NEW HERE? LEARN MORE
closeclose
Last visit was: 29 Apr 2024, 13:31 It is currently 29 Apr 2024, 13:31
Decision Tracker
My Rewards
New posts
New comers' posts

Close
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
Your Progress

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
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
Go to My Error Log Learn more
Close
Request Expert Reply
Confirm Cancel

12 Days of Christmas GMAT Competition - Day 2: If x, y, and z are pos

SORT BY:
Date
Kudos
Tags:
Find Similar Topics 
Show Tags
Hide Tags
User avatar
L
Bunuel
Math Expert
Joined: 02 Sep 2009
Posts: 93013
Own Kudos [?]: 619979 [8]
Given Kudos: 81634
Send PM
CAT Tests Forum Quiz Mode
12 Days of Christmas GMAT Competition - Day 2: If x, y, and z are pos [#permalink] New post   15 Dec 2022, 07:00
8
Bookmarks
Expert Reply
00:00
A
B
C
D
E

Difficulty:

95% (hard)

Question Stats:

51% (02:42) correct 49% (02:15) wrong based on 153 sessions
Hide Show timer Statistics
12 Days of Christmas GMAT Competition with Lots of Fun

If x, y, and z are positive integers, what is the units’ digit of \(34^{(2x)}*19^{(y+3)} + 17^{(2z)} + 15^{(x-y)}\) ?

(1) \(y^2*z – z\) is an odd integer
(2) \(y*z = x\)



 


This question was provided by GMATWhiz
for the 12 Days of Christmas Competition.

Win $30,000 in prizes: Courses, Tests & more

 

ShowHide Answer
Official Answer
E
Most Helpful Reply
User avatar
L
Bunuel
Math Expert
Joined: 02 Sep 2009
Posts: 93013
Own Kudos [?]: 619979 [3]
Given Kudos: 81634
Send PM
CAT Tests Forum Quiz Mode
12 Days of Christmas GMAT Competition - Day 2: If x, y, and z are pos [#permalink] New post   16 Dec 2022, 07:15
2
Kudos
1
Bookmarks
Expert Reply
Bunuel wrote:
12 Days of Christmas GMAT Competition with Lots of Fun

If x, y, and z are positive integers, what is the units’ digit of \(34^{(2x)}*19^{(y+3)} + 17^{(2z)} + 15^{(x-y)}\) ?

(1) \(y^2*z – z\) is an odd integer
(2) \(y*z = x\)



 


This question was provided by GMATWhiz
for the 12 Days of Christmas Competition.

Win $30,000 in prizes: Courses, Tests & more

 



GMATWhiz Official Explanation:

Step 1: Analyse Question Stem:
  • x, y and z are positive integers
  • We are asked the unit digit of \(34^{(2x)}*19^{(y+3)} + 17^{(2z)} + 15^{(x-y)}\)
  • To get the answer, we have to focus on the unit digit of each element individually
  • And unit digit of each element will depend on power of their unit digit
    \((34^{(2x)})_U*(19^{(y+3)})_U + (17^{(2z)})_U + (15^{(x-y)})_U=(4^{2x})_U*(9^{(y+3)})_U+(7^{2z})_U+(5^{(x-y)})_U\)
  • Cyclicity of 4 is 2 with unit digits 4 and 6
      o Thus \((4^{2x})_U=(4^{even})_U=6\)
         So, we don’t need to worry about x at all
  • Cyclicity of 9 is 2 with unit digits 9 and 1
      o Thus \((9^{(y+3)})_U\) will depend on the even-odd nature of y + 3
         So, we will need the odd-even nature of y

  • Cyclicity of 7 is 4 with unit digits 7, 9, 3 and 1
      o Thus \((7^{2z})_U=(7^{even})_U\) will depend on whether 2z stays a multiple of 2 or multiple of 4 i.e., if x is odd or even respectively
         If z is even, then \((7^{2z} )_U=1\)
         If z is odd, then \((7^{2z})_U=9\)
      o So, we need the odd-even nature of z

  • Cyclicity of 5 is 1 with unit digit 5
      o \((5^{(x-y)})_U=5\) if \(x – y > 0\).
      o If, \(x = y\) we will have \((5^{(x-x)})_U=1\).
      o Also, if x < y then the term become a fraction.
         So, we need the relation between x and y


Step 2: Analyse Statements Independently (And eliminate options) – AD/BCE

Statement 1: \(y^2*z – z\) is an odd integer
  • \(y^2*z – z = Odd\).
  • Or, \(z (y^2-1) = Odd\)
      o Now this can be odd when both z and \(y^2-1\) are odd.
      o Thus, z is odd and y is even

  • This statement does not give the relation between x and y
  • Thus, statement 1 alone is not sufficient and we can eliminate options A and D

Statement 2: \(y*z=x\)
  • This statement gives us no information on the even-odd nature of y and z
  • However, we have some information about the nature of x and y
      o x = y if z = 1
      o x > y if z > 1
     Thus x – y can be 0 or a positive integer.
  • Thus, statement 2 alone is also not sufficient


Step 3: Combining the Statements:
  • From statement 1, we got z = odd and y = even
  • Form statement 2, we got x – y = 0 or a positive integer.
  • We still do not have relation between x and y
      o If z = 1 (odd), then x = y
      o However, if z > 1 (odd) then x > y.
  • So even after combining the units digit of \(15^{(x-y)}\) can be 1 or 5.

