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When the positive integer x is divided by 4, is the remainder equal to
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10 Nov 2014, 06:53
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Re: When the positive integer x is divided by 4, is the remainder equal to
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17 Feb 2015, 05:10
Bunuel wrote: Tough and Tricky questions: Remainders. When the positive integer x is divided by 4, is the remainder equal to 3? (1) When x/3 is divided by 2, the remainder is 1. (2) x is divisible by 5. Kudos for a correct solution.Question: Is x = 4a+3? Statement 1) When x/3 is divided by 2, the remainder is 1i.e. x/3 is Odd number i.e. x/3 could be 1, 3, 5, 7 etc i.e. x = 3, 9, 15, 21 etc i.e. on Dividing x by 4 we get remainder = 3, 1, 3, 1 etc respectively i.e. remainders are not consistent therefore NOT SUFFICIENT Statement 2) x is divisible by 5i.e. x could be 5, 10, 15, 20 etc i.e. on Dividing x by 4 we get remainder = 1, 2, 3, 0 etc respectively i.e. remainders are not consistent therefore NOT SUFFICIENT Combining the two statements:x can be an odd multiple of 3 and 5 i.e. x can be only an odd multiple of 15 i.e. x can be 15, 45, 75 etc On dividing possible values of x by 4 the remainder are 3, 1, 3 etc. i.e. remainders are not consistent therefore NOT SUFFICIENT Hence, Answer: Option
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Re: When the positive integer x is divided by 4, is the remainder equal to
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10 Nov 2014, 07:21
The premise is asking if x/4 = something remainder 3. Essentially, does x = 3, 7, 11, 15, etc. Anything that leaves a remainder of 3 when divided by 4.
Statement 1 tells us that x/3 divided by 2 leaves remainder 1. So, x/3 must be 1, 3, 5, 7, etc. So, x = 3, 9, 15, 21, etc. This is insufficient as 3, 9 and leave a remainder of 3 when divided by 4, but 9 and 21 do not.
Statement 2 says x/5 is an integer. So, x = 5, 10, 15, 20, etc. This is also insufficient, leaving remainders of 1, 2, 3 and 0 when divided by 4.
Combining, the common terms for x are 15, 45, 75, etc. These leave a remainder of 3, 1, 3, 1, etc alternating.
Answer: E



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When the positive integer x is divided by 4, is the remainder equal to
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12 Nov 2014, 08:48
When the positive integer x is divided by 4, is the remainder equal to 3?
(1) When x/3 is divided by 2, the remainder is 1.
(2) x is divisible by 5.
1) something divided by 3, divided by 2, is basically saying divided by 6 (no?). if the remainder is 1 this can be 1, 7, 13, 19, etc. when you divide these by 4 you get r1, r3, r1, r3. Insufficient as it depends on the number X.
2) 5, 10, 15, 20. These give r1, r2, r3, r0. Insufficient as it depends on X.
combined. let's focus on 1 and get a number ending in 5 or 0.
1, 7, 13, 19, 25.. 25/6 = 4r1, and 25/5 =0. 25/4 = r1. doesn't give r3.
175/6 = 29r1. /5 works, /4 = 43r3. works.
conflicting answers therefore E.
Any quicker way? this took me too long.



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Re: When the positive integer x is divided by 4, is the remainder equal to
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12 Nov 2014, 09:25
revul wrote: When the positive integer x is divided by 4, is the remainder equal to 3?
(1) When x/3 is divided by 2, the remainder is 1.
(2) x is divisible by 5.
1) something divided by 3, divided by 2, is basically saying divided by 6 (no?). if the remainder is 1 this can be 1, 7, 13, 19, etc. when you divide these by 4 you get r1, r3, r1, r3. Insufficient as it depends on the number X.
2) 5, 10, 15, 20. These give r1, r2, r3, r0. Insufficient as it depends on X.
combined. let's focus on 1 and get a number ending in 5 or 0.
1, 7, 13, 19, 25.. 25/6 = 4r1, and 25/5 =0. 25/4 = r1. doesn't give r3.
175/6 = 29r1. /5 works, /4 = 43r3. works.
conflicting answers therefore E.
Any quicker way? this took me too long. When x/3 is divided by 2, the remainder is 1 should be expressed as x/3 = 2q + 1 > x = 6q + 3. So, x can be 3, 9, 15, ... For all those values, x/3 (1, 3, 5, ...) divided by 2 yields the remainder of 1.
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Re: When the positive integer x is divided by 4, is the remainder equal to
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17 Feb 2015, 03:33
Is this approach right about statement 1??
x/3 divided by 2, remainder is 1... That means x/3 is odd....
When x/3 is odd.... That means x is odd.
Quesiton asks... When x is divided by 4, is remainder=3???
So as we proved x is odd, it can be 3... 3/4 remainder is 3 it can be 5... 5/4 remainder is 1...
Hence insufficient.



