SAT Practice Exam

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Integer Sums

The sum of the positive odd integers less than 80 is subtracted from the sum of the positive even integers less than or equal to 80. What is the resulting difference?

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Answer
With problems like these, we need to consider the difference per each group of 2 terms

Display the first 80 even terms

Sum of the first 80 even terms = 2 + 4 + 6 + 8 + 10 + 12 + 14 + 16 + 18 + 20 + 22 + 24 + 26 + 28 + 30 + 32 + 34 + 36 + 38 + 40 + 42 + 44 + 46 + 48 + 50 + 52 + 54 + 56 + 58 + 60 + 62 + 64 + 66 + 68 + 70 + 72 + 74 + 76 + 78 + 80

Display the first 80 odd terms

Sum of the first 80 odd terms = 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21 + 23 + 25 + 27 + 29 + 31 + 33 + 35 + 37 + 39 + 41 + 43 + 45 + 47 + 49 + 51 + 53 + 55 + 57 + 59 + 61 + 63 + 65 + 67 + 69 + 71 + 73 + 75 + 77 + 79

The difference of the first 80 even terms minus the first 80 odd terms can be written as follows:

Group each set of odd and even terms

(2 - 1) + (4 - 3) + (6 - 5) + (8 - 7) + (10 - 9) + (12 - 11) + (14 - 13) + (16 - 15) + (18 - 17) + (20 - 19) + (22 - 21) + (24 - 23) + (26 - 25) + (28 - 27) + (30 - 29) + (32 - 31) + (34 - 33) + (36 - 35) + (38 - 37) + (40 - 39) + (42 - 41) + (44 - 43) + (46 - 45) + (48 - 47) + (50 - 49) + (52 - 51) + (54 - 53) + (56 - 55) + (58 - 57) + (60 - 59) + (62 - 61) + (64 - 63) + (66 - 65) + (68 - 67) + (70 - 69) + (72 - 71) + (74 - 73) + (76 - 75) + (78 - 77) + (80 - 79)

Simplify our term grouping

Since each term in parentheses equals 1, we have
1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1

Since we have 80 numbers, we have exactly ½ x 80 = 40 odd/even pairs
40 odd/even pairs x 1 = 40
On the exam, use this shortcut instead of listing out all terms, and you will get to your answer with lightning quickness.