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If x and y are positive integers, what is the remainder when 5^x is [#permalink]
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17 Jun 2015, 06:02
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Re: If x and y are positive integers, what is the remainder when 5^x is [#permalink]
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17 Jun 2015, 07:20
Bunuel wrote: 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. Question : what is the remainder when 5^x is divided by y?Statement 1: x is an even integerNo information about y Hence, NOT SUFFICIENTStatement 2: y = 3@x=1, 5^1 when divided by 3 gives remainder = 2 @x=2, 5^2 when divided by 3 gives remainder = 1 Hence, NOT SUFFICIENTCombining the two statementsRemainder [5^x divided by 3] = Remainder [(61)^x divided by 3] = Remainder [(1)^x divided by 3] For x to be even (1)^even = +1 hence Remainder = +1 e.g. @x=2, 5^2 when divided by 3 gives remainder = 1 @x=4, 5^4 when divided by 3 gives remainder = 1 Answer: Option
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If x and y are positive integers, what is the remainder when 5^x is [#permalink]
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17 Jun 2015, 08:18
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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. Kudos for a correct solution.[/quote] Solution  For x>=1, 5^x is 5, 25, 125, 625 ...... Stmt1  If x is even integer, then 5^x is 25, 625 .... We do not know the value of y. In sufficient. Stmt2  For y=3, remainders will repeat in 2, 1, 2, 1 .....Remainder is varying. In Sufficient. Stmt1+ Stmt2  x is even and y=3, the remainders are 1, 1, 1, ....... Sufficient. Thanks Kudos Please.
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Last edited by balamoon on 17 Jun 2015, 10:38, edited 3 times in total.



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Re: If x and y are positive integers, what is the remainder when 5^x is [#permalink]
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17 Jun 2015, 09:14
Bunuel wrote: 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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Re: If x and y are positive integers, what is the remainder when 5^x is [#permalink]
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17 Jun 2015, 09:23
vishwaprakash wrote: Bunuel wrote: 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 = 1Statement 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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Re: If x and y are positive integers, what is the remainder when 5^x is [#permalink]
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17 Jun 2015, 11:21
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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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Re: If x and y are positive integers, what is the remainder when 5^x is [#permalink]
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19 Jun 2015, 00:20
Bunuel wrote: 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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Re: If x and y are positive integers, what is the remainder when 5^x is [#permalink]
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22 Jun 2015, 06:07
Bunuel wrote: 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. 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.Attachment:
20150622_1705.png [ 16.06 KiB  Viewed 1697 times ]
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