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If pqr s is a four-digit number, where p, q, r and s are the digits, 06 Aug 2017, 01:02
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If pqrs is a four-digit number, where p, q, r and s are the digits, is the sum of digits of the number a multiple of 12?

(1) p + q = 2(r + s)
(2) p + r = 4(q + s)
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B
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If pqr s is a four-digit number, where p, q, r and s are the digits, 17 Aug 2017, 00:33
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Bunuel
If pqrs is a four-digit number, where p, q, r and s are the digits, is the sum of digits of the number a multiple of 12?

(1) p + q = 2(r + s)
(2) p + r = 4(q + s)
Show more


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
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If pqr s is a four-digit number, where p, q, r and s are the digits, 06 Aug 2017, 04:02
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Bunuel
If pqrs is a four-digit number, where p, q, r and s are the digits, is the sum of digits of the number a multiple of 12?

(1) p + q = 2(r + s)
(2) p + r = 4(q + s)
Show more

(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
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If pqr s is a four-digit number, where p, q, r and s are the digits, 16 Aug 2017, 21:26
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Can anyone explain why the answer is B? Thank you so much
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If pqr s is a four-digit number, where p, q, r and s are the digits, 16 Aug 2017, 22:30
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Can anyone explain why the answer is B? Thank you so much
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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.

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If pqr s is a four-digit number, where p, q, r and s are the digits, 17 Aug 2017, 01:10
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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.

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If pqr s is a four-digit number, where p, q, r and s are the digits, 15 Nov 2017, 13:14
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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)
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If pqr s is a four-digit number, where p, q, r and s are the digits, 21 Nov 2017, 08:19
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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.
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If pqr s is a four-digit number, where p, q, r and s are the digits, 26 Mar 2019, 14:19
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chetan2u
Bunuel
If pqrs is a four-digit number, where p, q, r and s are the digits, is the sum of digits of the number a multiple of 12?

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


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
Show more

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 :blushing
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If pqr s is a four-digit number, where p, q, r and s are the digits, 26 Mar 2019, 20:13
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Mahmoudfawzy83
chetan2u
Bunuel
If pqrs is a four-digit number, where p, q, r and s are the digits, is the sum of digits of the number a multiple of 12?

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


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

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 :blushing
Show more

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.
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If pqr s is a four-digit number, where p, q, r and s are the digits, 16 May 2021, 08:33
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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
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If pqr s is a four-digit number, where p, q, r and s are the digits, 07 Oct 2024, 01:35
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I had approached this problem bit differently,

1. Since, p + r = 2(r + s), sum of p + r = Even. Hence, p & r both should be an even number = 2,4,6,8 only.
So, if p + r = 16 (8+8) then r + s = 8 & total = 24 which is multiple of 12. However, if p + r = 6 (2+4) then total sum is not multiple of 12
2. p+q = 4(r + s ), sum of p+q = Even. Hence, p & q both should be an even number = 2,4,6,8 which should be multiple of 4 only & sum of r+s should not exceed 18. eg: p+q = 8+8 = 16 then r+s = 4. So no answer will be multiple of 12
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