Thus, we cannot answer the question even after combining statement 1 and 2.

Hence the right answer is Option E.
General Discussion
User avatar
L
Kinshook
GMAT Club Legend
GMAT Club Legend
Joined: 03 Jun 2019
Posts: 5344
Own Kudos [?]: 3974 [1]
Given Kudos: 160
Location: India
Schools: HBS '22 CBS '22 Wharton '22
GMAT 1: 690 Q50 V34
WE:Engineering (Transportation)
Send PM
Reviews Badge
Re: 12 Days of Christmas GMAT Competition - Day 2: If x, y, and z are pos [#permalink] New post   15 Dec 2022, 07:45
1
Kudos
Asked: If x, y, and z are positive integers, what is the units’ digit of \(34^{(2x)}*19^{(y+3)} + 17^{(2z)} + 15^{(x-y)}\) ?

Unit digit of 34^{2x} = 6
Unit digit of 19^{y+3} = 9 when y is even and y+3 is odd
Unit digit of 19^{y+3} = 1 when y is odd and y+3 is even

Unit digit of 17^{2z} = 9 when z is odd
Unit digit of 17^{2z} = 1 when z is even

Unit digit of 15^{x-y} = 1 when x=y
Unit digit of 15^{x-y} = 5 when x is not = y

(1) \(y^2*z – z\) is an odd integer
(y^2-1)z is odd integer
y^2 -1 is odd
y^2 is even;
y is even
z is odd

Unit digit of 34^{2x} = 6
Unit digit of 19^{y+3} = 9
Unit digit of 17^{2z} = 9
Unit digit of 15^{x-y} = 1 when x=y
Unit digit of 15^{x-y} = 5 when x is not = y
The units’ digit of \(34^{(2x)}*19^{(y+3)} + 17^{(2z)} + 15^{(x-y)}\) = 6*9 + 9 + 1 = 4 when x = y
The units’ digit of \(34^{(2x)}*19^{(y+3)} + 17^{(2z)} + 15^{(x-y)}\) = 6*9 + 9 + 5 = 8 when x is not = y
NOT SUFFICIENT

(2) \(y*z = x\)
4 cases when z is not = 1
Case 1: If y is odd and z is odd then x is odd
The units’ digit of \(34^{(2x)}*19^{(y+3)} + 17^{(2z)} + 15^{(x-y)}\) = 6*1 + 9 + 5 = 0
Case 2: If y is odd and z is even then x is even
The units’ digit of \(34^{(2x)}*19^{(y+3)} + 17^{(2z)} + 15^{(x-y)}\) = 6*1 + 1 + 5 = 1
Case 3: If y is even and z is odd then x is even
The units’ digit of \(34^{(2x)}*19^{(y+3)} + 17^{(2z)} + 15^{(x-y)}\) = 6*9 + 9 + 5 = 8
Case 4: If y is even and z is even then x is even
The units’ digit of \(34^{(2x)}*19^{(y+3)} + 17^{(2z)} + 15^{(x-y)}\) = 6*9 + 1 + 5 = 0

z = 1: x = y
Case 1: y is odd; z=1; x - y =0
The units’ digit of \(34^{(2x)}*19^{(y+3)} + 17^{(2z)} + 15^{(x-y)}\) = 6*1 + 9 + 1 = 6
Case 2: y is even; z=1; x - y = 0
The units’ digit of \(34^{(2x)}*19^{(y+3)} + 17^{(2z)} + 15^{(x-y)}\) = 6*9 + 9 + 1 = 4
NOT SUFFICIENT

(1) + (2)
(1) \(y^2*z – z\) is an odd integer
(y^2-1)z is odd integer
y^2 -1 is odd
y^2 is even;
y is even
z is odd
(2) \(y*z = x\)
Case 1: z=1; x = y
The units’ digit of \(34^{(2x)}*19^{(y+3)} + 17^{(2z)} + 15^{(x-y)}\) = 6*9 + 9 + 1 = 4
Case 2: z is odd but not = 1; x is not = y
The units’ digit of \(34^{(2x)}*19^{(y+3)} + 17^{(2z)} + 15^{(x-y)}\) = 6*9 + 9 + 5 = 8
NOT SUFFICIENT

IMO E
User avatar
B
AmanAngrish
Intern
Intern
Joined: 14 Jan 2018
Posts: 31
Own Kudos [?]: 24 [1]
Given Kudos: 51
Location: India
Schools: HBS '26 Stanford '26 Wharton '26 Kellogg '26 Sloan '26 Haas '26 Booth '26
GMAT 1: 600 Q47 V25
GPA: 4
WE:Operations (Other)
Send PM
Re: 12 Days of Christmas GMAT Competition - Day 2: If x, y, and z are pos [#permalink] New post   15 Dec 2022, 08:33
1
Kudos
Quote:
If x, y, and z are positive integers, what is the units’ digit of 34(2x)∗19(y+3)+17(2z)+15(x−y)34(2x)∗19(y+3)+17(2z)+15(x−y) ?

