The Ultimate Q51 Guide [Expert Level]

When n is divided by 4, the remainder is 3. If 3^n+2 is divided by 5, what is the remainder?

A. 1
B. 2
C. 3
D. 4
E. 0

==> You get n=4p+3, which becomes 3^n+2=3^{4p+3}+2=(3^4)^p(3^3)+2=(~1)^p(27)+2=(~1)(27)+2=(~7)+2=~9. Thus, the remainder when it is divided by 5 becomes 4.

The answer is D.
Answer: D

What is the sum of all digits of 2^{13}5^7?

A. 3
B. 4
C. 6
D. 8
E. 10

==> You get 2^{13}5^7=2^6(2^75^7)=64(10^7)=640,000,000. The sum of the 0's at the back always becomes 0, so you get 6+4=10.

The answer is E.
Answer: E

Is p/m>0?

1) p>m
2) pm>0.

==> If you modify the original condition and the question, from is p/m>0?, you multiply m^2 on both sides, you get (Squared number is always positive, so even if you multiply, the inequality sign doesn't change) p/m(m^2)>0(m^2)?, which becomes pm>0?.

The answer is B.
Answer: B

If a, b, and c are positive integers, is (a+b)c divisible by 3?

1) 2-digit integer ab is divisible by 3.
2) When c is divided by 3, the remainder is 0.

==> If you modify the original condition and the question, in order to have (a+b)c to be divided by 3, a+b or c has to be divided by 3. However, if you look at con 2), c is divisible by 3, hence it is yes and sufficient. For con 1), ab is also divisible by 3, and thus a+b is also divisible by 3, hence it is yes and sufficient. Therefore, the answer is D. This type of question is a 5051-level question which applies CMT 4 (B: if you get A or B too easily, consider D).

Answer: D

There are 5 apples in a bag. 4 apples are good but 1 apple is rotten. If you take out 2 apples from the bag, what is the probability that 1 apple selected is rotten?

A. 1/3
B. 1/4
C.2/5
D. 3/5
E.1/7

==> From probability=want/total, want=selecting one rotten apple and one good apple, and total=selecting 2 of 4 apples, so you get
probability=4C1*1C1/5C2=(4)(1)/(5*4/2!)=4/10=2/5.

The answer is C.
Answer: C

There is a sequence of r, t, s, 1, 4, 3, 8, 15, 26. Each number of the sequence is the sum of the 3 previous numbers. r=?

A. -2
B. -1
C. 0
D. 1
E. 2

==> You get r+t+s=1, from t+s+1=4, you get t+s=3, and from r+t+s=r+3=1, you get r=1-3=-2.

The answer is A.
Answer: A

When A works alone, it takes 14hrs, and when A works with B together, it takes 10hrs. How many hours does it take B to work alone?

A. 30hrs
B. 33hrs
C. 35hrs
D. 37hrs
E. 40hrs

==> For work rate questions, you solve “together and alone” reciprocally. It takes A 14hrs alone, and if you set the hours it took B to work alone as B hrs, from (1/14)+(1/B)=1/10 and 1/B=(1/10)-(1/14)=1/35, you get B=35.

The answer is C.
Answer: C

If y≠0, is |x/y|=x/y?
1) |xy|=xy
2) |x/y|=|x|/|y|

==> If you modify the original condition and the question, in order to get |x/y|=x/y, you get x/y≥0?, and if you multiply y^2 on both sides, you get xy≥0?. Then, for con 1), to satisfy |xy|=xy, you get xy≥0, hence yes, it is sufficient. For con 2), if x=y=2, yes, but if x=2 and y=-1, no, hence it is not sufficient.

The answer is A.
Answer: A

If x, y are positive integers, is xy=1?

1) x^2y^2=xy
2) 1/y=x.

==> 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 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), for con 2), you get xy=1, hence it is unique and sufficient. For con 1), from (xy)^2-xy=0, xy(xy-1)=0, you get xy=0,1, but since x and y are positive, you get xy=1, and thus con 1) = con 2).

