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(1) \(x−1\) is divisible by \(5\). --> \(x - 1 = 5k\), for any non negative integer k --> \(x^4 - 1 = (x - 1)*(x + 1)*(x^2 + 1) = 5k*(x + 1)*(x^2 + 1)\) --> \(x^4 - 1\) is a multiple of \(5\) --> Sufficient
(2) When \(x^2+1\) is divided by \(5\), the remainder is \(2\). --> \(x^2 + 1 = 5m + 2\), for some non negative integer \(m\) --> \(x^2 - 1 = 5m\) --> \((x - 1)*(x + 1) = 5m\)
So, \(x^4 - 1 = (x^2)^2 - 1 = (x - 1)*(x + 1)*(x^2 + 1)\) --> \(x^4 - 1 = (x^2)^2 - 1 = 5m*(x^2 + 1)\) --> \(x^4 - 1\) is a multiple of \(5\) --> Sufficient
Re: If x is positive integer, is x^4 - 1 divisible by 5?
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22 Nov 2019, 04:33
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When you encounter an expression which is a difference of two exponents, the question is testing you on your ability to apply the algebraic identities that you learnt way back in high school. And if the exponents are both even, there’s no doubt at all which identity the question wants you to apply. It has to be \(a^2 – b^2\) = (a-b) (a+b).
Let’s analyse the question data and the stem. x is a positive integer. We are trying to establish if \(x^4\) – 1 is a multiple of 5. In other words, we are trying to establish if the units digit of \(x^4\) – 1 is 5 or 0.
With this in mind, let us try to simplify the expression given in the question stem. \(x^4\) – 1= \((x^2)^2 – (1)^2\) = (\(x^2\) – 1) (\(x^2\)+1) = (x-1)(x+1)(\(x^2\)+1). If (\(x^4\)-1) has to be divisible by 5, at least one of the brackets of the expression should be divisible by 5. Note that we cannot simplify \(x^2\)+1 any further. If we try to calculate \(a^2+b^2\) from \((a+b)^2\) or \((a-b)^2\), we can’t be sure if it will simplify matters or does the opposite. So, better to retain \(x^2\) + 1 as it is.
From statement I alone, (x-1) is divisible by 5. This means, (x-1) multiplied by any integer will also be a multiple of 5. Since x is a positive integer, (x+1) and (\(x^2\)+1) will also be integers and when multiplied with (x-1), will give us a multiple of 5.
An alternative approach is by taking values. If (x-1) is divisible by 5, it only means that x should be a number with unit digit 6 or 1. As we know, for both of these numbers, the units digit cyclicity is always 1. So, if x ends with 6, \(x^4\) will also end with 6 and (\(x^4\) -1 ) will end with 5. This means (\(x^4\)-1) is divisible by 5. If x ends with 1, \(x^4\) also ends with 1 and (\(x^4\) – 1) is divisible by 5. Whichever way we look at it, we can conclusively say that (\(x^4\)-1) is divisible by 5. Statement I alone is sufficient. The possible answer options are A or D. Answer options B, C and E can be eliminated.
From statement II, \(x^2\) + 1 gives a remainder of 2 when divided by 5. This means, \(x^2\) + 1 can be written as 5k + 2 where k is an integer.
If k = 0, \(x^2\) + 1 = 2 which gives us \(x^2\) = 1 and thus x = 1 {remember, we do not have to worry about x = -1 since the question says that x is a positive integer}. If x = 1, (\(x^4\)-1) = 0, which is divisible by 5.
If k = 1, \(x^2\) + 1 = 7, which gives us \(x^2\) = 6. But \(x^2\) cannot be 6 because \(x^2\) HAS to be a perfect square since x is a positive integer {remember perfect squares are always squares of integers}. This leads us to taking values of k in such a way that \(x^2\) is a perfect square.
But, in doing so, we come to understand that we don’t have to try values for k, randomly. Instead, we need to breakdown the equation further.
\(x^2\) + 1 = 5k + 2 which can be simplified to \(x^2\) = 5k +1.
We know 5k is a multiple of 5 and any multiple of 5 always ends with 5 or 0. This only means that \(x^2\) will end with 6 or 1. This means that x will have 4 or 6 or 1 or 9 as the units digit.
If the units digit of x is 4 or 6, \(x^4\) – 1 will end with 5. This means that \(x^4\) -1 will be divisible by 5.
If the units digit of x is 1 or 9, \(x^4\) – 1 will end with 0. This means that \(x^4\) -1 will be divisible by 5.
Therefore, we can conclude that (\(x^4\)-1) will be divisible by 5 regardless of the value of k. Statement II is sufficient. Answer option A can be eliminated. The correct answer option is D.
