12 Days of Christmas GMAT Competition - Day 4: If x and y are prime

Question

12 Days of Christmas GMAT Competition with Lots of Fun

If x and y are prime numbers less than 20, then what is the unit digit of the product of x and y?

(1) The units digit  x*y +y  is a prime number
(2) The Units digits of  x^{(4z+4)} - y  is 5 (z is a positive integer)

Bunuel · Accepted Answer

GMATWhiz Official Explanation: Step 1: Analyse Question Stem: We know x and y are prime numbers that are less than 20. o Possible values of x and y are 2, 3, 5, 7, 11, 13, 17, 19. We need to find the units digit of x*y. Step 2: Analyse Statements Independently (And eliminate options) – AD/BCE Statement 1: The units digit (x*y + y) is a prime number The units digit of (x*y + y) can be 2, 3, 5, or 7. x*y + y = y(x + 1) o So, the units digit of y(x + 1) should be a prime. We can consider some cases to explore this. https://gmatclub.com/forum/download/file.php?id=69674 We can see couple of feasible cases of x and y. o For x = 5, y = 2, the units digit of xy = 0 o For x = 3, y = 3, the units digit of xy = 9 We do not get a unique answer, thus this statement is insufficient. We can eliminate options A and D. Statement 2: The Units digits of (x^(4z+4) - y) is 5 (z is a positive integer) The units digit of x^(4z+4) - y is 5. o We see 5 is an odd number and a difference of two numbers. o Hence, we can either have odd – even = 5 or even – odd = 5. Scenario 1: Odd – Even = 5 o Since both x and y are primes so y must be the even prime 2. o The units of digit x^(4z+4) - 2 = 5  So, the units digit of x^(4z+4) should be 7 o Now let us explore x^(4z+4).  We can rewrite it as x^(4(z+1))  The cyclicity of any number can be 1, 2, or 4. * Thus, we can say 4 is a common cyclicity. This essentially means x^4, x^8, x^12, and so on… will end with either 0, 1, 5, or 6.  For reference you can consider the table. https://gmatclub.com/forum/download/file.php?id=69675 o As we can see that x^(4(z+1)) can never end with 7.  Hence this scenario is not feasible. Scenario 2: Even – Odd = 5 o Since both x and y are primes so x must be the even prime 2. o The units of digit 2^(4z+4) - y = 5.  So, the units digit of 2^(4z+4) should be same as that of 2^4.  That is 6.  Thus, 6 – units digit of y = 5.  Units digit of y = 1, o In the range we have mentioned, y must be 11.  This is a valid scenario such that x = 2 and y = 11. o The units digit of x*y = 2. This statement is sufficient and we can eliminate options C and E. The answer is option B. 2022-12-18_18-05-46.png 2022-12-18_18-06-04.png

Archit3110 · Answer

given that x,y are prime numbers less than 20
possible values of x,y ( 2,3,5,7,11,13,17,19)

target what is the unit digit of the product of x and y

#1
The units digit  x*y +y  is a prime number
y * ( x+1)
to have prime number it has to be odd only
so value of x is '2' and y can be then 5, 11 
unit value of x*y will be 0, 2 
insufficient

#2
The Units digits of  x^{(4z+4)} - y  is 5 (z is a positive integer)
again value of x & y both cannot be odd 
x has to be even because for all odd integers raised to power of 4 the unit digit is '1' i.e. 3,11,17,19 , except '5'
here x will be '2' so if z is 1 then , unit digit of x will always be '6' as its raised to power multiple of '4'

2^(8) ; 256
256 - y ; to have unit digit as 5
y has to be 11
unit digit of x*y will be '2'
sufficient
 
 OPTION B

thisisit25 · Answer

x and y are prime numbers less than 20; x and y are from: 2, 3, 5, 7, 11, 13, 17, 19
Product can have several different unit digits. Let's look at the statements.

(1) The units digit x∗y+y is a prime number

x*y+y = (x+1)*y --> units digit for this is a prime. Possible values: 2, 3, 5, 7
To get 2, 3, 5, or 7 as unit digit, let's quickly test a few combinations to save time:
- 3x4 = 12 (unit digit is prime). x = 3, y = 3
- 2x6 = 12 (unit digit is prime). x=5, y=2

More than one possible values. Not sufficient.

Let's check statement 2.
(2) The Units digits of x^(4z+4)−y is 5 (z is a positive integer)
x^(4z+4) = x^4(z+4) --> fourth cyclical power of a prime.
eg: if x = 2, 2^1 = 2, 2^2 = 4, 2^3 = 8, 2^4 = 16. Last digit repeats every 4 powers. 2,4,8,6,2,4,8,6,...