The answer is D.
Answer: D

What is the remainder, when n(n+2) is divided by 24 for a positive integer x?

1) n is an even integer
2) n has remainder 0 or 1 when it is divided by 3.

=>Condition 1)
There are two kinds of even integers, which are n = 4k or n = 4k + 2 for some integer k. That means n could have 0 or 2 as a remainder when it is divided by n. If n = 4k, n(n+2) = 4k(4k+2) = 8k(2k+1) and n(n+1) is a multiple of 8. If n= 4k+2, n(n+2) = (4k+2)(4k+2+2) = 2(2k+1)*4(k+1) = 8(2k+1)(k+1) and n(n+1) is a multiple of 8. For both cases, n(n+1) is a multiple of 8. However, we can’t identify the remainder when it is divided by 3 from the condition 1).

Condition 2)
n = 3k or n = 3k +1. If n = 3k, n(n+2) = 3k(3k+2) is a multiple of 3. If n = 3k + 1, n(n+2) = (3k+1)(3k+3) = 3(3k+1)(k+1) is a multiple of 3. Thus, for both cases, n(n+1) is a multiple of 3. However, we can’t identify the remainder when it is divided by 8 from the condition 2).

Condition 1) & 2)
From the condition 1), n(n+1) is a multiple of 8. And n(n+1) is a multiple of 3 from the condition 2). Therefore, n(n+1) is a multiple of 24 from the both conditions 1) & 2) together.

Ans: C

Is x=y?

1) x^4+y^4=2x^2y^2
2) x^4+y^4=0

==> 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 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), for con 1), from x^4+y^4-2x^2y^2=0, (x^2-y^2)2=0, you get x^2=y^2, and from x=±y, yes and no coexists, hence it is not sufficient. For con 2), you only get x=y=0, hence yes, it is sufficient.

The answer is B.
Answer: B

If 10^n< 0.000025 <10^{n+1}, what is the value of an integer n?

A. -5 B. -4 C. -3 D. 4 E. 5

=>0.00001 < 0.000025 < 0.0001
10^{-5} < 0.000025 < 10^{-4}

Ans: A

Is the total profit from the sales of 3 products greater than $6?

1) The least profit from the sales of the 3 products is at least $2.1
2) The 2nd largest profit from the sales of the 3 products is at least $3.1.

=>Condition 1)
Since the minimum profit of 3 products is $2.1, the total profit of them is greater than or equal to $2.1 x 3 = $6.3, which is greater than $6.
Thus this is sufficient.

Condition 2)
Since the 3nd largest profit is at least $3.1, the sum of the first and the second products is at least $6.2, which is greater than $6.
Thus this is sufficient too.

Ans: D

In a certain conference, if all the n attendees shake hands 105 times, what is the value of n?

A. 10 B. 12 C. 15 D. 16 E. 18

=>
nC2 = 105.
n(n-1)/(1*2) = 105
n(n-1) = 210
n(n-1) = 15*14

Thus n = 15.

Ans: C

If x-7=√x+√7 , x=?

A. 8+2√7 B. 8-2√7 C. 8+√7 D. 8-√7 E. 7+2√8

=>x-7=√x+√7
(√x+√7 )(√x-√7)=√x+√7
√x-√7=1
√x = 1 +√7
x = (1 +√7)2 = 8 +2√7

Ans: A

If 11!/(11-r)!<1,000, what is the greatest possible value of r?

A. 1
B. 2
C. 3
D. 4
E. 5


=> 11! / (11-r)! = 11*(11-1)*…*(11-r+1)

r = 1 : 11 < 1,000
r = 2 : 11*10 = 110 < 1,000
r = 3 : 11*10*9 = 990 < 1,000
r = 4 : 11*10*9*8 = 7290 > 1,000

r = 3

Ans: C

Is a positive integer x an odd number?

1) The smallest prime factor of x is 7.
2) The greatest prime factor of x is 7.

=>Condition 1)
2 cannot be a factor of x, since the smallest prime factor of x is 7.

Condition 2)
Consider n = 2 and n =3.


Ans: A