Problem
Counting palindromic numbers
A palindrome reads the same forwards and backwards. Only the numbers from 10 to 999 count. Two-digit and three-digit numbers can be counted apart. Count the palindromes in that range.
Reasoning · Grade 4-1 Making Many Kinds of Numbers
Your answer
Watch out The first digit looks like ten choices, but it cannot be 0 — only nine.
How to solve
Strategy Make a Systematic List — Writing out all 990 numbers is hopeless, but the range splits cleanly into two easy subproblems: the two-digit numbers and the three-digit numbers. Inside each subproblem the palindromes have a fixed shape, so instead of listing numbers I can list the free digit choices and multiply. The two-digit case is small enough to write out in full, which also shows the pattern that makes the three-digit case safe.
1STEP 1
Split the range into two smaller problems
Split the range into two-digit and three-digit.
Numbers with a different number of digits behave differently when you reverse them, so separating by digit count turns one messy problem into two tidy ones.
4.NBT.A.2Identify Subproblems2STEP 2
List every two-digit palindrome
Two-digit gives 11 to 99, so 9.
11, 22, 33, 44, 55, 66, 77, 88, 99
Nine items is few enough to write out completely, so there is no chance of missing one or counting one twice.
4.NBT.A.2Make A Systematic List3STEP 3
See the shape of a three-digit palindrome
Three-digit ones are first and last the same, so two choices.
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Once you know only the first two digits are free to choose, the last digit is decided for you, so the pattern does the work instead of a list.
4.NBT.A.2Look For A Pattern4STEP 4
Count the three-digit palindromes by counting choices
9 × 10 = 90 three-digit palindromes.
9 × 10 = 90
Nine rows of ten is exactly what multiplication counts, so a picture of 9 groups of 10 replaces a list of 90 numbers.
3.OA.A.1Make A Systematic ListA three-digit palindrome is fixed by its first two digits, so nine choices for the first and ten for the second give ninety of them.
Why?
The first digit cannot be zero but the middle digit can be anything, and neither choice limits the other.
🧱Independent choices multiplyIf a first choice can go a ways and each of those leaves b ways for a second, the pair can be made a times b ways.
Why?
Nine rows of ten is nine equal groups of ten, which is exactly what multiplying counts, so no list of ninety numbers is needed.
🧱Multiplication as equal groupsa times b means a equal groups of b — it is just repeated adding of the same amount.
5STEP 5
Add the two counts
Adding gives 9 + 90 = 99.
9 + 90 = 99
Because the two groups do not overlap, plain addition combines them correctly.
4.NBT.B.4Identify SubproblemsAnswer
99 palindromes
9 + 9 × 10 = 99
The answer is a count of numbers, so it must be a whole number no bigger than the 990 numbers in the range, and 99 is comfortably smaller. A quick sanity test: among the hundred numbers 100 to 199 the palindromes are 101, 111, 121, ..., 191, which is 10 of them, and the same happens for the 200s, the 300s and so on up to the 900s, giving 90. Adding the nine two-digit ones gives 99, which matches.
Takeaway
In a palindrome the back half just copies the front half, so you only have to count the choices for the front, and that is small enough to count by groups.
- Split the range into two smaller problems
- List every two-digit palindrome
- See the shape of a three-digit palindrome
- Count the three-digit palindromes by counting choices
- Add the two counts