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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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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)
[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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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
C
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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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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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16 Aug 2017, 22:30
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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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Math Expert
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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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17 Aug 2017, 00:33
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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..

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

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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]

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15 Nov 2017, 13:14
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

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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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21 Nov 2017, 08:19
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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Re: If pqr s is a four-digit number, where p, q, r and s are the digits,   [#permalink] 21 Nov 2017, 08:19
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