12 Days of Christmas GMAT Competition - Day 5: x and y are positive in

Question

12 Days of Christmas GMAT Competition with Lots of Fun

x and y are positive integers and the sum of the digits of x*y is 1. If x and y are not multiples of 10, which of the following cannot be the remainder when |x - y| is divided by 10?

I. 2
II. 4
III. 5

(A) I only 
(B) I and II only 
(C) I and III only 
(D) II and III only 
(E) I, II and III

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M37-75

Bunuel · Accepted Answer

Official Solution:

x  and  y  are positive integers and the sum of the digits of  x*y  is 1. If  x  and  y  are not multiples of 10, which of the following cannot be the remainder when  |x - y|  is divided by 10?
 
I. 2 
 
II. 4 
 
III. 5

A. I only
B. I and II only
C. I and III only
D. II and III only
E. I, II and III

The sum of the digits of  x*y  is 1 means that  x*y  can be 10, 100, 1,000, 10,000, ... All these numbers have only 2 and 5 as primes, so those will also be the only primes of  x  and  y . But since we know that  x  and  y  are not multiples of 10, then  x  and  y  cannot have both 2 and 5 as their primes, meaning that  x  must have only 2's and  y  must have equal number of 5's, or vise-versa. So,  x=2^n  and  y=5^n , or vise-versa ( n  is a positive integer).
 
The question asks to find which of the options cannot be the remainder when  |x - y|=5^n-2^n  is divided by 10. When a number is divided by 10, the remainder is always the units digit of that number, so we need to find which numbers could be the units digit of  5^n-2^n .
 
The units digit of  5^n  is always 5.
 
The units digit of  2^n  can be 2, 4, 8, or 6.
 
Thus, the units digit of  5^n-2^n  can be  ...5-...2=...3 ,  ...5-...4=...1 ,  ...5-...8=...7 , or  ...5-...6=...9 .
 
So, the remainder when  |x - y|  is divided by 10 can only be 3, 1, 7, or, 9.

Answer: E

Deepakjhamb · Answer

x and y are positive integers and the sum of the digits of x*y is 1. If x and y are not multiples of 10, which of the following cannot be the remainder when |x - y| is divided by 10?

I. 2
II. 4
III. 5

(A) I only
(B) I and II only
(C) I and III only
(D) II and III only
(E) I, II and III

Solution :

answer is e)

x and y as per given conditions can be multiple of 5 and multiple of 2 excluding even multiples in case of 5 and excluding 5 being factor in case of multiple of 2's

like 25 , 4 ; 125 ,8 ; 625 ; 16 and so on as these give sum of digits of xy =1

now |x-y|/10 , |x-y| can never be a multiple of 5 as one of x and y is multiple of 5 and other one is not multiple of 5 
, also |x-y| can never be multiple of 2 one of them is multiple of 5 and other one is multiple of 2 so we can never get multiple of 2 so that means unit digit place cannot have 0,2,4,6,8 or 5 in it

only possible places in remainder can be 1,3 ,7 and 9 .

so 2, 4 and 5 as remainder when divided by 10 are ruled out

all of these cannot be the remainder

so answer is e )

harvardforsure · Answer

since x and y is not a multiple of 10 but sum of digits of x*y is 1. thus no can be 10,100,1000 etc with 2^1/2/3/4/5 as x and 5^1/2/3/4/5.. as y. since last two digits of 5 powers are always 5 and 2 follows 2,4,8,6 rule, remainders can be 7,9,3,1.

unraveled · Answer

Kudos to  Bunuel  for sharing such a nice questions with those types of layers lad out beautifully.
Although i took too much time, here's my take. Would love to see an easier approach.
Since each of x and y is not a multiple of 10, there multiple(x*y) can only be a multiple of 10(so that the products digits result in a sum of 1). This is possible only if either of x and y is a multiple of 5 and 2.

Note that 5 and 2 can't be multiplied with other as it would invalidate the condition that one of x and y would be then be a multiple of 10, additionally invalidating another condition that sum of digits of x*y would not be 1.

Now,  x*y = 10 = 5*2 
OR 
 x*y = 100 = 5^2*2^2 
 x*y = 1000 = 5^3*2^3 
..
..
so on.

Thus  x = 5^a  and  y = 2^a  or vice-a-versa, as we are only concerned about |x - y|(modulus being more important.
So we can check for remainders that would result from dividing |x - y| by 10 i.e. the last digit of x - y.
Hence, we have a pattern here
 5^a - 2^a = ..b 
 5^1 - 2^1 = ..3 
 5^2 - 2^2 = ..1 
 5^3 - 2^3 = ..7 
 5^4 - 2^4 = ..9 
 5^5 - 2^5 = ..3 
...

We can be sure of this because of the cyclicity of 2 which results in last digits of 2 as 2, 4, 8, 6.

Hence 3, 1, 7 and 9 are only possible remainders.
Hope this is helpful.

Answer E.

callmeDP · Answer

x and y are positive integers and the sum of the digits of x*y is 1. If x and y are not multiples of 10, which of the following cannot be the remainder when |x - y| is divided by 10?

the sum of the digits of x*y is 1 means x*y = 1,10,100,1000

Now, if x and y are not multiple of 10 that means x and y are neither even nor multiple of 5.

Thus, their product can never be equal to 10,100,100

Therefore, none of them can be the remainder.

IMO E

vv65 · Answer

The sum of the digits of x*y will be 1 only if x*y is a power of 10

We know that neither x nor y is a multiple of 10

For x*y=10: (x, y)= (2, 5)
For x*y=100: (x, y) = (4, 25)
For x*y=1000: (x, y) = (8, 125)
And so on...
For x*y=10**n, x=2**n and y=5**n

Since y is always odd, the difference |x-y| will always be odd
The remainder when we divide by 10 will also be odd.
So the remainder cannot be 2 or 4.

y is always a multiple of 5
For the remainder to be 5, x will need to be a multiple of 10 - but that is not possible.

So 2, 4, 5 : none of them can be the remainder
Hence the answer is E

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