If pqr s is a four-digit number, where p, q, r and s are the digits,

Hi..

With 100% answering it wrong in TIMER and confusion over the solution, this may help you...

PQRS is 4-digit number...
Logical inference - P cannot be ZERO..

Let's see the statements..

1) \(p+q=2(r+s)\)
So \(p+q+r+s=3(r+s)\)..
If r+s is MULTIPLE of 4 and rs is MULTIPLE of 4...... ans is yes.. possibility pq40 or pq04...
If rs is not MULTIPLE of 4 say 13,31 OR r+s is not MULTIPLE of 4, say R+s= 3.. ans is NO
Insufficient

2)\(P+R=4(Q+S)\)....
Q+S can not be 0, as then P and R will be zero and number will not be 4-digit number.

So max value of P and R can be 9 each or P+R is max 18..
So Q+S can be max \(\frac{18}{4}=4.5\) or INTEGER 4..
Now \(P+Q+R+S =5(Q+S)\)..
So Q+S has to be a MULTIPLE of 12, 0 is not possible
But Q+S can be max 4, so ans will be NO always
Sufficient

B
_________________

(1)
p + q = 2(r + s)
p + q + r + s = 3(r + s)

Now (r+s) can be any number ranging from 0 to 18 (inclusive). But only in case of (r+s) being 4,8,12,16 then answer will be yes and in all other cases it will be no.

Not Sufficient

(2)
p + r = 4(q + s)
p + q + r + s = 5(q + s)

Now (q+s) can be any number ranging from 0 to 18 (inclusive). But only in case of q+s being 12 then answer will be yes and in all other cases it will be no.

Not Sufficient

However on combining we have
3 (r +s) = 5 (q+s)
That can only happen when ((r+s),(q+s)) are (5,3), (10,6), (15,9)

Also since r+s can take 5,10,15 only value possible is 5 as p + q = 2(r + s) i.e. sum of two digits cannot go beyond 18
hence r+s = 5 and hence q+s = 3


Now
p + q + r + s = 3(r + s) = 3*5 = 15 not a multiple of 12
just validating p + q + r + s = 5(q + s) = 5*3 = 15 not a multiple of 12

Hence Sufficient
C
_________________

Can anyone explain why the answer is B? Thank you so much

B gives you a definite NO. No matter what number you use for the sum of q and s, the total value of p,q,r,s will only spit out multiples of 5. For example, (p+r) = 4 (q+s) when q+s = 8, will be 32 = 4 (8). The sum of the digits will be p+q+r+s or (p+r)+(q+s) = 32+8 = 40.

Sent from my SGH-M919V using GMAT Club Forum mobile app

I will only explain the 2nd part of the question since B is correct answer
p+r = 4(q+s)
So if
q+s p+r p+q+r+s
1 4 5
2 8 10
3 12 15
4 16 20
So in all the cases p+q+r+s is not divisible by 12. Hence B is sufficient.

Sent from my MotoG3 using GMAT Club Forum mobile app

To test sum -> \((p + q + r + s)\) is divisible by 12, we have to check if sum is divisible by factors of 12.
=>\({2,3,4,6}\) => check if the divisible by 3 AND 4.

Statement 1:
\(p + q = 2 (r+s)\)
let \(r + s = x\) , so \(p + q = 2x => p + q + r + s = 3x\)
=> so sum is multiple of 3, but we don't know if sum is multiple of 4 => Not Sufficient

Statement 2:
\(p + r = 4(q+s)\)
=> \(p + q + r + s => 4(q+s) + q+s = 5(q+s)\)
=> sum is divisible by 5
So for sum also to be divisible by 12, sum must be multiple of \(LCM (5,12) = 60.\)
But \(max(p+q+r+s) = 9 + 9 + 9 + 9 = 36\), so sum is never divisible by 60 => sum is never divisible by 12 (given sum is also divisible by 5)

Sufficient

Answer (B)

since pqrs is a four digit number, p > 0.

p,q,r and s can take a max value of 9 each.

Hence max value of p+q+r+s = 36 ---(i)

(1) p+q=2(r+s)
-->p+q+r+s = 3(r+s) where minimum value of r+s = 1 since the smallest possible value of pqrs is 1000.
--> 1<=(r+s)<=18
--> 3<=3*(r+s)<=54
--> 3<=(p+q+r+s)<=36 (see stmt (i) above)
Hence possible values of p+q+r+s = 3, 6, 9, 12,...,24,...,36.

Thus the (p+q+r+s) could be be divisible by 12. Not sufficient.

(2) p+r=4(q+s)
-->p+q+r+s=5(q+s),
but 1<=(q+s)<=18
--> 5<=5*(q+s)<=90
--> 5<=(p+q+r+s)<=36

Hence possible values of p+q+r+s = 5, 10, 15, 20, 25, 30, 35. i.e. No multiples of 12. Hence stmt (2) is sufficient.

Hi chetan2u
please can you confirm the assumption that "Q+S has to be a MULTIPLE of 12"?

I mean that to be divisible by 4, the last 2 digits are the only concern,
and to be divisible by 3, the sum should be divisible by 3 .... and (Q+S) can be a multiple of 3.

I assumed : P+R=4(Q+S) as 9 + 3 = 4 (1 + 2)
and generated a number PQRS as 9132 which is divisible by 3 and 4,
that makes B not sufficient.

please can you revise my method?
I am sorry for pulling you back to a 2017 discussion

Hi,
Happy to help, so no worries on getting back to some old discussion.

We ve to find the sum of digits to be divisible by 12. So the last two digits being divisible by 4 will not come into play as that would be the case is the 4-digit number as such is divisible by 12.
Here we are looking whether p+q+r+s is divisible by 12.
As far as , finding different possibilities of number pqrs is concerned, it can be time consuming and error prone.

If you are looking to show some statement as insufficient, it is ok to find an example that doesn't fit in in the statement. But if you are looking to something as sufficient, you have to try all possible combinations.
_________________

Please correct my approach if wrong,

Since the sume should be a multiple of 12, it can be only 12, 24 or 36 max.

Now statement a says that p+q= 2(r+s) so p+q+r+s= 3k where k is r+s which can be either 4,8,12 or any other number. So we cannot definitely say that the sum is a multiple of 12
Insufficient

Statement b says that p+q+r+s= 5k where k can be any integer. Since no integer multiple of k can be 12, 24, 36, this condition sufficiently says that the sum can never be a multiple of 12.

Hence option B

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________