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

 It is currently 18 Nov 2019, 02:19

### 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

# Algebra :: M26-02

Author Message
Director
Joined: 03 Sep 2006
Posts: 623

### Show Tags

02 Jan 2013, 06:53
2
25
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.
GMAT Tutor
Status: Tutor - BrushMyQuant
Joined: 05 Apr 2011
Posts: 622
Location: India
Concentration: Finance, Marketing
Schools: XLRI (A)
GMAT 1: 700 Q51 V31
GPA: 3
WE: Information Technology (Computer Software)

### Show Tags

02 Jan 2013, 07:07
4
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: 305
Location: United States
WE: Corporate Finance (Manufacturing)

### Show Tags

22 Jun 2013, 19:48
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: 130
Concentration: Strategy, Technology
GMAT Date: 08-01-2013
GPA: 3.7
WE: Consulting (Consulting)

### Show Tags

07 Mar 2014, 20:54
This is good question.

if the second number is smaller, then would it be 11-3 = 7?
Math Expert
Joined: 02 Sep 2009
Posts: 59110

### Show Tags

08 Mar 2014, 07:20
2
8
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: 180

### Show Tags

21 Apr 2014, 01:15
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: 59110

### Show Tags

21 Apr 2014, 01:32
2
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: 180

### Show Tags

21 Apr 2014, 01:47
Thanks Bunnel for your quick response.
Re: Algebra :: M26-02   [#permalink] 21 Apr 2014, 01:47
Display posts from previous: Sort by

# Algebra :: M26-02

Moderator: Bunuel