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1) n is the product of the 4 consecutive integers 2) n is a multiple of 12

==> In the original condition, there is 1 variable, and in order to match the number of variables to the number of equations, there must be 1 more equation. Since there is 1 for con 1) and 1 for con 2), D is most likely to be the answer. For con 1), the product of consecutive 4 number is always the multiple of 24, hence yes, it is sufficient. For con 2), if it is a multiple of 12, it is also a multiple of 6, hence yes, it is sufficient. Therefore, D is the answer.

==> In the original condition, there are 2 variables (x,y), and in order to match the number of variables to the number of equations, there must be 2 variables. Since there is 1 for con 1) and 1 for con 2), C is most likely to be the answer. By solving con 1) and con 2), from x>|y|≥0, it is x>0, so from x>|y|, it is |x|>|y|, which becomes x^2>y^2, hence yes, it is sufficient. The answer is C. However, this is an absolute value question, one of the key questions, so you apply CMT 4 (A: if you get C too easily, consider A or B). For con 1), you get (x,y)=(2,1) yes (2,-3) no, hence not sufficient. For con 2), from x>|y|≥0, you get x>0, which becomes |x|>|y|, then x^2>y^2, it is always yes and sufficient. Therefore, the answer is B.

If m and n are positive integers, is mn a multiple of 9?

1) m+n is a multiple of 3 2) mn is a multiple of 3

==> In the original condition, there are 2 variables (m,n) and in order to match the number of variables to the number of equations, there must be 2 equations. Since there is 1 for con 1) and 1 for con 2), C is most likely to be the answer. By solving con 1) and con 2), from the multiple of mn=3, it becomes a multiple of m or n=3. According to con 1), you always get a multiple of m=n=3, hence yes, it is sufficient.

We define the average (harmonic mean) as the reciprocal of the average (arithmetic mean) of reciprocals. What is the average (harmonic mean) of 2, 3, and 6?

A. 1/3 B. 1/2 C. 1 D. 2 E. 3

==> If you find the average of the reciprocals, you get \(\frac{++}{3}=13\) . Since it is the reciprocal of that, the answer is E.

==> In the original condition, there are 3 variables (x,y,z) and in order to match the number of variables to the number of equations, there must be 3 equations. Since there is 1 for con 1) and 1 for con 2), E is most likely to be the answer. By solving con 1) and con 2), if x=y=z=1, yes, but if x=y=x=-1, no, hence it is not sufficient.

Therefore, the answer is E. Answer: E
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Which of the following is the closest value of x such that \(x^{20}+x^2+0.0000019=0.09?\)

A. 0.1 B. 0.01 C. 0.3 D. 0.03 E. 0.003

==> From \(x^{20}+x^2+0.0000019=0.09, x^{20}\) and 0.0000019 are very close to 0, so you can ignore them, and the equation becomes \(x^2=0.09=0.3^2\). Thus, you get x=0.3.

==> In the original condition, there are 3 variables (a, b, c) and in order to match the number of variables to the number of equations, there must be 3 equations. Since there is 1 for con 1) and 1 for con 2), E is most likely to be the answer. By solving con 1) and con 2), if discriminant =b2-4ac<0, it doesn’t meet with the x-axis, and if a<0, you always get f(x)<0, hence yes, it is sufficient.

Therefore, the answer is C. Answer: C
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When a positive integer n is divided by 13, what is the remainder?

1) n+1 is divisible by 13 2) n+14 is divisible by 13

==> In the original condition, there is 1 variable, and in order to match the number of variables to the number of equations, there must be 1 equation. Since there is 1 for con 1) and 1 for con 2), D is most likely to be the answer. By solving con 1) and con 2), you get con 1) = con 2), and you use direct substitution for the rest, so you get n=12,25,38,… When you divide by 13, the remainder always becomes 12, hence it is unique and sufficient.

If the sum of the first 50 even numbers is 2,550, what is the sum of the first 50 odd numbers?

