Number Theory: Remainders
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1. If n is a positive integer and 18n+12 is divided by 9, what is the remainder?

(A) 2
(B) 3
(C) 2n
(D) 2n+3
(E) cannot be determined

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Dividing 18n by 9 doesn't produce a remainder. Dividing 12 by 9 produces a remainder of 3. The result will be 2n+1 with remainder 3 so the answer is (B).

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2. If today is Monday, what day will it be 100 days from now?

(A) Monday
(B) Tuesday
(C) Wednesday
(D) Thursday
(E) Friday

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In 7 days it will be Monday again. Dividing 100 by 7 gives 14 with a remainder of 2. So in 100 days you'll count 14 more Mondays and it will be Wednesday (2 days from Monday). The answer is (C).

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3. If n is divided by 90, the remainder is 1. What are the remainders, respectively if n+179 and n+182 are each divided by 90?

(A) 89 and 3
(B) 89 and 2
(C) 0 and 3
(D) 0 and 2
(E) 2 and 3

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Dividing n by 90 gives some number of 90's and a remainder of 1. Dividing 179 by 90 gives one 90 and a remainder of 89. When you put the remainders together you get another 90 so the remainder is zero (for n+179). In the case of n+182 you have a remainder of 1 and a remainder of 2 so the total remainder is 3. The answer is (C).

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4. If n=550 how many different remainders result when n, n+7, n+11, n+18, n+21, n+29, and n+32 are each divided by 11?

(A) 3
(B) 4
(C) 5
(D) 6
(E) 7

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All that matters is "how many 11's do you have?" You know that n+18 has exactly one more 11 than n+7 so they have the same remainder along with n+29 which has still another 11. Also, n+21 and n+32 have the same remainder because they are 11 apart. So there are three different remainders and the answer is (A).

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5. When positive integer n is divided by 5, the remainder is x. When 2n is divided by 5, the remainder is y. Which pair (x, y) is not possible?

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

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If you divide by 5, there are 5 possible remainders: 0, 1, 2, 3, or 4. If you double the number you double the number of 5's and you double the remainder to 0, 2, 4, 6, or 8. But wait, if the remainder is 6 or 8 you get another 5 so the remainders for 2n are actually: 0, 2, 4, 1, and 3. So the pair of remainders could be: (0,0) or (1,2) or (2,4) or (3,1) or (4,3). Choice (E) is not possible so this is the answer.

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Questions 6-8: The modulus is defined as follows:
j mod k = R where R is the remainder when positive integer j is divided by positive integer k.

6. If j mod k = R which statement is true for any j and k?

(A) 0 ≤ R ≤ j–1
(B) R ≥ k–1
(C) R ≥ j–1
(D) 0 ≤ R ≤ k–1
(E) k–1 ≤ R ≤ j–1

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You are dividing by k. If k were 5, the possible remainders would be 0, 1, 2, 3, or 4. The remainder R is anywhere from 0 to k–1. The answer is (D).

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7. If j mod k = R and j < k then:

(A) R = j
(B) R = k
(C) R = k–j
(D) R = k+j
(E) R = 0

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Now you are dividing a small number (j) by a big number (k). Suppose you do 7 divided by 25. The answer in "remainder language" is 0 with remainder 7. Whenever you divide a small number by a big number the remainder is always the small number. The answer is (A).

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8. If j mod k = R and k < j < 2k then:

(A) R = j
(B) R = k
(C) R = j–k
(D) R = j+k
(E) R = 0

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Now you are doing something like 40 (j) divided by 25 (k) (j is bigger than k but smaller than 2k). There is one 25 in 40 and the remainder is 15 or j-k. The answer is (C). Note that it is very useful on SAT problems to work with a concrete example. You are NOT writing your Ph.D. thesis in mathematics; you just want the answer.

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9. A positive integer N is divided by 23 and the remainder is 11. What is the remainder if 4N is divided by 23?

(A) 11
(B) 15
(C) 19
(D) 21
(E) 22

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When you divide N by 23 you get some number of 23's with 11 leftover. If you divide 4N by 23 you get 4 times as many 23's with 44 (11 times 4) leftover. But the 44 has a 23 in it so your actual remainder is 21. The answer is (D).

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10. The integer m is between 0 and 100 inclusive. When m is divided by 6, the remainder is 5. How many possible values of m are there?

(A) 15
(B) 16
(C) 17
(D) 18
(E) 19

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For the smallest value of m, when you divide by 6, you get "0 R 5" (that is, zero sixes with a remainder of five). The largest value of m (under 100) when divided by 6 gives "15 R 5." You can't go further because 16 sixes with a remainder of 5 would be 101. Counting from 0 R 5 to 15 R 5, you get 16 numbers. So the answer is (B).

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