Author 
Message 
TAGS:

Hide Tags

Math Expert
Joined: 02 Sep 2009
Posts: 62464

If x is positive integer, is x^4  1 divisible by 5?
[#permalink]
Show Tags
22 Nov 2019, 01:40
Question Stats:
68% (02:13) correct 32% (01:48) wrong based on 105 sessions
HideShow timer Statistics
Competition Mode Question If x is positive integer, is \(x^4  1\) divisible by 5? (1) \(x1\) is divisible by 5. (2) When \(x^2 + 1\) is divided by 5, the remainder is 2. Are You Up For the Challenge: 700 Level Questions
Official Answer and Stats are available only to registered users. Register/ Login.
_________________



SVP
Joined: 20 Jul 2017
Posts: 1508
Location: India
Concentration: Entrepreneurship, Marketing
WE: Education (Education)

Re: If x is positive integer, is x^4  1 divisible by 5?
[#permalink]
Show Tags
22 Nov 2019, 02:13
\(x^4  1 = (x^2)^2  1 = (x^2  1)*(x^2 + 1) = (x  1)*(x + 1)*(x^2 + 1)\)
(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
IMO Option D



GMAT Club Legend
Joined: 18 Aug 2017
Posts: 6049
Location: India
Concentration: Sustainability, Marketing
GPA: 4
WE: Marketing (Energy and Utilities)

Re: If x is positive integer, is x^4  1 divisible by 5?
[#permalink]
Show Tags
22 Nov 2019, 02:29
#1 x−1 is divisible by 5 x=6,11,16,21... sufficient #2 When x^2+1 is divided by 5, the remainder is 2 x=6,9,11,16,21 sufficient IMO D
If x is positive integer, is x^4−1 divisible by 5?
(1) x−1 is divisible by 5.
(2) When1x2+1 is divided by 5, the remainder is 2.



VP
Joined: 24 Nov 2016
Posts: 1359
Location: United States

Re: If x is positive integer, is x^4  1 divisible by 5?
[#permalink]
Show Tags
22 Nov 2019, 03:22
Quote: If x is positive integer, is x^4−1 divisible by 5?
(1) x−1 is divisible by 5.
(2) When x^2+1 is divided by 5, the remainder is 2. \(x^41…(x^2)^21^2…(x^2+1)(x^21)…(x^2+1)(x+1)(x1)\) (1) x−1 is divisible by 5. sufic\(\frac{(x^2+1)(x+1)(x1)}{5}…\frac{(x^2+1)(x+1)(5)}{5}…(x^2+1)(x+1)=integer\) (2) When x^2+1 is divided by 5, the remainder is 2. sufic\((x^2+1)/5=m5+2=(2,7,12,17,22,27,32,37…)\) \(x^2=(1,6,11,16,21,26,31,36…)\) \(x=integer:x=(1,4,6,14…)\) \(x=1:(x^2+1)(x+1)(x1)…(2)(2)(0)/5=0\) \(x=4:(x^2+1)(x+1)(x1)…(17)(5)(4)/5=17*4\) \(x=6:(x^2+1)(x+1)(x1)…(37)(7)(5)/5=37*7\) \(x=14:(x^2+1)(x+1)(x1)…(197)(15)(13)/5=197*13*3\) Ans (D)



CrackVerbal Quant Expert
Joined: 12 Apr 2019
Posts: 478

Re: If x is positive integer, is x^4  1 divisible by 5?
[#permalink]
Show Tags
22 Nov 2019, 03:33
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 \(a^2 – b^2\) = (ab) (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) = (x1)(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 \((ab)^2\), 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, (x1) is divisible by 5. This means, (x1) 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 (x1), will give us a multiple of 5. An alternative approach is by taking values. If (x1) 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. Hope that helps!
_________________



Senior Manager
Joined: 01 Mar 2019
Posts: 485
Location: India
Concentration: Strategy, Social Entrepreneurship
GPA: 4

Re: If x is positive integer, is x^4  1 divisible by 5?
[#permalink]
Show Tags
22 Nov 2019, 05:14
(1) x−1 is divisible by 5........then x=5n+2 whereN=0,1,2..... i.e x becomes 2,7,12.......
for whom x^41 divisible by 5..............SUfficient
(2) When x2+1 is divided by 5, the remainder is 2.
then x can take values.....1,4,6....
for whom x^41 divisible by 5............SUfficient
OA:D