So for each of the primes, last digit will repeat in a cyclical manner.
2 - 2,4,8, 6 
3 or 13 - 3,9,7, 1 
5 - 5,5,5, 5 
7 or 17 - 7,9,3, 1 
11 - 1,1,1, 1 
19 - 9,1,9, 1

Unique 4th digit variations for x: 6, 1, 5
We have to subtract y from these to get 5 in units place.

last digit of x to the 4k-th power - 5 = units digit of y
6-5 = 1 -->  y would be 11. x has to be 2. 
5-5 = 0 --> not possible as y never ends in a zero (no prime does)
1-5 = 11-5 (take carry over) = 6 --> no prime ends in 6

Only one possible value.
Sufficient.

Answer: B

Note: Attempted this as a part of the Christmas challenge (answers not revealed yet). Let me know if there is an error.

Elite097 · Answer

We need to know the UD of x and y.

1. In order to be a prime number the resulting answer has to be odd and hence one of x or y must be even because only even + odd = odd. Also x must be even since because if y is even the resulting sum will be even. Thus x=2 but y can be 5 or 11 thus NS

2. By cyclicity, we can determine the fourth power of all the primes <20. Since the given difference is 5, the only way to get this difference is if x=2 and y=11 as cyclicity of 2 only gives a 4th power >5 (i.e 6) so in order to get a difference of 5, we need UD of y as 1 which is only possible if y=11. Sufficient

Ans B

Lizaza · Answer

To tackle this, let's start with looking at all the possibilities for our X and Y:
2 | 3 | 5 | 7 | 11 | 13 | 17 | 19
Also, we don't know for sure that the variables are different, so potentially X and Y can be the same number. 
To solve the task, we need to find the last digit of the product XY - and for that, we need to at least know the last digits of both X and Y.
As the general task doesn't leave us much more work to do, let's look at the additional conditions:

Let's transform the formula into looking like a product:
 y*(x+1) 
Now, we know that two numbers - one prime, and another 'prime+1' - have a product that ends in: 2 | 3 | 5 | 7

Let's look into some options of which numbers we can multiply to get the results with these endings:
 _2:    6*2 | 3*4 | 6*7 | 11*2 
 _3:    11*3 | 7*9 
 _5:   5*  any odd number 
 _7:    3*9 | 11*7

Okay, so let's see, what pairs of PRIME * (PRIME+1) (or simply  (X+1)*Y ) fit some of these products:
 6*2 = (5+1)*2
3*4 = (2+1)*3
6*7 = (5+1)*7
11*3 = (2+1)*11  etc.
We already have quite a few different options for  X*Y , so unfortunately by itself  Condition 1 is  INSUFFICIENT .

So this seems a bit more difficult, but let's change the power formula into pure X to the power of Z:
 x^{4z+4} = (x^{4})^{z+1} 
So now, we have  (x^{4})^{z+1} = Y + 5 
Here I suggest we do a bit of good ol' picking the number: let's look at all the possible endings of the numbers for  PRIME + 5 , as well as for  PRIME^4  from our initial list (for now, not thinking about the value of Z). Also, we don't need to look at repeating endings (13 as 3 and 17 as 7), so we exclude them from the list:

X/Y | Y+5 | X^4:
...2 | ....7 | _ 6 
...3 | ....8 | _1
...5 | .._0 | _5
...7 | .._2 | _1
..11 | _ 6  | _1
..19 | _4 | _1

Interestingly, when you put any of the numbers under  x^{4}  into any further fraction (for instance,  Z  or  Z+1 ) - it doesn't matter, because the last digit remains as 1, 5 or 6 regardless of the power.
Now, the only ending of Y+5 in this case which aligns with any of the possible powers of X is  6 , which we get when  Y = 11 , and  X=2 .
Therefore, those two are the only possible values, and the product  X*Y = 22 , so the last digit is  2 . Therefore,  Condition 2 by itself is  SUFFICIENT .

Answer option B.

HarshaBujji · Answer

Statement (1) The units digit  x*y +y  is a prime number

x,y E {2,3,5,7,11,13,17,19}

We need the product of x,y's unit digit.

Unit's digit of y(x+1) should be prime.

If x=odd prime, y=Any.. 13(3+1), 7(5+1) .... this is always 2 which is prime..

13*3 = 39. 7*5=35 Unit's digit is not unique

Hence we can eliminate A,D

Statement (2) The Units digits of  x^{(4z+4)} - y  is 5 (z is a positive integer)

The cyclicity of the given primes is 4{2,3,7},1{5,11}, 2{19};

2^4 =6, 3^4=1, 7^4=1 , 9^2=1 , 5^1 =5 , 11^1 =1

So  x^{(4z+4)}  possible unit's digits are 6,1,5 of all of then only 6 is possible

so x=2, y=11 only possible pair

Either x or y has to be odd. So no other alternate

IMO B

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12 Days of Christmas GMAT Competition - Day 4: If x and y are prime

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