For some integer q, q^2 - 5 is divisible by all of the follo

Question

For some integer q, q^2 - 5 is divisible by all of the following EXCEPT

(A) 29
(B) 30
(C) 31
(D) 38
(E) 41

dimitri92 · Accepted Answer

q^2 - 5 = (q^2 - 1) - 4
= (q + 1)(q - 1) - 4

Now q-1, q, q+1 are three consecutive integers , one of them should be a multiple of 3

If q-1 or q+1 is a multiple of 3; q^2-5 is not because 4 is not a multiple of 3
if q is a multiple of 3, q^2 is a multiple of 3 but q^2 -5 is not a multiple of 3

So q^2 -5 is never a multiple of 3, so it will not be a multiple of 30

Bunuel · Answer

Hint: q^2-5 (q is an integer) is never multiple of 3 (try to prove this), hence 30 is out.

Answer: B.

cipher · Answer

I did this by POE and the crude way, not sure whether we have an elegant way of solving it

IMO B,

For 29, we have option q = 11 Q^2=121 and 121-5 =116 = 29*4

for  31 q= 6 q^2=36 
for  38 q= 9 q^2=81 
for  41 q= 13 q^2 =169

gurpreetsingh · Answer

q^2 - 5 = 30k + r

q^2 =30k +5 + r = 5(6k+1) + r

now if we take r =0 then q can never be a perfect square as 6k+1 is never a multiple of 5.

Hence B

shrouded1 · Answer

There is a systematic way to prove this.

That is not true, Eg. 25,55,85,.... are all multiples of 5 of the form 6k+1

Proof that  q^2-5  cannot be a multiple of 30

q^2 = 30k + 5 = 6*(5k) + 5 
So q^2 leaves remainder 5 when divided by 6.

Consider the various cases for the remainder that q leaves when divided by 6 :
q=6k ... q^2 will leave remainder 0 when divided by 6
q=6k+1 ... q^2 will leave remainder 1 when divided by 6
q=6k+2 ... q^2 will leave remainder 4 when divided by 6
q=6k+3 ... q^2 will leave remainder 3 when divided by 6
q=6k+4 ... q^2 will leave remainder 2 when divided by 6
q=6k+5 ... q^2 will leave remainder 1 when divided by 6

So it is not possible to have a perfect square that leaves remainder 5 when divided by 6.

So 30k+5 is not a perfect square for any choice of k.

gurpreetsingh · Answer

For some integer Q, (Q^2)-5 is divisible by all of the following except
1. 29
2. 30
3. 31
4. 38
5. 41

One way is there -> POE

31+5 = 36 = 6^2 -> out

41*4 -> 164 -. 164+5 -> 13^2 -> out

38*2 = 76 -> 76+5 = 81 = 9^2 -> out

29*4 = 116 -> 116 + 5 = 121 = 11^2 -> out

Multiple all the values with 1,2,3,4 -> you will get the answer. It can be solved in 2 minutes using POE.

LM · Answer

Only approach I could figure out to solve it quickly ( without going into details of proving wheter 30...or something )

29*1 = 29 +5 = 34
29*2= 58+5= 63
29*3 = 87+5 = 92
29*4= 116+5 = 121 ( 11^2)

41*1 = 41 +5 = 46
41*2= 82+5= 87
41*3 = 123+5 = 128
41*4= 164+5 = 169 ( 13^2)

38*1 = 38 +5 = 43
38*2= 76+5= 81 (9^2)

31*1 = 31+5 = 36 (6^2)

Left with only one choice and that is the answer.

Be strong with multiplication tables and also with the square of numbers. Then you can approach this question faster using plug and play.

KarishmaB · Answer

The way I would approach this question:

So q^2 - 5 is divisible by all of the following except: 
29, 31, 41 - big prime numbers, don't know any divisibility rules for these, forget them for the time being.. 38 = 19*2. (q^2 - 5) can be divisible by 2 (e.g. when q^2 ends with a 5, q^2 - 5 ends with a 0). As for 19, again a big prime number. Leave it for the time being.

(If the question is anywhere close to an actual GMAT question, they will not expect you to do many calculations with 29, 31, 41 etc. I see these big prime numbers and am quite convinced that they are just a smokescreen.Try and focus on what they could ask you like divisibility by 2, 3 etc. )

As for 30, q^2 - 5 is divisible by 10 (using the logic shown above). What about 3?
 q^2 - 5 = q^2 - 1 - 4 = (q - 1)(q + 1) - 4 
In any 3 consecutive numbers, (e.g.  (q - 1), q, (q + 1) ), one and only one number will be divisible by 3.
If either (q - 1) or (q + 1) is divisible by 3, (q - 1)(q + 1) is divisible by 3, which means  (q - 1)(q + 1) - 4  cannot be divisible by 3. If q is divisible by 3, then q^2 will be divisible by 3 and q^2 - 5  will not be divisible by 3.

subrataroy0210 · Answer

Taking lead from Bunuel's post...