Director
Joined: 07 Mar 2019
Posts: 910
Location: India
WE: Sales (Energy and Utilities)

Re: If x is positive integer, is x^4  1 divisible by 5?
[#permalink]
Show Tags
22 Nov 2019, 06:39
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 \(x1\) 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 \(x1\) 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



Manager
Joined: 06 May 2019
Posts: 103
Location: India
Concentration: Leadership, Finance
GPA: 4
WE: Information Technology (Computer Software)

Re: If x is positive integer, is x^4  1 divisible by 5?
[#permalink]
Show Tags
22 Nov 2019, 07:52
If x is positive integer, is x4−1 divisible by 5?
(1) x−1 is divisible by 5.
(2) When x2+1 is divided by 5, the remainder is 2.
Hey, Here is explanation.
Correct Answer is D.
1: x1 is divisible by 5.
we knows  x^4 1 = (x1)(x+1)(x2+1) hence x^4 1 is divisible by 5
A is sufficient.
2: When x2+1 is divided by 5, the remainder is 2.
x^41 = (x^2+1)(x^21) given x2+1 is divided by 5, the remainder is 2 , that mean x^2+1 2 = x^21 is divisible by 5.
and x^41 = (x^2+1)(x^21)
so, x^41 is divisible by 5
B is also sufficient
Hence D is answer.



Director
Joined: 25 Jul 2018
Posts: 642

Re: If x is positive integer, is x^4  1 divisible by 5?
[#permalink]
Show Tags
22 Nov 2019, 13:34
x — positive integer Is \(x^{4}—1\) divisible by 5?
(Statement1): (x—1 ) is divisible by 5. —> \(x^{4}—1= (x^{2}+ 1)(x—1)(x+1)\) is divisible by 5
Sufficient
(Statement2): When (\(x^{2}+ 1\))is divided by 5, the remainder is 2.
—> In order the remainder to be equal to 2, the units digit of number should end with 2 or 7.
—>\( x^{2} +1 =...2 —> x^{2}= ...1\) x—positive integer —> x could be ..1 or ...9
—> \(x^{2}+1= ...7 —> x^{2}= ...6\) x —positive integer —> x could be ...4 or ...6
Let’s say that x =21 —> \(x^{4}—1= (x^{2}+ 1)(x—1)(x+1)—> 21^{2}*20*22\) (Yes)
Let’s say that x =24 —> 24^{2}*23*25(Yes) ...... Yes ......Yes
Sufficient
The answer is D
Posted from my mobile device



Manager
Status: Student
Joined: 14 Jul 2019
Posts: 244
Location: United States
Concentration: Accounting, Finance
GPA: 3.9
WE: Education (Accounting)

Re: If x is positive integer, is x^4  1 divisible by 5?
[#permalink]
Show Tags
22 Nov 2019, 14:57
If x is positive integer, is x4−1x4−1 divisible by 5?
(1) x−1x−1 is divisible by 5.
(2) When x2+1x2+1 is divided by 5, the remainder is 2
Simplifying x^4 1, we get = (x^21) (x^2+1) = (x+1)(x1)(x^2+1)
(1) x1 is divisible by 5, so whole expression x^41 will be divisible by 5. Sufficient.
(2) (x^2 +1) = 5x + 2, so x^2 1 = 5x, since x^21 is divisible by 5, x^4 1 will also be divisible by 5. sufficient.
D is the answer.



CR Forum Moderator
Joined: 18 May 2019
Posts: 800

Re: If x is positive integer, is x^4  1 divisible by 5?
[#permalink]
Show Tags
23 Nov 2019, 05:56
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^21)(x^2+1) = (x1)(x+1)(x^2+1)
Statement 1: (x1) is divisible by 5 Statement 1 is sufficient, since we know that (x1) is a factor of (x^4  1). Hence x^41 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^41=255 is divisible by 5. When x=1, x^2+1 =2 which leaves a remainder of 2 when divided by 5, and x^41 =0 leaves a remainder of zero, since zero is divisible by 5. Similarly, x=6 also satisfy the condition in statement 2 and x^41=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. 5n1) or a multiple of 5 +1 (i.e. 5n+1), which given the simplification above will always yield a value of x^41 to be a multiple of 5. Hence statement 2 is also sufficient.
Both statements are sufficient on their own to determine whether x^41 is divisible by 5.
The answer therefore options D.



