the same mistake, but the answer must be 2. One of the traps here is that second term is larger than first term, so the second term will "go on top" in standard subtraction. Think of this way too: if you multiply $$-1$$ through the expression, then the result will also be 2.

I hope this helps
Manager
Joined: 14 Jan 2013
Posts: 156
Concentration: Strategy, Technology
GMAT Date: 08-01-2013
GPA: 3.7
WE: Consulting (Consulting)
Followers: 2

Kudos [?]: 114 [0], given: 29

Re: Algebra :: M26-02 [#permalink]  07 Mar 2014, 19:54
This is good question.

if the second number is smaller, then would it be 11-3 = 7?
_________________

"Where are my Kudos" ............ Good Question = kudos

"Start enjoying all phases" & all Sections

__________________________________________________________________
http://gmatclub.com/forum/collection-of-articles-on-critical-reasoning-159959.html

percentages-700-800-level-questions-130588.html

700-to-800-level-quant-question-with-detail-soluition-143321.html

Math Expert
Joined: 02 Sep 2009
Posts: 29203
Followers: 4743

Kudos [?]: 50178 [0], given: 7537

Re: Algebra :: M26-02 [#permalink]  08 Mar 2014, 06:20
Expert's post
2
This post was
BOOKMARKED
Mountain14 wrote:
This is good question.

if the second number is smaller, then would it be 11-3 = 7?

It would be 11-3=8.

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 that same as that of $$(7^3)^4$$ and the units digit of $$1973^{3^2}$$ is that 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 (cyclicity): 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=9 (last digit is 9)
3. 7^3=3 (last digit is 3)
4. 7^4=1 (last digit is 1)

5. 7^5=7 (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 th units digit of $$7^{12}$$ will be 1 (4th in pattern, as 12 is a multiple of cyclicty 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 then 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.

_________________
Manager
Joined: 10 Mar 2014
Posts: 200
Followers: 1

Kudos [?]: 34 [0], given: 13

Re: Algebra :: M26-02 [#permalink]  21 Apr 2014, 00:15
Bunuel wrote:
Mountain14 wrote:
This is good question.

if the second number is smaller, then would it be 11-3 = 7?

It would be 11-3=8.

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 that same as that of $$(7^3)^4$$ and the units digit of $$1973^{3^2}$$ is that 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 (cyclicity): 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=9 (last digit is 9)
3. 7^3=3 (last digit is 3)
4. 7^4=1 (last digit is 1)

5. 7^5=7 (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 th units digit of $$7^{12}$$ will be 1 (4th in pattern, as 12 is a multiple of cyclicty 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 then 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.

Bunnel,

Could you please explain following statement?

11-13=-2, which gives the final answer that the units digit of (17^3)^4-1973^{3^2} is 2

Thanks
Math Expert
Joined: 02 Sep 2009
Posts: 29203
Followers: 4743

Kudos [?]: 50178 [1] , given: 7537

Re: Algebra :: M26-02 [#permalink]  21 Apr 2014, 00:32
1
KUDOS
Expert's post
PathFinder007 wrote:
Bunuel wrote:
Mountain14 wrote:
This is good question.

if the second number is smaller, then would it be 11-3 = 7?

It would be 11-3=8.

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 that same as that of $$(7^3)^4$$ and the units digit of $$1973^{3^2}$$ is that 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 (cyclicity): 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=9 (last digit is 9)
3. 7^3=3 (last digit is 3)
4. 7^4=1 (last digit is 1)

5. 7^5=7 (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 th units digit of $$7^{12}$$ will be 1 (4th in pattern, as 12 is a multiple of cyclicty 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 then 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.

Bunnel,

Could you please explain following statement?

11-13=-2, which gives the final answer that the units digit of (17^3)^4-1973^{3^2} is 2

Thanks

(positive number ending with 1) - (greater number ending with 3) = (negative number ending with 2)

11-23=-12
31-133=-102
41-123=-82
....

Hope it's clear.
_________________
Manager
Joined: 10 Mar 2014
Posts: 200
Followers: 1

Kudos [?]: 34 [0], given: 13

Re: Algebra :: M26-02 [#permalink]  21 Apr 2014, 00:47
Thanks Bunnel for your quick response.
Re: Algebra :: M26-02   [#permalink] 21 Apr 2014, 00:47
Similar topics Replies Last post
Similar
Topics:
Quantitative :: Problem solving :: Algebra :: D01-44 0 14 Dec 2013, 19:53
Algebra II 1 12 Feb 2012, 10:35
algebra II qn8 3 23 Oct 2009, 17:11
29 payments - algebra (m08q17) 23 27 Dec 2007, 12:50
27 chickens - algebra (m08q13) 30 27 Dec 2007, 11:33
Display posts from previous: Sort by

# Algebra :: M26-02

Moderators: WoundedTiger, Bunuel

 Powered by phpBB © phpBB Group and phpBB SEO Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®.