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What is the value of positive integer q?

(1) When positive integer p is divided by q, the remainder is 5.
St 1: p=qK + 5, where k = constant p can be 5 and q can be any value>5 so no sufficient

(2) When q is divided by positive integer p, the remainder is 7.
St 1: q=pK + 7, where k = constant q can be 7 and 7 can be any value>7
also what if q = 15 and p = 8 so two value not sufficient

After combining both st.
we get to know that possible value for p = 12 and q = 7 so both sufficient

IMO C
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Quote:
What is the value of positive integer q?

(1) When positive integer p is divided by q, the remainder is 5.
-- Considering statement 1 alone
=> p = qx + 5
from this we know q is greater than 5. But don't know the exact value.

-- Not sufficient.

(2) When q is divided by positive integer p, the remainder is 7.
=> q = py + 7
p is greater than 7.

-- Not sufficient.

On combining both the statement,
Case 1 : P =5, then remained for q will not be 7, it should become 2.
Case 2: q = 7, then P can take any value for which we will get remainder 5.


On combining we will get answer. Therefore, C is the answer.
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Asked: What is the value of positive integer q?

(1) When positive integer p is divided by q, the remainder is 5.
pmodq = 5
p = qk + 5; where k Is an integer.
q > 5;
Since p is unknown, the value of q can not be found
NOT SUFFICIENT

(2) When q is divided by positive integer p, the remainder is 7.
qmodp = 7
q = pm + 7; where m Is an integer.
p > 7;
q can not be determined.
NOT SUFFICIENT

(1) + (2)
(1) When positive integer p is divided by q, the remainder is 5.
(2) When q is divided by positive integer p, the remainder is 7.
q = (qk + 5)m + 7 = qkm + 5m + 7
q(1-km) = 5m + 7
q = (5m+7)/(1-km) > 5
Since 5m + 7 > 0 ; 1-km > 0 ; km < 1
It is only possible when km=0
If k=0; p=5; not possible since p>7
If m=0; q = 7; p = 7k + 5; possible
q = 7
SUFFICIENT

IMO C
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Stat 1- p/q = 5, not sufficient to get value of q.

Stat 2- q/p= 7, not sufficient to get value of q, but p>7

Combining both, from stat 2, p= 8, 9, 11, 12, 13...then making stat 1 holds true, q= 3, 4, 6, 7, 8...
Now, to hold stat 2 true as well (p,q)= (12,7). Sufficient to get q value.

So, I think ans. C.
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A:
p= qx+5
q can be any number
not sfficient p can be 5, and when divide by 6,7,8,9,10- all give remainder 5

B.
q= py+ 7
p and k can be any number, so insufficient
when q= 7 is divided by 8,9,10,11,12 --all give reminder 7
when q= 19 is divided by 12- we get remainder 7
when q= 18 is divided by 11 - we get remainder 7
not sufficient


A+B:

when 19 is divided by 7 , we get reamidner 5
when q=7 is divided by 19 , we get reminder 7

when 19 is divided by 7 , we get reamidner 5
when q=7 is divided by 19 , we get reminder 7


add 5 in series of 7,14,21,28
12,19,26, 33 ...divide by 7 give 5 remainder
when q is divided by 12,19,26,33;
we can get 7 remainder if q is ( 5, 12,19 etc)

still not sufficient
hence E
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imSKR
A:
p= qx+5
q can be any number
not sfficient p can be 5, and when divide by 6,7,8,9,10- all give remainder 5

B.
q= py+ 7
p and k can be any number, so insufficient
when q= 7 is divided by 8,9,10,11,12 --all give reminder 7
when q= 19 is divided by 12- we get remainder 7
when q= 18 is divided by 11 - we get remainder 7
not sufficient


A+B:

when 19 is divided by 7 , we get reamidner 5
when q=7 is divided by 19 , we get reminder 7

when 19 is divided by 7 , we get reamidner 5
when q=7 is divided by 19 , we get reminder 7


add 5 in series of 7,14,21,28
12,19,26, 33 ...divide by 7 give 5 remainder
when q is divided by 12,19,26,33;
we can get 7 remainder if q is ( 5, 12,19 etc)

still not sufficient
hence E

Hi Sir IanStewart

can you please suggest why the numbers can not be 12,19,
where did i do wrong?
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imSKR

Hi Sir IanStewart

can you please suggest why the numbers can not be 12,19,
where did i do wrong?

I don't really follow what you're doing in the last paragraph of your solution, but when you divide 12 by 19, the remainder is 12 (the quotient is zero). So the numbers cannot be 12 and 19.

When you divide a smaller number by a larger number, the smaller number is the remainder. So when we use both Statements here, since either p < q or q < p, then it must be true that either p = 5 or q = 7. But it can't be true that p = 5, because we know we can get a remainder of 7 when we divide by p, and you can't get a remainder of 7 when you divide by 5. So q = 7 must be true.
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Bunuel
What is the value of positive integer q?

(1) When positive integer p is divided by q, the remainder is 5.
(2) When q is divided by positive integer p, the remainder is 7.


 

This question was provided by Veritas Prep
for the Heroes of Timers Competition

 

The answer is E not C.
p=qi+5
q=pj+7
all positive needless to say.
q(1-ij)=5j+7
LHS>0 and q>0
1-ij>0
1>ij>=0
ij=0
So
i=0, j=1
i=0, j=0
i=1,j=0
following cases are only valid
p=5, q=12
p=12, q=7
Answer inconclusive.
Hence E.
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thakurarun85

The answer is E not C.
p=qi+5
q=pj+7

The answer is C. You've used the same "j" as the quotient in both divisions, but the quotient cannot be the same in both divisions unless p = q, which cannot be true here. One of the two quotients must be zero, in fact, because either p < q or q < p, and when you divide a smaller number by a larger number, the quotient is always zero.

thakurarun85

following cases are only valid
p=5, q=12

When you divide 12 by 5, the remainder is 2, not 7, so this case is not valid.
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IanStewart
thakurarun85

The answer is E not C.
p=qi+5
q=pj+7

The answer is C. You've used the same "j" as the quotient in both divisions, but the quotient cannot be the same in both divisions unless p = q, which cannot be true here. One of the two quotients must be zero, in fact, because either p < q or q < p, and when you divide a smaller number by a larger number, the quotient is always zero.

thakurarun85

following cases are only valid
p=5, q=12

When you divide 12 by 5, the remainder is 2, not 7, so this case is not valid.

i and j were different. sure out of 3 only one pair is good. Hence C. Thanks for correction.
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thakurarun85

i and j were different.

Oh right, I see that now, my vision must be going. :) Sorry about that!
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