Sums of numbers in a row
Tonight's idea
\(9 = 4 + 5\), and also \(9 = 2 + 3 + 4\). Call such a sum a run: two or more whole numbers in a row, starting at 1 or higher.
Most numbers have one: \(3 = 1 + 2\), \(6 = 1 + 2 + 3\), \(7 = 3 + 4\). A computer check up to 200 finds only these failures: 1, 2, 4, 8, 16, 32, 64, 128. Start at 1 and keep doubling. These are the powers of 2.
Why? A run with an odd number of terms has a middle. Move one from the top to the bottom and it levels out, so the sum is (how many terms) × (the middle).
Turn that around. \(15 = 3 \times 5\) has the odd factor 3, so use three numbers with 5 in the middle: \(4 + 5 + 6\). An odd factor gives the run a centre.
Powers of 2 have no odd factor bigger than 1. No odd factor, no centre, no run.
Watch one: every run for 30
The odd factors of 30 bigger than 1 are 3, 5 and 15.
3: centre \(30 \div 3 = 10\). Run: \(9 + 10 + 11\).
5: centre \(30 \div 5 = 6\). Run: \(4 + 5 + 6 + 7 + 8\).
15: centre \(30 \div 15 = 2\). Fifteen numbers centred on 2 go from \(-5\) to 9. But \(-5\) cancels 5, \(-4\) cancels 4, and so on: everything from \(-5\) to 5 adds to zero, leaving \(6 + 7 + 8 + 9\).
Each odd factor bigger than 1 gives one run, and there are no others.
Your turn
- Write each number as a run: (a) 21 using three numbers; (b) 35 using five numbers; (c) 49 using seven numbers.
- 18 has two odd factors bigger than 1. Find them, and find both runs for 18.
- (a) \(22 = 11 \times 2\). Write out the eleven numbers centred on 2, cancel what you can, and give the run that is left. (b) Do the same for \(26 = 13 \times 2\).
- (a) Exactly one number from 60 to 70 has no run. Which one, and how do you know without trying? (b) 17 is prime. Explain why it has exactly one run, and find it.
Stretch: runs with no middle
A run with an even number of terms, like \(3 + 4 + 5 + 6\), has no middle number. Pair the first term with the last, the second with the second-last, and so on. Explain why the sum must still have an odd factor bigger than 1. What does that prove about powers of 2?
Answers
- (a) \(21 \div 3 = 7\): \(6 + 7 + 8\). (b) \(35 \div 5 = 7\): \(5 + 6 + 7 + 8 + 9\). (c) \(49 \div 7 = 7\): \(4 + 5 + 6 + 7 + 8 + 9 + 10\).
- The odd factors are 3 and 9. With 3: centre 6, so \(5 + 6 + 7\). With 9: centre 2, nine numbers from \(-2\) to 6. Cancel \(-2\) to 2 and \(3 + 4 + 5 + 6\) is left. Both add to 18.
- (a) \(-3, -2, \dots, 7\). Cancel \(-3\) to 3, leaving \(4 + 5 + 6 + 7 = 22\). (b) \(-4, -3, \dots, 8\). Cancel \(-4\) to 4, leaving \(5 + 6 + 7 + 8 = 26\).
- (a) 64. It is a power of 2 (\(2 \times 2 \times 2 \times 2 \times 2 \times 2\)), so it has no odd factor bigger than 1. Every other number from 60 to 70 has one. (b) The only odd factor of 17 bigger than 1 is 17 itself, so there is one run. Centre \(17 \div 17 = 1\): seventeen numbers from \(-7\) to 9. Cancel \(-7\) to 7, leaving \(8 + 9\).
Stretch. In a run of even length the first and last terms are one odd and one even, so they add to an odd number, at least \(1 + 2 = 3\). Every pair adds to that same odd number. So the sum is (number of pairs) × (an odd number bigger than 1). With an odd number of terms, the sum is (number of terms) × (middle), and the number of terms is odd and at least 3. Either way a run has an odd factor bigger than 1. A power of 2 has none, so it can never be a run.