Problem
Use odd and even to rule out totals
Six bowling pins are worth 5, 5, 5, 3, 3 and 1. Exactly four of them are knocked down. The score is the sum of those four. Pick the scores that cannot happen.
Reasoning · Grade 5-1 Making a Target Number
Your answer
Watch out Adding odd numbers suggests an odd total, but four odds always make an even.
How to solve
Strategy Look for a Pattern — Choosing 2 pins to leave standing instead of 4 to knock down keeps the counting tiny, but even before counting there are two quick tests that throw out impossible scores wholesale. First, every pin number here is odd, and adding four odd numbers always lands on an even total — that is a pattern that kills any odd target instantly. Second, the smallest and largest four-pin totals fence the answer into a narrow range, killing anything outside it. What survives those two tests I then confirm by actually building it with a guess-and-check combination, and finally I write the complete systematic list of four-pin totals so nothing is left to chance.
1STEP 1
Read the pin values off the picture
The pins read 5, 5, 5, 3, 3, 1.
{5, 5, 5, 3, 3, 1}
Sorting the six numbers before doing anything else makes the shared feature — they are all odd — jump out, and that single feature does most of the work in this problem.
2.OA.C.3Make A Systematic List2STEP 2
Fence the score in: smallest and largest possible totals
The score sits between 12 and 18.
1+3+3+5 = 12 ≤ score ≤ 5+5+5+3 = 18
Knocking down four pins that are all worth at least 1 point cannot give a tiny score, and checking the two extreme choices is enough to pin down the whole range.
2.NBT.B.6Guess And Check3STEP 3
Use odd and even to rule out 17
Four odd pins make the score always even.
(odd+odd) + (odd+odd) = even + even = even
Odd numbers are 'one more than a pair', and four of those extra ones make two more pairs, so nothing is left over — that is why four odds always total an even number.
2.OA.C.3Look For A PatternA score of 17 is impossible, because four pins with these values can only ever add to an even number.
Why?
Adding numbers of a fixed parity settles the parity of the total, so the total's oddness or evenness is decided before any pin is chosen.
🧱Even and odd behave predictablyTwice any whole number is even and adding an odd amount always swaps even with odd, so a total's parity is settled by its parts before anything is computed.
Why?
A single target of the wrong parity is refuted on the spot, without any need to try arrangements.
🧱One counterexample refutes an always-claimA claim about every case dies the moment you exhibit one case where it fails, and exhibiting is easier than arguing.
4STEP 4
Build the three scores that survive
So 12, 14, 16 and 18 are all reachable.
5+3+3+1 = 12, 5+5+3+1 = 14, 5+5+5+3 = 18
Showing a score is possible needs only one working example, so a single well-chosen guess for each target settles it — and each one must be checked against the pins that actually exist, since there are only three 5s and two 3s.
2.NBT.B.6Guess And Check5STEP 5
List every four-pin total to be sure
The impossible scores are 10 and 17.
22 - 10 = 12, 22 - 8 = 14, 22 - 6 = 16, 22 - 4 = 18
Turning 'pick 4 of 6' into 'pick 2 of 6' shrinks the list to a handful of cases a young solver can write out in full, which converts a clever argument into a complete check.
3.OA.D.9Make A Systematic ListAnswer
10, 17 points
12 to 18
Scores here are whole numbers of points, and the complete list of reachable totals 12, 14, 16, 18 is a set of even numbers spread between the smallest and largest possible four-pin totals, exactly as the two arguments predicted. The two rejected scores fail for two different and independent reasons: 10 is simply too small to reach even with the four cheapest pins, while 17 is the right size but the wrong parity. The three accepted scores 12, 14 and 18 were each demonstrated with an actual set of four pins, so nothing was accepted on a hunch.
Takeaway
Before hunting for combinations, ask how big the total can be and whether it must be odd or even — four odd pins always add to an even score, so 17 was never possible.
- Read the pin values off the picture
- Fence the score in: smallest and largest possible totals
- Use odd and even to rule out 17
- Build the three scores that survive
- List every four-pin total to be sure