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Bunuel
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nelliegu
Hi! If x>7 and y>11, isn't it possible to say that x+y>18? Then, the answer could also be 19.
I understand that if we consider each equation separately, the answer has to be 20, but I wanted to check if we're allowed to combine the equations to get "19".

The point is, we are told that x and y are positive integers. For positive integers x and y, where x > 7 and y > 11, it's not possible that x + y = 19.
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I had missed that, thanks.
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m = xa+7 and x>7
n = yb + 11 and y>11
Eliminate I. and II.
Answer C or E

Ex. for III.
m=15; a=1; x=8
n=23; b=1; y=12
Answer C
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I think this is a high-quality question and I agree with explanation.
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I think this is a high-quality question and I agree with explanation.
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Reminder always < Divisor.
Therefore, 7<X, 11<Y.
Least value of X = 8
Least value of Y = 12
So, lest value of X + Y = 8 + 12 = 20.
Hence, C
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Bunuel
When positive integer \(m\) is divided by positive integer \(x\), the reminder is 7 and when positive integer \(n\) is divided by positive integer \(y\), the reminder is 11. Which of the following is a possible value for \(x + y\)?

I. 18

II. 19

III. 20


A. I only
B. II only
C. III only
D. II and III only
E. None

Bunuel,

But I thought, diving any positive integers between 1-6 by 7 will lead to a remainder of 7. can you please tell me what I am missing?

Best Regards,
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Bunuel
When positive integer \(m\) is divided by positive integer \(x\), the reminder is 7 and when positive integer \(n\) is divided by positive integer \(y\), the reminder is 11. Which of the following is a possible value for \(x + y\)?

I. 18

II. 19

III. 20


A. I only
B. II only
C. III only
D. II and III only
E. None

Bunuel,

But I thought, diving any positive integers between 1-6 by 7 will lead to a remainder of 7. can you please tell me what I am missing?

Best Regards,

No. When divisor is more than dividend then the reminder equals to the dividend. For example, 1 divided by 7 gives the remainder of 1. 2 divided by 7 gives the remainder of 2...

Check the links below for more:

6. Remainders



For other subjects:
ALL YOU NEED FOR QUANT ! ! !
Ultimate GMAT Quantitative Megathread
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what is the proof from the equation that the remainder is always less than the divisor?

eq: m/x = q + 7/x
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what is the proof from the equation that the remainder is always less than the divisor?

eq: m/x = q + 7/x


The remainder should always be less than the divisor because that is how we define the remainder in the context of division:

If \(x\) and \(y\) are positive integers, there exist unique integers \(q\) and \(r\), called the quotient and remainder, respectively, such that \(y =divisor*quotient+remainder= xq + r\) and \(0\leq{r}<x\).

When we divide one number (the dividend) by another number (the divisor), we are essentially asking how many times the divisor can fit into the dividend. The quotient is the whole number answer to this question, and the remainder is the amount that is left over after we've taken out as many whole divisors as we can.

For example, if we divide 13 by 3, we get a quotient of 4 and a remainder of 1. This means that 3 times 4 can fit into 13, with 1 left over. The remainder is the "leftover" amount, which must be less than the divisor (in this case, less than 3) because if it were equal to or greater than the divisor, then we could fit another whole divisor into the dividend.

Therefore, the remainder must always be less than the divisor in order for the result of the division to make sense and be consistent with the way we define the operation.
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I have edited the question and the solution by adding more details to enhance its clarity. I hope it is now easier to understand.
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I think this is a high-quality question and I agree with explanation.
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