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.

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) $\ \ 36$

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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).

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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) $\ \ 32$

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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).

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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: $$\begin{array}{c}&(7, 24), &(16, 63), &(20, 21), &(28, 45), &(44, 117)? \end{array}$$

(A) $\ \ 1$
(B) $\ \ 2$
(C) $\ \ 3$
(D) $\ \ 4$
(E) $\ \ 5$

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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).

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Incorrect Choices:

(A), (C), (D), and (E)
See the solution for why these choices are wrong.

• SAT General Tips