If x is a positive integer, what is the remainder when x is divided by

Question

If x is a positive integer, what is the remainder when x is divided by 5?

(1) x^2 has a remainder of 4 when divided by 5
(2) x^3 has a remainder of 2 when divided by 5

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firas92 · Accepted Answer

Since this is a division by 5, we only need to know the units digit to determine the remainder.

(1) x^2 has a remainder of 4 when divided by 5

For a remainder of 4, the units digit of x^2 must be 4 or 9

So x can be any integer with units digit 2, 3, 7 or 8 and so the remainder when x is divided by 5 can be 2 or 3

1 is insufficient

(2) x^3 has a remainder of 2 when divided by 5

x^3 must have units digit 2 or 7

So, x can be any integer with units digit 8 or 3

The remainder in both cases is 3

2 is sufficient

Answer is  (B)

Archit3110 · Answer

#1 x^2 has a remainder of 4 when divided by 5
x=2,7 x^2 ; remainder is 4 and x /5 is 2
x=8 , remainder is 3 insufficient
#2
x^3 has a remainder of 2 when divided by 5
x=3,13,23
so x divided by 5 ; remainder is 3
IMO B sufficient

OC2910 · Answer

If x is a positive integer, what is the remainder when x is divided by 5?

we need a unique remainder value

(1) x^2 has a remainder of 4 when divided by 5 :we know x^2 has remainder 4 so min x^2 can be 4,9,64,49.....
from this x can be 2 or 3 etc : remainder 2/5 = 2 and 3/5 = 3 so  Insufficient

(2) x^3 has a remainder of 2 when divided by 5 : we know x^3 has remainder 2 when divided by 5 so units digit of the cube ends in either 2 or 7 : multiple of 5 ends in 0 and 5 so remainder of 2 can be obtained if 
number is A0+ 2 or A5+2 
having inferred that the cubes will end in 7 or 2, we see x can be 3,8,13,18.... 
diving each of them by 5 given remainder of 3 a unique answer :  sufficient  
 (B)

CrackverbalGMAT · Answer

If x is a positive integer, what is the remainder when x is divided by 5?

(1)   x^2  has a remainder of 4 when divided by 5

We can conclude that the last digit of  x^2  should be either 4 or 9 , then only you will get a remainder of 4 when divided by 5.
Therefore, the last digit of x can be 2,3,7 or 8.

Case 1:  If the last digit of X is 2 or 7, then remainder when divided by 5 will be 2.
Example: 12 , 27 , 147 . When you divide these numbers by 5, the remainder is 2.

Case 2:  If the last digit of X is 3 or 8, then remainder when divided by 5 will be 3. 
Example: 13 , 28 , 148 . When you divide these numbers by 5, the remainder is 3.

As per the points mentioned above, when X is divided by 5, two remainders are possible i.e. 2 or 3.
Hence Statement 1 alone is insufficient.
 
(2)   x^3  has a remainder of 2 when divided by 5

That means the last digit of  x^3  should be 2 or 7.From this, we can conclude that the last digit of X should be 3 or 8.

When you take a cube of a number ending with 3, the last digit of the cube will be 7
Example:  13^3  = 13*13*13 The last digit will be same as last digit of (3*3*3) i.e. 7

Similarly when you take a cube of a number ending with 8, the last digit of the cube will be 2.

Case 1 : Last digit of X is 3
In this case, the remainder when divided by 5 will be 3

Case 2 : Last digit of X is 8
In this case also , the remainder when divided by 5 will also be 3

So we can conclude that in both cases , the remainder is 3. Hence Statement 2 alone is sufficient.

Option B is the answer.

Thanks,
  Clifin J Francis
GMAT SME

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If x is a positive integer, what is the remainder when x is divided by

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If x is a positive integer, what is the remainder when x is divided by

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