Director
Joined: 12 Dec 2015
Posts: 506

Re: If x is positive integer, is x^4  1 divisible by 5?
[#permalink]
Show Tags
23 Nov 2019, 06:20
If x is positive integer, is x^4−1 divisible by 5?
(1) x−1 is divisible by 5.
(2) When x^2+1 is divided by 5, the remainder is 2.
Answer: D
x^4−1 = (x^2+1)(x^21) eq(i) ' = (x^2+1)(x+1)(x1) eq(ii)
(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+12=5d=> x^21=5d > so from eq(ii) x^4−1 is divisible by 5 suff



Math Expert
Joined: 02 Aug 2009
Posts: 8300

Re: If x is positive integer, is x^4  1 divisible by 5?
[#permalink]
Show Tags
24 Nov 2019, 03:55
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.. \(x^4  1=(x^21)(x^2+1)=(x1)(x+1)(x^2+1)\) (1) \(x1\) is divisible by 5. So x is NOT divisible by 5. hence \(x^41\) is divisible by 5 ALSO \(x^4  1=(x^21)(x^2+1)=(x1)(x+1)(x^2+1)\), and since x1 is divisible by 5, x^41 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 \(x^41\) is divisible by 5 ALSO \(x^4  1=(x^21)(x^2+1)=(x1)(x+1)(x^2+1)\), and since x^2+1 is divisible by 5, x^41 is also divisible by 5. D
_________________



Intern
Joined: 12 Jan 2018
Posts: 38
Location: India
Concentration: General Management, Marketing
WE: Engineering (Energy and Utilities)

Re: If x is positive integer, is x^4  1 divisible by 5?
[#permalink]
Show Tags
24 Nov 2019, 05:09
The answer is D.
Starting with 1, substitute and find answers



Manager
Joined: 31 Oct 2015
Posts: 95

Re: If x is positive integer, is x^4  1 divisible by 5?
[#permalink]
Show Tags
24 Nov 2019, 10:16
Is \(x^41\) divisible by 5?
lets factorise the question =\((x^21)(x^2+1)\) =(x1)(x+1)\((x^2+1)\)
If any of the factors is divisible by 5, then the term \(x^41\) is divisble by 5.
(1) x−1 is divisible by 5. Therefore sufficient
(2) When \(x^2+1\) is divided by 5, the remainder is 2. This means \(x^21\) is divisible by 5. Therefore sufficient.
D is the answer IMO



CEO
Joined: 03 Jun 2019
Posts: 2506
Location: India
GMAT 1: 690 Q50 V34
WE: Engineering (Transportation)

Re: If x is positive integer, is x^4  1 divisible by 5?
[#permalink]
Show Tags
29 Mar 2020, 22:29
Bunuel wrote: Competition Mode Question If x is positive integer, is \(x^4  1\) divisible by 5? (1) \(x1\) is divisible by 5. (2) When \(x^2 + 1\) is divided by 5, the remainder is 2. Are You Up For the Challenge: 700 Level QuestionsAsked: If x is positive integer, is \(x^4  1\) divisible by 5? x^41 = (x^2+1)(x+1)(x1) (1) \(x1\) is divisible by 5. Since (x1) is factor of x^41 = (x^2+1)(x+1)(x1) If x1 is divisible by 5 , x^41 = (x^2+1)(x+1)(x1) is also divisible by 5 SUFFICIENT (2) When \(x^2 + 1\) is divided by 5, the remainder is 2. x^2 + 1 mod 5 = 2 x^2  1 mod 5 = 0 Since (x^21) is factor of x^41 = (x^2+1)(x^21) If x^21 is divisible by 5 , x^41 = (x^2+1)(x^21) is also divisible by 5 SUFFICIENT IMO D




Re: If x is positive integer, is x^4  1 divisible by 5?
[#permalink]
29 Mar 2020, 22:29






