Patterns & Reasoning

Depth · one concept from first principles · step 2 of 4

Regroup a long sum of differences

Compute the following.

(110−3)+(108−5)+(106−7)+⋯+(4−109)+(2−111)(110-3)+(108-5)+(106-7)+\cdots+(4-109)+(2-111)

Number system
Your answer
How to solve
Strategy Organize Information in More Ways — Adding brackets one by one is slow. Instead reorganize: remove the brackets and collect all the even numbers being added and all the odd numbers being subtracted. Each odd number is just one more than an even number on the other list, and that pattern makes almost everything cancel.
1STEP 1

Count the brackets

The first numbers are the even numbers 2, 4, …, 110.

110 ÷ 2 = 55
2STEP 2

Regroup the terms

Addition can be done in any order, so gather the added numbers and the subtracted numbers.

(2+4+…+110) - (3+5+…+111)
3STEP 3

Match each odd number with an even one

Write each odd number as the even number just below it, plus 1: 3=2+1, 5=4+1, and so on up to 111=110+1.

3+5+…+111 = (2+4+…+110) + 55
4STEP 4

Let the lists cancel

Subtracting the odd list takes away the whole even list and 55 more.

(2+4+…+110) - (2+4+…+110) - 55 = -55
Answer
-55
Each bracket's value drops by 4: 107, 103, 99, …, ending at 2-111=-109. The values are evenly spaced, so the average is (107+(-109))/2 = -1, and 55 brackets give 55 × (-1) = -55.
Takeaway

When a sum is long, reorder it so matching pieces cancel.

  • Count the brackets
  • Regroup the terms
  • Match each odd number with an even one
  • Let the lists cancel
Where next?
Another one like thissuggested

A parent dashboard for the family lives at sensimlab.com.

▶ Practice — 12 problems

The same shape shows up later on the AMC, as Arithmetic Expression Decomposition, Lattice / Grid Pattern from Small Cases, Pair-Sum Invariant, Small-Domain Constrained Search. Why the shapes repeat →