Number Theory: Divisibility
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1. The integer n is a factor of the integer x and x>n>1. Which of the following must be true?

I. n is a factor of x + 15n
II. 5 is a factor of x + 15n
III. x is a factor of x + 15n

(A) none
(B) I, only
(C) II, only
(D) II and III, only
(E) I, II, and III

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We know n is a factor of x and n is a factor of 15n. Therefore n is a factor of x+15n. Since 5 may not be a factor of x and x may not be a factor of 15n, (II) and (III0 are not necessarily true. the answer is (B).

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2. The positive integer x is a product of three different prime numbers, p, q, and r. If r>q>p, which of the following must be true?

I. The greatest prime number that is a factor of x is r.
II. If p > 5 then x is not divisible by 5.
III. x2 is divisible by p, q, and r.

(A) III, only
(B) I and II, only
(C) I and III, only
(D) II and III, only
(E) I, II, and III

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Since x = pqr and p, q, and r are prime, there are no other prime factors of x besides p, q, and r (this is called the fundmental theorem of arithmetic). So (I) is true. Since p is the smallest prime factor, if it is bigger than 5 then none of the prime factors can be 5 so (II) is true. If x = pqr then x2=p·p·q·q·r·r which is obviously also divisible by p, q, and r. So (III) is also true and the answer is (E).

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3. If h, j, and k are prime numbers and h = 2 and x = hjk + 1 then which of the following is true?

(A) x is divisible by k
(B) x is divisible by 3
(C) x cannot be prime
(D) x must be prime
(E) none of these

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Choice A is not true because when you divide x by k, you get h·j with remainder 1. Choice B is not true because we don't know whether x is divisible by 3. We also don't know whether or not x is prime. You know that x is not divisible by h, j, or k but it could be divisible by some other prime. Or not. If h, j, and k are 2, 3, and 5 then x is 31 which is prime. If h, j, and k, are 2, 5, and 11 then x is 111 which is divisible by 3. There is no known formula that is guaranteed to produce a prime number. The answer is (E).

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4. The product of 3 positive integers, x, y, and z is 105. All three integers are greater than 1. The product of a fourth integer w, and x is 36. Therefore, x =

(A) 2
(B) 3
(C) 5
(D) 12
(E) 18

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If you factor 105 you get 3·5·7. Since wx = 36, we know that x can't be 5 or 7. So x is 3. The answer is (B).

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5. A number divisible by 1, 2, 3, 4, 5, and itself but not divisible by any other integers is called a "simple number." How many simple numbers below 100 are there?

(A) 0
(B) 1
(C) 2
(D) 3
(E) 4

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Trick question. You can't have a number divisible by 1, 2, 3, 4, 5 and itself and no other numbers. The number would have to be divisible by 6 also and by 10 and by 12 and by 15 and by 20. For example, 60 is divisible by 3, 4, and 5 (and therefore also by 2) but 60 has lots of factors. The answer is (A).

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6. If j and k are positive integers and j>k and both j and k are divisible by 2, 3, and 5, then the quantity (j–k) must be greater than or equal to:

(A) 15
(B) 30
(C) 45
(D) 60
(E) 65

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The smallest value for k is 30. Since j also has a 2, 3, and 5 in it, the smallest value for j is 60 (one extra 2). This means the smallest value for the difference is 30 or (B).

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7. If j and k are positive integers and j>k and both j and k are divisible by 2, 3, and 8, then the quantity (j–k) must be greater than or equal to:

(A) 8
(B) 16
(C) 24
(D) 48
(E) 56

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Now the smallest value for k is something with three 2's and one 3 or 24. Note that if j is divisible by 8 then it is automatically divisible by 2. Again, if you throw in an extra 2 you'll get the smallest value of j which is 48. The difference is 24 or (C).

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8. The quantity 12100 is NOT divisible by which of the following?

(A) 18
(B) 24
(C) 27
(D) 30
(E) 36

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The number 12 is 2·2·3 (when in doubt, factor). If you repeat this 100 times you get a long chain of 200 2's and 100 3's. But there are no 5's. Now 18 is made of 2's and 3's as is 24, 27, and 36. But 30 has a 5 in it which would never cancel with a string of 200 2's and 100 3's. The answer is (D).

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9. If k is a positive integer and y = 2k2+6k, then y is divisible by:

(A) 3
(B) k2
(C) (k+3)
(D) (2k+3)
(E) 12

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Factor to get 2·k·(k+3). So (k+3) is a factor and the answer is (C).

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10. If p, q, x, and y are positive integers and x>p>q>y>1 and p and q are prime, then which of the following must be true?

I. p/q is not an integer
II. x/p is not an integer
III. p/y is not an integer

(A) I, only
(B) II, only
(C) III, only
(D) I and II, only
(E) I and III, only

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Prime numbers like 17 and 5 never divide each other so (I) is true. Since p is smaller than x and x is not necessarily prime, (x could be 20 and p could be 5) x/p might be an integer so (II) is NOT true. Prime numbers like p never have any divisors so (III) is true. The answer is (E).

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