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 theSAT 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)
为什么这些解决方案的选择是错误的。

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)
为什么这些解决方案的选择是错误的。

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)
为什么这些解决方案的选择是错误的。

• SAT General Tips