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In regards to this thread only.
Don't post random SAT questions for the sake of posting them. Only post problems you are having trouble solving/solving quickly. I'll try my best to explain a simple approach. Any question will do. So long as it's not a passage-based reading question.
You're on your own for those.
DO NOT I REPEAT DO NOT POST ANYTHING ELSE BESIDES QUESTIONS OR ATTEMPTED SOLUTIONS TO SAT PROBLEMS.
Also, do not post duplicate solutions unless you have something new to offer or spot a mistake that needs to be corrected.
If you get ninja'd please spoiler your post so this page doesn't get cluttered.
That is all. Thanks.
 
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United States24698 Posts
When an integer H is divided by 2 times the sum of its factors, the quotient is 5 and the remainder is 1. What is the value of H?
10, 11, 17, 20, 24
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Korea (South)1068 Posts
H/2 (x+y) Q = 5 R = 1
The answer is 11.
The way you solve it is.
H / (2 (x+y)) = 5
(x y) = 5
(x y) / (2 (x + y) )
(x/2) + (y/2) = 5
(x + y) / 2 = 5
x + y = 10
x + y = H
H = 10 + 1
You want to team up mister in order to solve these problems? 
DONE REWRITING NOTATION
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On September 05 2010 15:48 micronesia wrote:
When an integer H is divided by 2 times the sum of its factors, the quotient is 5 and the remainder is 1. What is the value of H?
10, 11, 17, 20, 24
Prime factors or not O_o
Not sure I'm getting the question, but I would guess 11 (thank I god I am done with SAT) simply because it is the only number than can have a remainder of 1 with a quotient of 5 (talking about integers here)
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Korea (South)1068 Posts
On September 05 2010 16:02 Loser777 wrote:Show nested quote +On September 05 2010 15:48 micronesia wrote:
When an integer H is divided by 2 times the sum of its factors, the quotient is 5 and the remainder is 1. What is the value of H?
10, 11, 17, 20, 24 Prime factors or not O_o Not sure I'm getting the question, but I would guess 11 (thank I god I am done with SAT) simply because it is the only number than can have a remainder of 1 with a quotient of 5 (talking about integers here)
GUESSING IS BAD RAWR!
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United States24698 Posts
On September 05 2010 16:00 kineSiS- wrote:
H/2 (x+y) Q = 5 R = 1
The answer is 11.
The way you solve it is.
H / 2 (x+y) = 5
x y = 5
x y / 2 (x + y)
x/2 + y/2 = 5
x + y / 2 = 5
x + y = 10
x + y = H
H = 10 + 1
You want to team up mister in order to solve these problems?
Can you make your notation less confusing with parentheses or more clear spacing?
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On September 05 2010 15:48 micronesia wrote:
When an integer H is divided by 2 times the sum of its factors, the quotient is 5 and the remainder is 1. What is the value of H?
10, 11, 17, 20, 24
By factors, you mean its prime factors, or just non-repeating positive factors (including 1 and itself)? As it is printed, I don't think this is a valid SAT question since it leaves too much room for interpretation.
Anyway, I'll assume it's "unique factors that aren't self", since the traditional definition of "factors" doesn't make sense here. In fact that contrived definition is the only definition that makes sense for this problem as we'll soon see.
We know the basic problem here is that there is some number H, and another number, let's call it J that is the sum of the factors of H.
They have the relationship as follows:
1.) floor(H / 2J) = 5 -> H > 10J
*Floor simply means to round down
From this, we know that the number H is at least 10 times larger than the sum of its "factors." Thus, the immediate conclusion is that the number itself can't be one of its own factors (thus my previous definitional conclusion)
If we do a brief factorization of the options we get the following:
10 -> 1, 2, 5. Sum is 8. 10 / 16 is clearly not 5.
11 -> 1 Sum is 1. 11/2 floors to 5. Nice.
17 -> 1 Sum is 1. 17/2 floors to 8. Booo.
20 -> 1, 2, 4, 5, 10. Sum is clearly way too big.
24 -> 1, 2, 3, 4, 6, 8, 12 And we're clearly way too big.
So immediately we know the answer is 11.
But just to be sure, we check the second property:
2.) H % (or mod) 2J = 1, which simply says the remainder when H is divided by 2J is 1.
The fact that the remainder after the division by an even number (2J) is 1 tells us that H must be odd. The principle behind this is rather obvious... Assume an even divisor and a number X.