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Re: When the positive integer x is divided by 4, is the remainder equal to
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17 Feb 2015, 08:31



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When the positive integer x is divided by 4, is the remainder equal to
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31 Dec 2015, 18:38
took a little over 2 mins just because I wanted to double check if everything is all right.
1. x can be 1, 7, 13, etc. so remainder is different each time. 1 alone is insufficient. A and D  out. 2. x can be 5, 10, 15, 20, etc  so remainder is different each time. 2 alone is insufficient. B  out.
1+2 x can be 25 or 55, and remainder thus 1 or 3. 1+2 insufficient, and C is out.
E is the answer.



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Re: When the positive integer x is divided by 4, is the remainder equal to
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14 Jul 2017, 16:26
When the positive integer x is divided by 4, is the remainder equal to 3?
Question: x ≡ 3 (mod 4) ?
(A) When x/3 is divided by 2, the remainder is 1.
i.) x∕3 ≡ 1 (mod 2)
therefore x ≡ 1 (mod 2)
INSUFFICIENT
(B) x is divisible by 5.
ii.) x ≡ 0 (mod 5)
INSUFFICIENT
(C)
x ≡ 1 (mod 2) x ≡ 0 (mod 5)
thus
x ≡ 5 (mod 10)
INSUFFICIENT.
Answer is "E."



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Re: When the positive integer x is divided by 4, is the remainder equal to
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05 Nov 2017, 12:41
grimm1111 wrote: When the positive integer x is divided by 4, is the remainder equal to 3?
Question: x ≡ 3 (mod 4) ?
(A) When x/3 is divided by 2, the remainder is 1.
i.) x∕3 ≡ 1 (mod 2)
therefore x ≡ 1 (mod 2)
INSUFFICIENT
(B) x is divisible by 5.
ii.) x ≡ 0 (mod 5)
INSUFFICIENT
(C)
x ≡ 1 (mod 2) x ≡ 0 (mod 5)
thus
x ≡ 5 (mod 10)
INSUFFICIENT.
Answer is "E." i got the ques correct but want to learn your mod concept can u explain your answer? x/3 = 1(mod2) => when x/3 is div by 2 rem is 1 then how u wrote x= 1(mod2) last doubt when u combined 1 and 2 => x = 5 (mod 10) how u deduced?
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Re: When the positive integer x is divided by 4, is the remainder equal to
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20 Oct 2018, 04:35
Bunuel wrote: revul wrote: When the positive integer x is divided by 4, is the remainder equal to 3?
(1) When x/3 is divided by 2, the remainder is 1.
(2) x is divisible by 5.
1) something divided by 3, divided by 2, is basically saying divided by 6 (no?). if the remainder is 1 this can be 1, 7, 13, 19, etc. when you divide these by 4 you get r1, r3, r1, r3. Insufficient as it depends on the number X.
2) 5, 10, 15, 20. These give r1, r2, r3, r0. Insufficient as it depends on X.
combined. let's focus on 1 and get a number ending in 5 or 0.
1, 7, 13, 19, 25.. 25/6 = 4r1, and 25/5 =0. 25/4 = r1. doesn't give r3.
175/6 = 29r1. /5 works, /4 = 43r3. works.
conflicting answers therefore E.
Any quicker way? this took me too long. When x/3 is divided by 2, the remainder is 1 should be expressed as x/3 = 2q + 1 > x = 6q + 3. So, x can be 3, 9, 15, ... For all those values, x/3 (1, 3, 5, ...) divided by 2 yields the remainder of 1. Hi Bunnel, I understood the solution to this problem.However,I have a query. When I solved statement I got numbers as 3,9,15,21,27,33...and statement 2 being multiples of 5 Do we really have to go further till 45 as second multiple..How do you figure that out..I didnt check for further mutiples post 33 and so ended up with wrong option. (Since it left me with 15 as the only possible element after combing statemnt 1 and 2) Please help!!




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