SOLUTIONS TO ARITHMETIC PROBLEMS

Problem 16

What is the greatest integer that will always evenly divide the sum of three consecutive even integers?A) 2B) 3C) 4D) 6E) 12

• 2n + (2n+2) + (2n + 4) = 6n + 6 = 6(n + 1).• 6 is the greatest integer that will always evenly divide the

sum of three consecutive even integers.

Problem 17

The sum of three consecutive integers is 312. What is the sum of the next three consecutive integers?A) 315B) 321C) 330D) 415E) 424

• Let the first three consecutive integers be n, (n+1) and (n+2). The sum of them is n + (n+1) + (n+2) =3n + 3 = 312.

• The next three consecutive integers are (n+3), (n+4) and (n+5).

• So, their sum is (n+3) + (n+4) + (n+5) = 3n + 12. Since 3n + 12 is 9 greater than 3n + 3, and 312 + 9 = 321, 3n + 12 = 321.

Problem 18

The integer P is greater than 7. If the integer P leaves a remainder of 4 when divided by 9, all of the following must be true EXCEPT

A) The number that is 4 less than P is a multiple of 9.B) The number that is 5 more than P is a multiple of 9.C) The number that is 2 more than P is a multiple of 3.D) When divided by 3, P will leave a remainder of 1.E) When divided by 2, P will leave remainder of 1.

• Choice (A) is always true. Since P has a remainder of 4, P is 4 greater than some multiple of 9. So, P – 4 is a multiple of 9.

• Therefore, the next multiple of 9 would be (p – 4) + 9 , or P + 5; thus, choice (B) is also correct.

• With choice (C), we know that since P – 4 is a multiple of 9, it is also a multiple of 3. So, by adding 3s to P – 4, we know that we will get other multiples of 3. So, (P – 4) + 3 = P – 1, and (P – 4) + 3 + 3 = P + 2, are also multiples of 3, so choice (C) must be true.

• And, since P – 1 is a multiple of 3, when P is divided by 3, it will have a remainder of 1, and so, Choice (D) is always true.

Problem 20

How many positive integers less than 100 have a remainder of 4 when divided by 8?

• To leave a remainder of 4 when divided by 8, an integer must be 4 more than a multiple of 8. So, we take the multiples of 8 and add 4. (Don’t forget that 0 is a multiple of 8).

• 0 x 8 + 4 = 4• 1 x 8 + 4 = 12• 2 x 8 + 4 = 20• 3 x 8 + 4 = 28• ….…• Instead of continuing this procedure, you can add 8 to each

of the numbers that you get to obtain the next number. So, after 28, we will get 36, 44, 52, 60, 68, 76, 84, and 92.

• If we count all of these numbers (i.e., 4, 12, 20, 28, 36, 44, 52, 60, 68, 76, 84, and 92) we get 12. So, the answer is 12.

Problem 22

If q is an odd integer greater than 1, what is the value of + 1?

A) – 2B) – 1C) 0D) 2E) It cannot be determined from the information given.

• Since q is always an odd integer, is always equal to -1. So, the answer is Choice (C), or 0.

Problem 23

Problem 27

Since is greater than , (B) is correct.

QA = = =

QB = = =

SOLUTIONS TO GEOMETRY PROBLEMS

Problem 14A line segment joining two points on the circumference of a circle is one inch from the center of the circle at its closest point. If the circle has a two-inch radius, what is the length of the line?A) 1B) √2C) 2D) 2√2E) 2√3

Sketch a diagram:

Since the radius of a circle is 2, the end-points of the line are both 2 inches from the center. The line can be seen as the legs of two right triangles, each of which has a hypotenuse of 2 and a leg of 1. Each of the legs that make up the line must have a length equal to -, or 3. The total length of the line is twice this, or 23.

Problem 15

In the figure below, O is the center of the circle. If AB has a length of 16 and OB has a length of 10, what is the length of CD?A) 2B) 4C) 2√3D) 8 - √35E) 8 - √39

• Note that OC is a radius of the circle, and if we knew the length of OC and OD, we could find CD, since CD = OC-OD. We are given that OB is 10, so the radius is 10, and therefore, OC is 10.

• Now we need to find OD. If we connect O and A, then we create two right triangles, ADO and BOD.

• Since both of these triangles have a radius as their hypotenuse, and both have a leg in common (OD), then they must be equal in size. Therefore, the other legs, AD, and DB, must also be equal. That means that D is the midpoint of AB, and so DB is (16) = 8.

• Considering right triangle BDO, we have a hypotenuse of 10 and a leg of 8; thus, the other leg has length 6. (It’s a 6-8-10 Pythagorean Triplet). So, OD has length 6, and CD = 10 – 6 = 4.

Problem 16

Problem 19

In the figure below, the smaller circle is inscribed in the square and the square is inscribed in the larger circle. If the length of each side of the square is s, what is the ratio of the area of the larger circle to the area of the smaller circle?A) 2 : 1B) 2 : 1C) : 1D) 2s : 1E) s : 1

• Let the radius of the smaller circle be . Then, the area of the smaller circle is =

• The diameter of the smaller circle (that is 2) is equal to the side of the square, or s.

• So, s = 2. Also, the diagonal of the square is =2.• Since the diameter of the larger circle is equal to the

diagonal of the square, the diameter of the larger circle is = 2 = =2. So, = = = .

• The area of the larger circle is = = =2 • = = • So, the ratio is 2:1.