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

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

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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)
[Reveal] Spoiler: OA

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Re: If pqr s is a four-digit number, where p, q, r and s are the digits, [#permalink]

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New post 06 Aug 2017, 04:02
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Bunuel wrote:
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)


(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
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Re: If pqr s is a four-digit number, where p, q, r and s are the digits, [#permalink]

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New post 16 Aug 2017, 21:26
Can anyone explain why the answer is B? Thank you so much

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Re: If pqr s is a four-digit number, where p, q, r and s are the digits, [#permalink]

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pclawong wrote:
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.

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Re: If pqr s is a four-digit number, where p, q, r and s are the digits, [#permalink]

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Bunuel wrote:
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 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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Re: If pqr s is a four-digit number, where p, q, r and s are the digits, [#permalink]

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New post 17 Aug 2017, 01:10
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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Re: If pqr s is a four-digit number, where p, q, r and s are the digits,   [#permalink] 17 Aug 2017, 01:10
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