A. 1,275 B. 2,550 C. 2,500 D. 2,600 E. 3,000

==> You get 2+4+…..+98+100=2,550, then 1+3+….+97+99=?. If you compare each numbers, 1 is 1 less than 2, 3 is 1 less than 4, 97 is 1 less than 98, and 99 is 1 less than 100. Since there are 50 numbers in total, it is 50 less than 2,550.

Therefore, the answer is C. Answer: C
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==> In the original condition, there are 2 variables (a,b) and in order to match the number of variables to the number of equations, there must be 2 equations. Since there is 1 for con 1) and 1 for con 2), C is most likely to be the answer. By solving con 1) and con 2), you get |a^2-b^2|=|a+b||a-b|=20, 10|a-b|=20, |a-b|=2.

==> In the original condition, there are 2 variables (a, b) and in order to match the number of variables to the number of equations, there must be 2 equations. Since there is 1 for con 1) and 1 for con 2), C is most likely to be the answer. By solving con 1) and con 2), from -1<b<1, if a=-0.1 and b=-0.2, no, but if b=0.1 and a=-0.1, yes.

Hence, the answer is E. Answer: E
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If x and y are integers, what are the values of x and y?

1) 2^x3^y=16/27 2) x+y=1

==> For con 1), you get 2^x*3^y=16/27=2^4*3^-3, then x=4 and y=-3. Therefore, the answer is A. In other words, it is CMT 4(A), in which A and C are both the answers. In the original condition, there are 2 variables, so the answer is C, but since it is an integer question, you apply CMT 4(A) and get the final answer as A.

This is a 4950 level question. Answer: A
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If x and y are positive integers and y(2^x)=24, x=?

1) x 2 2) y is even

==> In the original condition, there are 2 variables (a,b), and in order to match the number of variables to the number of equations there must be 2 equations as well. Since there is 1 for con 1) and 1 for con 2) C is most likely to be the answer. By solving con 1) and con 2), you get 24=6(2^2).

==> In the original condition, there are 2 variables (x,y) and 1 equation (x+y=integer). In order to match the number of variables to the number of equations, there must be 1 more equation. Since there is 1 for con 1) and 1 for con 2), D is most likely to be the answer. For con 1), from x=integer and x+y=integer, integer+y=integer, you get y=integer, which is yes and sufficient. For con 2), from x+2y=x+y+y=integer, integer+y=integer, you get y=integer, which is also yes and sufficient.

Therefore, the answer is D. Answer: D
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==> In the original condition, there are 2 variables (m,r) and in order to match the number of variables to the number of equations, there must be 2 equations. Since there is 1 for con 1) and 1 for con 2), C is most likely to be the answer. By solving con 1) and con 2), you get con 1) = con 2), so from m/r+1=3, you get m/r=2, hence it is unique and sufficient.

Therefore, the answer is D. Answer: D
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==> If you modify the original condition and the question, when xyz≠0, you get xz<0?. There are 3 variables (x,y,z), and in order to match the number of variables to the number of equations, there must be 2 equations. Since there is 1 for con 1) and 1 for con 2), E is most likely to be the answer. By solving con 1) and con 2), you get x^2y<0, y<0 and yz<0, z>0. Since x is unknown, it is not sufficient.

Therefore, the answer is E. Answer: E
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If x>0, y<0, and a>b>0, then which of the following is (are) positive? Ⅰ. ax+by Ⅱ. ax-by Ⅲ. by-ax

A. Ⅰonly B. Ⅱ only C. Ⅲ only D.Ⅰ& Ⅱ only E. Ⅱ & Ⅲ only

==> You getⅠ. ax+by a=2, b=x=1, y=-10, which is negative. (X) From Ⅱ. ax-by by<0, you get –by>0 and ax>0, ax-by>0, hence it is always positive. (O) From Ⅲ. by-ax a=2, b=1, x=1, y=-10, it is negative. (X)

Therefore, the answer is B. Answer: B
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