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Bunuel
When 900 is divided by positive integer d, the remainder is r. For some integer N > 5000, when N is divided by positive integer D, the remainder is R. Is R > d?

(1) r = 1

(2) D = 23


Kudos for a correct solution.

MAGOOSH OFFICIAL SOLUTION:

This is a tricky one about remainders remainders.

Statement #1: If r = 1, then we divide 900 by d, and the remainder is 1. This means that d is a factor of 899. That’s interesting, but at the moment, we know zilch about R, which could be anything. This statement, alone and by itself, is not sufficient.

Statement #2: If D = 23, then when we divide by 23, the remainder has to be smaller than the divisor. We know R < 23. But, now, the only thing we know about d is that it’s not a factor of 900: d could be 7 or 97. We have no idea of its size, so we can’t compare it to R. This statement, alone and by itself, is not sufficient.

Combined:

From the second statement, we know R < 23. From the first, we know d must be a factor of 899. What are the factors of 899? For this we will use an advanced factoring technique. Notice that 899 = 900 – 1. This means, we can express 899 as the Difference of Two Squares, because 900 is 30 squared. We can use that algebraic pattern to factors numbers.

899 = 900 - 1 = 30^2 - 1^1 = (30 + 1)(30 - 1) = 31*29.

So, it turns out that 899 is the product of two prime numbers, 29 and 31. This means that 899 has four factors: {1, 29, 31, and 899}. Those are the candidate values for d. Obviously, d cannot equal 1, because when we divide any integer by 1, we never get a remainder of any sort: 1 goes evenly into every integer. That means, d could be 29 or 31 or 899. Well, if R < 23, this means that R must be less than d. We can give a definitive “yes” answer to the prompt question. Combined, the statements are sufficient.

Answer = (C)
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The tricky part : factors of 899 = 29*31 (took me a while to figure this out, started with the approx values around 30 as it is close to 900)
this means d could be either 29 or 31

It can be useful to remember that a^2 - b^2 = (a+b)(a-b). Whenever you see a number that is close to a perfect square, like 899, think of it as (30-1)(30+1)
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Bunuel
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When 900 is divided by positive integer d, the remainder is r. For some integer N > 5000, when N is divided by positive integer D, the remainder is R. Is R > d?

(1) r = 1

(2) D = 23


Kudos for a correct solution.


Statement #2: If D = 23, then when we divide by 23, the remainder has to be smaller than the divisor. We know R < 23. But, now, the only thing we know about d is that it’s not a factor of 900: d could be 7 or 97. We have no idea of its size, so we can’t compare it to R. This statement, alone and by itself, is not sufficient.



Hi,

Statement 2 does not state anything regarding "r" and we know remainder can be greater than or equal to zero i.e r>=o
So, if we assume r = 0 then "d" can also be a factor of 900.

Am I wrong?
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Bunuel
Bunuel
When 900 is divided by positive integer d, the remainder is r. For some integer N > 5000, when N is divided by positive integer D, the remainder is R. Is R > d?

(1) r = 1

(2) D = 23


Kudos for a correct solution.


Statement #2: If D = 23, then when we divide by 23, the remainder has to be smaller than the divisor. We know R < 23. But, now, the only thing we know about d is that it’s not a factor of 900: d could be 7 or 97. We have no idea of its size, so we can’t compare it to R. This statement, alone and by itself, is not sufficient.



Hi,

Statement 2 does not state anything regarding "r" and we know remainder can be greater than or equal to zero i.e r>=o
So, if we assume r = 0 then "d" can also be a factor of 900.

Am I wrong?

If I got what you're saying you're not wrong, but it wouldn't make statement 2 sufficient either.
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And Another one of those Veritas Prep Out of Bound Question...!!
Definitely solvable..
But you would need a PHD in Mathematics to actually do it in under 2 minutes.
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Though took 3 mins to solve it but here is my solution,

From information given in the question we can write,
900 = dq + r (q- quotient)
N = DQ + R (Q - quotient)

S1: r = 1

From this we came to know that dq = 899. (since dq +r = 900 => dq +1 =900 => dq = 899)
and 899 = 29 *31 *1
so d could be 29, 31 or 899 (d can't be 1 since in that case there won't be any remainder).
no information available for D and R so ignore this statement

S2: D=23
If D - 23, the remainder(R) could be anything between 1 to 22.
But this statement doesn't tell anything about d and r so ignore it.

S1 + S2
From S1, we know d could be 29, 31 or 899
From s2, we know R could be between 1 to 22.

So d is always greater than R.
Hence option C.
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