Find all possible ways of expressing 100 as the sum of consecutive natural numbers.
Step-by-step solution
Idea: The sum of k consecutive numbers starting at m is (average of first and last) × (number of terms) = k(2m + k − 1)2. Set it equal to 100 and look for whole-number solutions.
- Take k consecutive numbers m, m + 1, …, m + k − 1. Sum = first + last2 × k = k(2m + k − 1)2 = 100, so k(2m + k − 1) = 200.1 mark
- k and 2m + k − 1 multiply to 200, the second factor is larger than k, and one of them is odd (they differ by 2m − 1, an odd number). The factor pairs of 200 with one odd factor: 1 × 200, 5 × 40, 8 × 25.1 mark
- k = 5: 2m + 4 = 40 ⇒ m = 18: 18 + 19 + 20 + 21 + 22.½ mark
- k = 8: 2m + 7 = 25 ⇒ m = 9: 9 + 10 + … + 16.½ mark
- k = 1 just gives 100 itself (one number). So there are exactly two ways using two or more consecutive natural numbers.1 mark
Check: 18 + 22 = 40, 19 + 21 = 40, plus 20: 100 ✓. 9 + 16 = 25, four such pairs: 100 ✓.
Answer to write in the exam
k consecutive numbers from m: sum = k(2m + k − 1)2 = 100
k(2m + k − 1) = 200; one factor odd, second factor > k
Pairs: 1 × 200, 5 × 40, 8 × 25
k = 5: 2m + 4 = 40 ⇒ m = 18 ∴ 18 + 19 + 20 + 21 + 22 = 100
k = 8: 2m + 7 = 25 ⇒ m = 9 ∴ 9 + 10 + … + 16 = 100
∴ Two ways (besides 100 itself).
Common mistakes that cost marks
- Trying only a few starting numbers by guesswork and missing the 8-term sum.
- Including 0 (e.g. 0 + 1 + … ). 0 is not a natural number.
- Allowing k = 25 (the pair 25 × 8): it would need 2m + 24 = 8, so m would be negative.
How this can come in the exam
Which of these is a sum of consecutive natural numbers equal to 45?
- 5 + 6 + 7 + 8 + 9 + 10
- 6 + 7 + 8 + 9 + 10
- 8 + 9 + 10 + 11 + 12
- 13 + 14 + 15
Show answer
(A) 5 + 6 + 7 + 8 + 9 + 10
5 + 6 + … + 10 = 6 × 5 + 102 = 45. The others give 40, 50 and 42.
Try one yourself
Express 30 as a sum of consecutive natural numbers in all possible ways (two or more numbers).
Show answer
k(2m + k − 1) = 60: 9 + 10 + 11, 6 + 7 + 8 + 9, 4 + 5 + 6 + 7 + 8.
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