When the positive integer x is divided by 4, is the remainder equal to

Question

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.

GMATinsight · Accepted Answer

Question: Is x = 4a+3?

Statement 1) When x/3 is divided by 2, the remainder is 1 
i.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 5 
i.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  E

intheend14 · Answer

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

revul · Answer

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.

Bunuel · Answer

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.

gmathopeful90 · Answer

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.

Bunuel · Answer

x cannot be 5. Check here: when-the-positive-integer-x-is-divided-by-4-is-the-remainder-equal-to-188357.html#p1441810

mvictor · Answer

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.

grimm1111 · Answer

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."

sahilvijay · Answer

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?

prabsahi · Answer

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!!

ricardorr04 · Answer

Great question - took me 2:30

A) Not sufficient, clearly because doesnt give us info about x/4 (could be 1/6 or 7/6, for example, in which these cases would yield different remainders for 4)

B) obviously not sufficient

C) Not sufficient either, since you can test cases (ie. 25/6 and 25/4 both yield remainder of 1, but 55/6 and 55/4 both yield remainders of 1 and 3, respectively, this we cannot conclude and answer thus insufficient.

answer is E

QuantMadeEasy · Answer

(1) When x/3 is divided by 2, the remainder is 1. 
That mean x/3 = odd integer
Hence, x = 3*odd integer
Possible values of x are 3, 9, 15, 21,...
Possible values of remainders are 3, 1, 3, 1,...
Insufficient

(2) x is divisible by 5. 
Possible values of x are 5, 10,15, 20,...
Possible values of remainders are 1,2,3,0,...
Insufficient

(1)+(2); 
still possible values of remainders are 1 and 3
Insufficient

E is correct

bumpbot · Answer

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When the positive integer x is divided by 4, is the remainder equal to

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