Problem
Iterate a rule to a far term
A list starts at 16. A one-digit number gets doubled. A two-digit number is replaced by the sum of its digits. Find the 100th number.
Reasoning · Grade 5-1 Finding the Hidden Rule
Your answer
Watch out Listing all 100 looks necessary, but it repeats every nine — just divide.
How to solve
Strategy Look for a Pattern — Writing out 100 numbers by hand would work but gives a hundred chances to slip. Instead I answer the easier question first — what are the first dozen or so numbers? — by making a systematic list, one rule application at a time. Because each number is decided entirely by the one before it, the moment a number I have already seen comes back, the whole list must start repeating from there. So I only need to find how long that repeating block is, and then work out where position 100 falls inside a block.
1STEP 1
Generate the list one rule at a time
Following the rules, 16 comes back.
16 → 7 → 14 → 5 → 10 → 1 → 2 → 4 → 8 → 16 → 7 → 14 → …
Each move is either doubling a one-digit number or adding the two digits of a two-digit number, and both are things a Grade 4 solver does in their head — so a careful list of ten terms costs almost nothing.
4.OA.C.5Make A Systematic List2STEP 2
Say why the list must repeat forever from here
So a block of nine repeats forever.
block = (16, 7, 14, 5, 10, 1, 2, 4, 8), length 9
This is the whole trick behind repeating patterns: once a step lands back on an earlier number, everything after it repeats too, so a nine-term list already describes all one hundred terms.
4.OA.C.5Look For A PatternThe list has to repeat forever, because a value that has appeared once will lead to the same continuation again.
Why?
There are only finitely many values the rule can produce, so sooner or later one of them has to turn up a second time.
🧱More items than boxes forces a shared boxn boxes can hold at most n items one apiece, so item number n plus one has nowhere new to go and must share.
Why?
Once a value repeats, everything after it repeats too, so the list settles into a block that comes round again and again.
🧱A cycle returns after a full periodA fixed repeating cycle of length p comes back to the same position every p steps.
3STEP 3
Find which positions hold the number 16
16 sits at positions 1, 10, 19, and so on.
positions holding 16: 1, 10, 19, 28, 37, …
Counting in nines is just what happens when you lay the list out in rows of nine — each column always holds the same number, so a position's column is all you need.
4.OA.C.5Look For A Pattern4STEP 4
Divide 100 by the block length
100 divided by 9 leaves 1.
100 ÷ 9 = 11 remainder 1, 9 × 11 = 99
Division with a remainder is built for exactly this job: the quotient counts the whole blocks and the remainder points at the slot inside the next one.
4.NBT.B.6Look For A Pattern5STEP 5
Read off the answer
So the 100th number is 16.
100th term = 1st term = 16
Once the block is known, finding any term — the 100th, even the 1000th — costs one division instead of a thousand doublings.
4.OA.C.5Look For A PatternAnswer
16
100 ÷ 9 = 11 R 1
The answer has to be one of the nine numbers that actually appear in the list — 16, 7, 14, 5, 10, 1, 2, 4, 8 — and 16 is one of them, so nothing has drifted outside the pattern. The bookkeeping checks out too: 9 × 11 = 99, so term 99 is the last term of the eleventh block, which is 8, and the next term must restart the block; doubling 8 does give 16, exactly what the rule says should happen at that step. The numbers also stay small the whole way — adding the digits of a two-digit number can never give more than 18, and doubling a one-digit number can never give more than 18 either, so the list can never run away to large values.
Takeaway
When a rule brings back a number you have already seen, the whole list starts over — so one division by the block length jumps you straight to the 100th term.
- Generate the list one rule at a time
- Say why the list must repeat forever from here
- Find which positions hold the number 16
- Divide 100 by the block length
- Read off the answer