SAT Reasoning Perfect Score
To get a perfect score on SAT Math, you need to
- get every single problem correct;
- have complete mastery of all of the SAT skills;
- remember the tips and use them;
- figure out your common mistakes and avoid them.
SAT Hardest Problems
How many triples of integers \((a, b, c)\) are there such that \[ a \times b \times \sqrt{c} = 6 ?\]
(A) \(\ \ 12\)
(B) \(\ \ 18\)
(C) \(\ \ 24\)
(D) \(\ \ 30\)
(E) \(\ \ 36\)
Correct Answer: B
Solution:
We check all possible cases as follows:
- If \( a = 6 \), then \((b, c)=(1, 1),\) which gives \(1\) solution.
- If \( a = 3 \), then \((b, c)=(2, 1)\) or \((1, 4),\) which gives \(2\) solutions.
- If \( a = 2 \), then \((b, c)=(3, 1)\) or \((1, 9),\) which gives \(2\) solutions.
- If \( a = 1 \), then \((b, c)=(6, 1)\) or \((3, 4)\) or \((2, 9)\) or \( (1, 36),\) which gives \(4\) solutions.
- If \( a = -1 \), then \((b, c)=(-1, 36)\) or \((-2, 9)\) or \((-3, 4)\) or \( (-6, 1),\) which gives \(4\) solutions.
- If \( a = -2 \), then \((b, c)=(-1, 9)\) or \( (-3, 1),\) which gives \(2\) solutions.
- If \( a = -3 \), then \((b, c)=(-1, 4)\) or \( (-2, 1),\) which gives \(2\) solutions.
- If \( a = -6 \), then \((b, c)=(-1, 1),\) which gives \(1\) solution.
Hence, there are a total of \(1+2+2+4+4+2+2+1=18 \) solutions.
Thus, the correct answer is (B).
Incorrect Choices:
(A), (C), (D), and (E)
See the solution for why these choices are wrong.
How many triples of integers \((a, b, c)\) are there such that \[ a \times \sqrt{b \times c} = 6 ?\]
(A) \(\ \ 24\)
(B) \(\ \ 26\)
(C) \(\ \ 28\)
(D) \(\ \ 30\)
(E) \(\ \ 32\)
Correct Answer: E
Solution:
We check all possible cases as follows:
- If \( a = 6 \), then \((b, c)=(1, 1)\) or \((-1, -1),\) which gives \(2\) solutions.
- If \( a = 3 \), then \((b, c)=(4, 1)\) or \((2, 2)\) or \((1, 4)\) or \((-1, -4)\) or \((-2, -2)\) or \((-4, -1),\) which gives \(6\) solutions.
- If \( a = 2 \), then \((b, c)=(9, 1)\) or \((3, 3)\) or \((1, 9)\) or \((-1, -9)\) or \((-3, -3)\) or \((-9, -1),\) which gives \(6\) solutions.
- If \( a = 1 \), then \((b, c)=(36, 1)\) or \((18, 2)\) or \((12, 3)\) or \((9, 4)\) or \((6, 6)\) or \((4, 9)\) or \((3, 12)\) or \((2, 18)\) or \((1, 36)\) or \((-1, -36)\) or \((-2, -18)\) or \((-3, -12)\) or \((-4, -9)\) or \((-6, -6)\) or \((-9, -4)\) or \((-12, -3)\) or \((-18, -2)\) or \((-36, -1),\) which gives \(18\) solutions.
Hence, there are a total of \(2+6+6+18=32 \) solutions.
Thus, the correct answer is (E).
Incorrect Choices:
(A), (B), (C), and (D)
See the solution for why these choices are wrong.
If \(a^2+b^2=c^2,\) where \(a, b\) and \(c\) are positive integers, then \(c\) has at most \(3\) positive divisors.
How many of the following pairs \((a, b)\) are counter-examples to the above claim: \[\begin{array} &(7, 24), &(16, 63), &(20, 21), &(28, 45), &(44, 117)? \end{array}\]
(A) \(\ \ 1\)
(B) \(\ \ 2\)
(C) \(\ \ 3\)
(D) \(\ \ 4\)
(E) \(\ \ 5\)
Correct Answer: B
Solution:
We get \(c\) for each pair of \((a, b)\) and factorize the integer \(c\) as follows to see how many divisors it has:
- For \((7, 24),\) we have \(7^2+24^2=25^2,\) which implies \(c=25.\) Since \(25=5^2\) has \(3\) divisors \(1, 5\) and \(25,\) this is not a counter-example.
- For \((16, 63),\) we have \(16^2+63^2=65^2,\) which implies \(c=65.\) Since \(65=5\times 13\) has \(4\) divisors \(1, 5, 13\) and \(65,\) this is a counter-example.
- For \((20, 21),\) we have \(20^2+21^2=29^2,\) which implies \(c=29.\) Since \(29\) is a prime and has \(2\) divisors \(1\) and \(29,\) this is not a counter-example.
- For \((28, 45),\) we have \(28^2+45^2=53^2,\) which implies \(c=53.\) Since \(53\) is a prime and has \(2\) divisors \(1\) and \(53,\) this is not a counter-example.
- For \((44, 117),\) we have \(44^2+117^2=125^2,\) which implies \(c=125.\) Since \(125=5^3\) has \(4\) divisors \(1, 5, 25\) and \(125,\) this is a counter-example.
Thus, \((16, 63)\) and \((44, 117)\) are counter-examples, and the correct answer is (B).
Incorrect Choices:
(A), (C), (D), and (E)
See the solution for why these choices are wrong.