A) 29 - Prime
B) 30 - 2 * 3 * 5
C) 31 - Prime
D) 38 - 2 * 19
E) 41 - Prime

Let's start with simple numbers.

1) 2

Remainder when 5/2 is 1. And, perfect squares such as 81 produce remainder 1 when divided by 2. Thus, the overall remainder (1-1) is 0. 2 might divide. Park aside.

2) 3

Remainder when 5/3 is 2. The perfect squares when divided by 3, produce either 0 or 1 remainder. 
Case 1) When the remainder is 0
The remainder value of the expression is 0-2 = -2. Not divisible by 3

Case 2) When the remainder is 1
The remainder value of the expression is 1-2 = -1. Not divisible by 3

Thus, the expression will not be divisible by 3 or any multiple of it. Thus, option B.

JeffTargetTestPrep · Answer

We need to find the answer choice that does not divide 5 less than a perfect square. Let’s analyze each answer choice:

A) 29

Since 11^2 - 5 = 116, which is divisible by 29, answer A is not correct.

B) 30

It doesn’t seem that we can find an integer q such that q^2 - 5 is divisible by 30. However, let’s make sure we can find an integer q such that q^2 - 5 is divisible by 31, 38, and 41.

C) 31

Since 6^2 - 5 = 31, which is divisible by 31, answer C is not correct.

D) 38

Since 9^2 - 5 = 76, which is divisible by 38, answer D is not correct.

E) 41

Since 13^2 - 5 = 164, which is divisible by 41, answer E is not correct.

Answer: B

Kinshook · Answer

Bunuel

A systematic way to solve such questions is needed. Please guide.

Robo_123 · Answer

I am unable to understand this. Can someone please elaborate?

RAldertree · Answer

We only need to be concerned with the answers, so we can check each of them individually. But rather than checking the answers themselves, we can check their factors because all the factors of a number, n, must divide some other number, m, in order for n to divide m. Example: if 5 does not divide some number, then it is impossible for 10 to divide that number.

So let's start with rewriting all of the numbers for each answer as their prime factorization:
a) 29 (prime)
b) 2 * 3 * 5
c) 31 (prime)
d) 2 * 19
e) 41 (prime)

Let's start by checking the smallest prime numbers first, because they will be easier to calculate. 2 must divide some q^2 - 5 because it is present in multiple answers and there must be a single correct answer, so we can ignore 2. The next smallest number is 3.

Explanation using Modular Arithmetic:
In the modular space of 3, there are 3 possible values: 0, 1, 2. So these are the only relevant values of q so far as divisibility by 3 is concerned. If q is 0, q^2 - 5 is 1 (mod 3). If q is 1, q^2 - 5 is 2 (mod 3). if q is 2, q^2 - 5 is 2 (mod 3). Of all 3 possibilities, none are 0, so q^2 - 5 must never be divisible by 3. Thus the answer has to be b) 30.

Explanation using Remainders:
There are 3 possibilities for the number q: Either q is divisible by 3, q is one less than a number divisible by 3, or q is two less than a number divisible by 3. In other words, q takes one of these three forms ONLY: 3k, 3k - 1, 3k - 2. If we square these (q ^2) we get:
9k^2 -> 3c,
9k^2 - 6k + 1 -> 3c + 1,
9k^2 - 12k + 4 -> 3c + 4 -> 3c + 1
So q^2 is either a multiple of 3, or 1 more than a multiple of 3. A multiple of 3 minus 5 is never a multiple of 3. One more than a multiple of 3 minus 5 is the same as a multiple of 3 minus 4, so it is never a multiple of 3. Therefore, q^2 - 5 can never be a multiple of 3, so it can also never be a multiple of 30. The answer must be b) 30.

PriyanshuR · Answer

q^2 - 5 = (q^2 - 1) - 4 could be written as (q + 1)(q - 1) - 4

Now q-1, q, q+1 are three consecutive integers , one of them should be a multiple of 3

If q-1 or q+1 is a multiple of 3; q^2-5 is not because 4 is not a multiple of 3
if q is a multiple of 3, q^2 is a multiple of 3 but q^2 -5 is not a multiple of 3

So q^2 -5 is never a multiple of 3. Thus, answer is 30.

For some integer q, q^2 - 5 is divisible by all of the follo

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For some integer q, q^2 - 5 is divisible by all of the follo

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