Sequences: Counting
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1. What is the last digit in 7503?

(A) 1
(B) 3
(C) 5
(D) 7
(E) 9

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The last digit in powers of 7, starting with 71 are as follows: 7, 9, 3, 1, 7, 9, 3, 1 . . . etc. You don't have to calculate the full number to know what the last digit is. This pattern repeats in groups of 4 and 500/4 = 125 groups. At 7501 we start over again (with the 126th group) and the last digit is a 7. The last digit in 7503 is 3. The answer is (B).

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2. Bob and Carol and Ted and Alice see 3 silver dollars lying on the street. They run for the money and a fight ensues. At the end of the battle, all three silver dollars are spoken for. One possible outcome is Bob has 2 of the silver dollars, Carol has 1, and Ted and Alice have none. How many such possible outcomes are there?

(A) 12
(B) 20
(C) 24
(D) 64
(E) 81

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Let's say 3000 means Bob gets 3 and everyone else gets none. For Bob gets 2 you have: 2100, 2010, 2001. For Bob gets 1 you have: 1200, 1110, 1101, 1020, 1011, 1002. For Bob gets none you have: 0300, 0210, 0201, 0120, 0111, 0102, 0030, 0021, 0012, 0003. The total is 20 or (B).

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3. One four-person team will be created from 4 boys and 4 girls. If the team must have two boys and two girls, how many different possible teams might be created?

(A) 16
(B) 32
(C) 36
(D) 48
(E) 64

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When picking a team from a group you don't want to count the team of John and Bob and the team of Bob and John as two seperate teams. Picking 2 people from 4 you multiply 4 by 3 (4 choices for the first team member and 3 choices for the second) and divide by 2 so you don't double count. So there are 6 teams of 2 that you can pick from 4 people (if you are picking a president and a vice president instead of a team, there would be 12 ways to pick because Bob-John and John-Bob would be different). There are also 6 teams you can pick from the 4 girls. Next you have to pick from two groups. Any one of the 6 boy teams can be paired with any one of the six girl teams for a total of 6 times 6 = 36 different teams. The answer is (C).

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4. Alice, Bob, Carol, David, and Ted are going to sit down in five chairs numbered 1 - 5. The chairs are bolted to the floor and cannot be moved. Ted is superstitious and will not sit in chair #3 since he notes that both ETS and SAT have three letters and has concluded, possibly erroneously, that 3 is an unlucky number. If each person chooses one chair and Ted does not choose chair #3, in how many ways may the five people arrange themselves among the five chairs?

(A) 60
(B) 75
(C) 90
(D) 96
(E) 105

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You do the problem 4 times, once for each chair Ted is willing to sit in. If Ted picks the first chair, there are 4! (4 factorial or 4·3·2·1) = 24 ways the remaining people can sit. The same is true if Ted picks the second, fourth, or fifth chair. So the answer is 24 times 4 or 96 ways or (D).

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5. A person's birthday could be on any of 366 days including leap year, February 29th. How many people must gather in a room in order for you to be 100% sure that at least two of them have the same birthday? Assume that you don't know the birthday of anyone in the room and that you are not part of the group.

(A) 183
(B) 184
(C) 366
(D) 367
(E) more than 367

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SATan throws in the occasion puzzle like this one. The worst you could do is pick 366 people each of whom had a different birthday. Your 367th person would have to match one of the previous people. The answer is (D).

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6. The fraction 1/7 is equal to 0.142857142857. . . The first digit after the decimal point is 1. What is the sum of the first 100 digits after the decimal point?

(A) 432
(B) 435
(C) 440
(D) 445
(E) 447

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Each group of 6 has a sum of 1+4+2+8+5+7=27. There are 16 full groups in 100 digits and 16 times 27 is 432. The last four digits, 1+4+2+8 add an extra 15 so the answer is 447 or (E).

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7. In the sequence 1, 2, 3, 1, 2, 3, 1, 2, 3, . . . the first digit is a 1, the second digit is a 2, the third digit is a 3, the fourth digit is a 1, etc. A subset S of this sequence consists of every tenth digit starting with the 10th digit, the 20th digit, and the 30th digit up to and including the 1000th digit. There are 100 digits in subset S. What is the sum of the digits in subset S?

(A) 190
(B) 191
(C) 199
(D) 209
(E) 210

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The 10th digit is a 1, the 20th digit is a 2, the 30th digit is a 3, the 40th digit is a 1 again. So you are just adding 1+2+3+1+. . . for 100 digits. Groups of 3 repeat and the sum of each group is 6. There are 33 complete groups of 3 for a total of 33 times 6 is 198. That's 99 digits. The final digit is a 1 so the sum is 199. The answer is (C).

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8. From the (fictional) ETS headquarters, there are three underground roads that lead directly to the big boss's office. From this office, there are three roads that lead further underground to the ETC (Eternal Testing Center) where bad people spend eternity taking the SAT. How many different routes are there from ETS headquarters to the ETC and back if no route can use the same road twice?

(A) 9
(B) 18
(C) 27
(D) 36
(E) 45

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To get from ETS to the boss's office (it is NOT air-conditioned so dress lightly) you have 3 choices of route. From the boss's office to the Eternal Testing Center you have another 3 choices. On the way back you have 2 choices (since you can't repeat roads) and then 2 choices again. The total number of routes is 3·3·2·2 = 36 or (D).

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9. A teacher at the TSA (Testing School of America) finally gets fed up with giving meaningless tests. So she makes a deal with her twenty-student class. There will be no tests. At the end of the year, students will pick their grade out of a hat which will contain 20 slips of paper each with a different number between 81 and 100 (inclusive) written on it. The piece of paper you pick will be your grade for the year. Pieces of paper will NOT be returned to the hat after they are picked. You have a friend in the class whose name is Joe. What is the probability that you and Joe will be within one point of each other after you have each picked a grade?

(A) 1/10
(B) 1/19
(C) 1/20
(D) 1/21
(E) 1/30

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There are 20 times 19 divided by 2 equals 190 different pairs of grades assuming we don't count (83,84) and (84,83) as different. Nineteen of these pairs are within one point of each other. So the chances that you and Joe will be within one point is 1 in 10 or (A).

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10. Positive integer N is the product of four different prime numbers w, x, y, and z. How many different factors does N have (including itself and 1)?

(A) 6
(B) 12
(C) 16
(D) 20
(E) cannot be determined

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The factors are: 1, w, x, y, z, wx, wy, wz, xy, xz, yz, wxy, wxz, wyz, xyz, and wxyz. That's a total of 16 or (C). It's useful to be good at counting systematically. You don't do it much in regular math courses, but SATan loves counting.

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