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

Question

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

Bunuel · Accepted Answer

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.

Kinshook · Answer

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

AmanAngrish · Answer

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

UtkarshAnand · Answer

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.

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