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Bunuel
If x and y are positive integers, what is the remainder when 5^x is divided by y?

(1) x is an even integer.
(2) y = 3.

Kudos for a correct solution.

My Attempt :

Given X,Y > 0
Unit digit of 5^X will always be 5. So Remainder of (5^x)/y will depend on Y
Say for 5^2/5 will give remainder as 0, where as 5^2/3 will give remainder as 1

Statement 1 :
x is an even integer.
Say for 5^2/5 will give remainder as 0, where as 5^2/3 will give remainder as 1
Not sufficient

Statement 2:
Clearly sufficient to give the reminder value.
Example :
5^2 / 3 leaves remainder = 1
5^3/3 leaves remainder = 1
Statement 2 is sufficient to answer


Answer Option B
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vishwaprakash
Bunuel
If x and y are positive integers, what is the remainder when 5^x is divided by y?

(1) x is an even integer.
(2) y = 3.

Kudos for a correct solution.

My Attempt :

Given X,Y > 0
Unit digit of 5^X will always be 5. So Remainder of (5^x)/y will depend on Y
Say for 5^2/5 will give remainder as 0, where as 5^2/3 will give remainder as 1

Statement 1 :
x is an even integer.
Say for 5^2/5 will give remainder as 0, where as 5^2/3 will give remainder as 1
Not sufficient

Statement 2:
Clearly sufficient to give the reminder value.
Example :
5^2 / 3 leaves remainder = 1
5^3/3 leaves remainder = 1
Statement 2 is sufficient to answer


Answer Option B


Hi vishwprakash,

You seem to have made a mistake here. Check the highlighted part

5^3/3 leaves remainder = 2 whereas you have taken it as 1

So this statement will NOT be Sufficient as it leads to inconsistent values of Remainders.

I hope It clears you mistake! :)
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clearly any statement alone is not enough , as it gives info about only 1 variable at a time

so combining 1 and 2 ,
5^2 / 3 = 25/3 so reminder 1
5^4 /3 = 125/3 so reminder 1. and so on.

So both statement toghether are enough

Answer = C
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Bunuel
If x and y are positive integers, what is the remainder when 5^x is divided by y?

(1) x is an even integer.
(2) y = 3.

Kudos for a correct solution.

5^x will be a multiple of 5, and so it will have a units digit of 5.

1: tells us nothing about y. Insufficient.
2: 5/3 = 1 R 2. 25 /3 = 8 R 2. So insufficient.

Together: you can do examples and see that 5^2=25 = 3*8+1, 5^4 = 625 = 208*3+1.
But algebraically:
5 = 3 + 2. so 5*5 = (3+2)(3+2) = 3*3+2*2*3+2*2=3*(3+2*2)+2*2. 2*2 = 4 so it is really 3*(3+2*2+1)+1. This can be simplified to 3*I+1, where I is some integer.
Because 5^even will be a multiple of 5^2, it is basically (3I+1)^some exponent. (3I+1)^2=(3I)^2+2*3*I+1^2, but again (3I)^2+2*3*I = 3*some integer. So the remainder will be 1 again.
Therefore, (3I+1)*(3I+1)=3I+1, so (3I+1)^any exponent will be 3I+1, so it will leave a remainder of 1. So sufficient.
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If x and y are positive integers, what is the remainder when 5^x is divided by y?

(1) x is an even integer.
(2) y = 3.

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MANHATTAN GMAT OFFICIAL SOLUTION:

(1) INSUFFICIENT: 5^(even integer means) that 5^x = 25, 625, 15,625, etc. All of these numbers end in 25. If y = 5, then the remainder equals 0. If y = 4, then the remainder is 1. Therefore we cannot determine the answer just by knowing this pattern of x.

(2) INSUFFICIENT: Let's test some different values for x:


The pattern is clear: when 5 is raised to an odd power, the remainder is 1, but when 5 is raised to an even power, the remainder is 2. However, with only Statement 2, we don't know whether x is even or odd.

Combining the two statements, we know the pattern for the remainder when 5^x is divided by 3, and we know which term in that pattern applies. When 5^(even integer) is divided by 3, the remainder is always 1.

Proving this theoretically is not trivial, but we don't need to do a theoretical proof.

The correct answer is (C): BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.

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