(as a proof, Assume the result is D and the remainder is R, and we have the divisor 2J from before. Then we have D * 2J + R = X. We know D * 2J is even, and we know R (which is 1) is odd. Even + odd = odd. )
This, would immediately leave us with 11 and 17.
11 works for both properties, so it is the answer we must pick.
On September 05 2010 16:11 micronesia wrote:Show nested quote +On September 05 2010 16:00 kineSiS- wrote:
H/2 (x+y) Q = 5 R = 1
The answer is 11.
The way you solve it is.
H / 2 (x+y) = 5
x y = 5
x y / 2 (x + y)
x/2 + y/2 = 5
x + y / 2 = 5
x + y = 10
x + y = H
H = 10 + 1
You want to team up mister in order to solve these problems? Can you make your notation less confusing with parentheses or more clear spacing?
As written this solution is also rubbish. It literally makes no sense.
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On September 05 2010 16:10 kineSiS- wrote:Show nested quote +On September 05 2010 16:02 Loser777 wrote:On September 05 2010 15:48 micronesia wrote:
When an integer H is divided by 2 times the sum of its factors, the quotient is 5 and the remainder is 1. What is the value of H?
10, 11, 17, 20, 24 Prime factors or not O_o Not sure I'm getting the question, but I would guess 11 (thank I god I am done with SAT) simply because it is the only number than can have a remainder of 1 with a quotient of 5 (talking about integers here) GUESSING IS BAD RAWR!
DO NOT I REPEAT DO NOT POST ANYTHING ELSE BESIDES QUESTIONS OR ATTEMPTED SOLUTIONS TO SAT PROBLEMS.
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United States24698 Posts
Yeah this question I took from a site but it seems ridiculously ambiguous... it makes sense for a very specific definition of factor... oh well.
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On September 05 2010 16:10 kineSiS- wrote:Show nested quote +On September 05 2010 16:02 Loser777 wrote:On September 05 2010 15:48 micronesia wrote:
When an integer H is divided by 2 times the sum of its factors, the quotient is 5 and the remainder is 1. What is the value of H?
10, 11, 17, 20, 24 Prime factors or not O_o Not sure I'm getting the question, but I would guess 11 (thank I god I am done with SAT) simply because it is the only number than can have a remainder of 1 with a quotient of 5 (talking about integers here) GUESSING IS BAD RAWR!
My scores disagree
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DO NOT I REPEAT DO NOT POST ANYTHING ELSE BESIDES QUESTIONS OR ATTEMPTED SOLUTIONS TO SAT PROBLEMS.
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On September 05 2010 16:22 love1another wrote:
DO NOT I REPEAT DO NOT POST ANYTHING ELSE BESIDES QUESTIONS OR ATTEMPTED SOLUTIONS TO SAT PROBLEMS.
what he said.
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On September 05 2010 15:48 micronesia wrote:
When an integer H is divided by 2 times the sum of its factors, the quotient is 5 and the remainder is 1. What is the value of H?
10, 11, 17, 20, 24
When you say "2 times the sum of its factors", are you ignoring the number H as a factor of itself (1xH = H)? I think you must, or else "H being divided by 2 times the sum of its factors" will necessarily be a fraction smaller than 1 :-/
Assuming you are, the answer is 11. 11/(2x1) = 5R1. 1 is the only other factor of 11 besides 11, and that gives you a quotient of 5 with a remainder of 1.
By the way, I don't think this would appear on an actual SAT... and I've taught SAT Math prep for 3 years now. Did you get this from an actual exam? (It's far too ambiguous to be considered a "good" question, IMO.)
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United States24698 Posts
On September 05 2010 16:38 DarkPlasmaBall wrote:Show nested quote +On September 05 2010 15:48 micronesia wrote:
When an integer H is divided by 2 times the sum of its factors, the quotient is 5 and the remainder is 1. What is the value of H?
10, 11, 17, 20, 24 When you say "2 times the sum of its factors", are you ignoring the number H as a factor of itself (1xH = H)? I think you must, or else "H being divided by 2 times the sum of its factors" will necessarily be a fraction smaller than 1 :-/ Assuming you are, the answer is 11. 11/(2x1) = 5R1. 1 is the only other factor of 11 besides 11, and that gives you a quotient of 5 with a remainder of 1. By the way, I don't think this would appear on an actual SAT... and I've taught SAT Math prep for 3 years now. Did you get this from an actual exam? (It's far too ambiguous to be considered a "good" question, IMO.)
From some dumb sat website lol... so you are probably right.
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EPT