(1) y2∗z–z is an odd integer
(2) y∗z=x


Cyclicity of 4: 2
Possible units places 4, 6

Cyclicity of 9: 2
Possible units places 9,1

Cyclicity of 7: 4
Possible units places 7,9,3,1

Cyclicity of 5: 1
Possible units places 5

From question stem units digit of:
34^(2x) = 4^(2x) = 4^(even no) = 6

19^(y+3) = 9^(y+3) = 9^(odd/even)
if 9^(odd no) = units digit is 9
if 9^(even no) = units digit is 1

17^(2z) = 7^(2z) = 7^(even no)
Therefore units digit can be 9(7^2, 7^6 and so on) or 1(7^4,7^8 and so on)

15^(x-y) = 5^(x-y) = 5^(any no) = units digit is 1

Therefore stem becomes: 6*(9 or 1) + (9 or 1) + 5
From statement A:
(y^2)*z - z is odd
=> (y^2)*z is even/odd and z is odd/even
But if z is even, then (y^2)*z cannot be odd. Neglecting that case we get
y is even and z is odd.
Since y is even; 9^(y+3) = 9^(odd no) = Units digit is 9
z is odd
=> 7*(2z) can have units digit 9 or 1.
Therefore statement A is insufficient

From statement B:
yz = x
=> Does not provide necessary information. Y and z van be odd/even.
Therefore statement B is insufficient

Combining both the statements also we do not get a response.
IMO E is the answer
avatar
S
UtkarshAnand
Manager
Manager
Joined: 08 May 2021
Posts: 147
Own Kudos [?]: 82 [1]
Given Kudos: 1156
Location: India
Schools: Booth '24 Harvard '24 LBS '24 Stanford '24 Wharton '24
Send PM
CAT Tests
Re: 12 Days of Christmas GMAT Competition - Day 2: If x, y, and z are pos [#permalink] New post   15 Dec 2022, 15:52
1
Kudos
Bunuel wrote:
12 Days of Christmas GMAT Competition with Lots of Fun

If x, y, and z are positive integers, what is the units’ digit of \(34^{(2x)}*19^{(y+3)} + 17^{(2z)} + 15^{(x-y)}\) ?

(1) \(y^2*z – z\) is an odd integer
(2) \(y*z = x\)



 


This question was provided by GMATWhiz
for the 12 Days of Christmas Competition.

Win $30,000 in prizes: Courses, Tests & more

 



Solution:

Analysing the question statement: 34^{(2x)}*19^{(y+3)} + 17^{(2z)} + 15^{(x-y)}

Here, 4,9,7 and 5 raised to their respective powers will decide the units digit of the entire expression.

Cyclicity of 4 is 2. Here, 4 raised to even power will always give the units digit as 6.

Cyclicity of 9 is 2. If y+3 is odd (i.e. y is even) then units digit is 9 otherwise if y+3 is even (i.e. y is odd) the. units digit is 1.

Cyclicity of 7 is 4. But 7 is raised to an even power so the units digit will either be 9 (if z is not a multiple of 4) or 1 (if z is a multiple of 4).

Cyclicity of 5 is 1. But it is raised to power (x-y). If x=y then 5^0 will give units digit as 1 and if x is not equal to y then units digit is 5.

So, things we need to know to solve the question is:
A) Whether y is even or odd.
B) Whether z is a multiple of 4 or not.
C) Whether x=y or not.

Statement 1: y^2*z – z is an odd integer.
On solving,
z(y+1)(y-1) = odd integer.
This implies that z, y+1, y-1 are odd.
So, we get z as odd and y as even. We got answer to A and B but not C.
So statement 1 is insufficient.

Statement 2: y*z = x.
Case 1: If z=1 then x=y.
Case 2: If z is not equal to 1 then x is not equal to y.
We do not get a clear answer for A, B or C.
So statement 2 is insufficient.

Combining statement 1 and 2, we get: z as odd and y as even. We have answers for A and B but not for C as again z could be or could be not equal to 1 which will determine C i.e. x=y or not. Hence, clearly insufficient.

ANSWER E.

Posted from my mobile device
User avatar
bumpbot
Non-Human User
Joined: 09 Sep 2013
Posts: 32740
Own Kudos [?]: 823 [0]
Given Kudos: 0
Send PM
Re: 12 Days of Christmas GMAT Competition - Day 2: If x, y, and z are pos [#permalink] New post   15 Apr 2024, 23:35
Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
GMAT Club Bot
gmatclubot
Re: 12 Days of Christmas GMAT Competition - Day 2: If x, y, and z are pos [#permalink]
15 Apr 2024, 23:35
Moderator:
Bunuel
Math Expert
93014 posts

Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne
Main navigation
GMAT Resources
Partners
MBA Resources
www.gmac.com|www.mba.com | | GMAT Club Rules | Terms and Conditions