In a question on divisibility by 5, it’s a good idea to work with the unit digit concept; this will help you narrow down the options you have to try.
Re: If x is positive integer, is x^4 - 1 divisible by 5?
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22 Nov 2019, 07:39
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If x is positive integer, is \(x^4−1\) divisible by 5? \(x^4−1\) = \((x^2)^2−(1^2)^2\) = \((x^2−1)*(x^2+1)\) = \((x−1)*(x+1)*(x^2+1)\)
So, if any of the factors is divisible by 5, then \(x^4−1\) is divisible by 5
(1) x−1 is divisible by 5. Since \(x-1\) is divisible by 5 \(x^4−1\) is divisible by 5.
SUFFICIENT.
(2) When \(x^2+1\) is divided by 5, the remainder is 2. This implies that \(x^2\) has its units digit either 6 or 1. So, x has its units digits either 4, 6 or 9 since square of numbers having unit digit either of these would only result in either 6 or 1. (4^2 and 6^2 has unit digit as 6 and 9^2 has unit digit as 1) Thus, x can be 4, 14, 6, 16, 9, 19 and so on..
Now, since only 4, 6 or 9 are possible unit digit of x, either of \(x-1\) or \(x+1\) would result in a number that would have unit digit as either 5 or 0. Hence \(x^4−1\) would always be divisible by 5.
SUFFICIENT.
Answer D.
_________________
Ephemeral Epiphany..!
GMATPREP1 590(Q48,V23) March 6, 2019 GMATPREP2 610(Q44,V29) June 10, 2019 GMATPREPSoft1 680(Q48,V35) June 26, 2019
Re: If x is positive integer, is x^4 - 1 divisible by 5?
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23 Nov 2019, 06:56
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We are given that x is a positive integer, and we are to determine if x^4 - 1 is divisible by 5.
Now, we know that x^4 - 1 = (x^2-1)(x^2+1) = (x-1)(x+1)(x^2+1)
Statement 1: (x-1) is divisible by 5 Statement 1 is sufficient, since we know that (x-1) is a factor of (x^4 - 1). Hence x^4-1 is divisible by 5.
Statement 2: When x^2+1 is divided by 5, the remainder is 2. Statement 2 is sufficient. This is because when x=4, x^2+1=17 which leaves a remainder of 2, and x^4-1=255 is divisible by 5. When x=1, x^2+1 =2 which leaves a remainder of 2 when divided by 5, and x^4-1 =0 leaves a remainder of zero, since zero is divisible by 5. Similarly, x=6 also satisfy the condition in statement 2 and x^4-1=1295 which is divisible by 5. It can be inferred that all three numbers that satisfy statement 2 are either 1 less than a multiple of 5 (i.e. 5n-1) or a multiple of 5 +1 (i.e. 5n+1), which given the simplification above will always yield a value of x^4-1 to be a multiple of 5. Hence statement 2 is also sufficient.
Both statements are sufficient on their own to determine whether x^4-1 is divisible by 5.
(1) x−1 is divisible by 5--> so from eq(i) x^4−1 is divisible by 5 -suff (2) When x^2+1 is divided by 5, the remainder is 2. --> x^2+1=5d+2 => x^2+1-2=5d=> x^2-1=5d --> so from eq(ii) x^4−1 is divisible by 5 -suff
Re: If x is positive integer, is x^4 - 1 divisible by 5?
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24 Nov 2019, 04:55
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If x is positive integer, is \(x^4 - 1\) divisible by 5?
\(x^4\) will either leave a remainder 1, when the units digit of x is 1, 2, 3, 4, 6, 7 ,8 or 9, and would be divisible by 5 when when the units digit of x is 5 or 0. All ODD digits except 5 have the units digit 1, when raised to power 4, while all EVEN digits except 0 have the units digit 6, when raised to power 4.
So, the question is whether x is 5 or 0... OR other way.. \(x^4 - 1=(x^2-1)(x^2+1)=(x-1)(x+1)(x^2+1)\)
(1) \(x-1\) is divisible by 5. So x is NOT divisible by 5. hence \(x^4-1\) is divisible by 5 ALSO \(x^4 - 1=(x^2-1)(x^2+1)=(x-1)(x+1)(x^2+1)\), and since x-1 is divisible by 5, x^4-1 is also divisible by 5.
(2) When \(x^2 + 1\) is divided by 5, the remainder is 2. Again x is NOT divisible by 5. hence \(x^4-1\) is divisible by 5 ALSO \(x^4 - 1=(x^2-1)(x^2+1)=(x-1)(x+1)(x^2+1)\), and since x^2+1 is divisible by 5, x^4-1 is also divisible by 5.
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68% (02:10) correct 
