Number Play · Lesson 3 of 8
Palindromes, Kaprekar, and Calendar Patterns
“Explore what happens when digits are mirrored, rearranged, and repeated in familiar settings.”
• Identify and construct palindromes using matching digit positions. • Follow reverse-and-add and distinguish an observed pattern from a general conclusion. • Apply the four-digit Kaprekar procedure correctly, including leading zeros. • Recognise valid repeating and palindromic clock times and dates. • Explain the conditions needed for two yearly calendars to match.
The same digits can make different patterns
Compare 123, 321, and 121. They use familiar digits, but their arrangements tell different stories. Reversing 123 produces a different number, whereas reversing 121 leaves it unchanged. In this lesson, we will examine reversal, digit rearrangement, and repetition as precise procedures. A procedure gives instructions that another person can follow and check.
A written number that reads the same from left to right and from right to left.
66, 848, 575, 797, and 1111 are palindromes. Matching digits are equally far from the ends. In a three-digit palindrome, the first and third digits must match; the middle can be any allowed digit. In a five-digit palindrome, the first matches the fifth and the second matches the fourth. The centre has no partner to match.
Problem
List every three-digit palindrome using only 1, 2, and 3, with repetitions allowed.
- 1.Choose the first digit. The last digit must be the same, while the middle has three choices.
- 2.With outside digit 1, the choices are 111, 121, 131. With outside digit 2, they are 212, 222, 232.
- 3.With outside digit 3, they are 313, 323, 333. There are 3 × 3 = 9 possibilities because each outside choice allows three middle choices.
Problem
A five-digit palindrome is odd. Its tens digit is twice its units digit, and its hundreds digit is twice its tens digit. Find it.
- 1.Let the units digit be an odd digit, because the whole number is odd. The possible units digits are 1, 3, 5, 7, 9.
- 2.The hundreds digit is four times the units digit. Any units digit 3 or more would make that value at least 12, which is not a single digit.
- 3.Thus the units digit is 1, the tens digit is 2, and the hundreds digit is 4. Reflection supplies the first two digits as 1 and 2.
- 4.The number is 12,421, read as twelve thousand four hundred twenty-one. It satisfies all three clues.
Reverse, add, and check again
Another way to search for palindromes begins with a two-digit number. Reverse its digits and add the two numbers. Check whether the sum is a palindrome. If it is not, apply the same instructions to the new sum. Each round contains an addition and a check; stopping depends on the result, not on a fixed number of rounds.
| Start | First addition | Further addition if needed | Palindrome obtained |
|---|---|---|---|
| 34 | 34 + 43 = 77 | None | 77 |
| 29 | 29 + 92 = 121 | None | 121 |
| 48 | 48 + 84 = 132 | 132 + 231 = 363 | 363 |
| 76 | 76 + 67 = 143 | 143 + 341 = 484 | 484 |
Problem
Starting with 57, how many additions are needed to obtain a palindrome?
- 1.Reverse 57 to get 75 and add: 57 + 75 = 132. This is not a palindrome because its outside digits differ.
- 2.Reverse 132 to get 231 and add: 132 + 231 = 363.
- 3.363 reads the same in both directions. We stop after two additions. The number of rounds counts additions, not the original starting value.
Try different two-digit starts and record how long they take. Some require many more rounds than these short examples. Two-digit starting numbers do eventually produce palindromes under this process, but checking a few examples would not by itself justify the word all. For larger starting numbers, the question is much harder. The familiar starting number 196 is an example for which eventual arrival at a palindrome has not been established; do not claim that your own trial proves it can never happen.
Reversing 132 gives 231, not 312. A palindrome test compares the written order in the two directions. If a reversal begins with zero, its numerical value omits that leading zero; for example, the reverse of 120 has value 21.
Kaprekar’s procedure rearranges digits in two orders
D. R. Kaprekar, a mathematics teacher in Devlali, Maharashtra, enjoyed discovering numerical patterns. In 1949 he found a remarkable procedure for four-digit numbers. Begin with a four-digit number whose digits are not all the same. Arrange those digits in descending order to form the largest arrangement, A, and in ascending order to form the smallest arrangement, B. Subtract B from A and use the four digits of the result for the next round.
Four positions must be retained throughout the procedure. If an ascending arrangement begins with zero, keep that zero in the arrangement even though its numerical value is smaller than 1000. A result such as 999 is written as 0999 before the next rearrangement. This is different from choosing the smallest ordinary four-digit number, where a leading zero would be forbidden.
Problem
Apply Kaprekar’s procedure to 6382.
- 1.Arrange 6, 3, 8, 2 in descending and ascending order: 8632 − 2368 = 6264.
- 2.Use the digits of 6264: 6642 − 2466 = 4176.
- 3.Use the digits of 4176: 7641 − 1467 = 6174. This has taken three subtraction rounds.
- 4.Repeat once more: 7641 − 1467 = 6174 again. The procedure now returns the same number. This repeating value is the Kaprekar constant.
| Round | Input | Descending A | Ascending B | A − B |
|---|---|---|---|---|
| 1 | 5683 | 8653 | 3568 | 5085 |
| 2 | 5085 | 8550 | 0558 | 7992 |
| 3 | 7992 | 9972 | 2799 | 7173 |
| 4 | 7173 | 7731 | 1377 | 6354 |
| 5 | 6354 | 6543 | 3456 | 3087 |
| 6 | 3087 | 8730 | 0378 | 8352 |
| 7 | 8352 | 8532 | 2358 | 6174 |
The table illustrates why zeros matter. For 5085, the ascending order is 0, 5, 5, 8, so B has value 558. Choosing 5058 would break the procedure and produce the wrong sequence. Try other eligible four-digit starting numbers and compare the number of rounds. They reach 6174 when this four-position convention is followed. A number with four identical digits, such as 4444, instead subtracts to 0000 and is excluded.
Problem
Try the same idea with 321, keeping three positions at every round.
- 1.321 − 123 = 198; then 981 − 189 = 792.
- 2.972 − 279 = 693; then 963 − 369 = 594.
- 3.954 − 459 = 495. Rearranging 495 gives 954 − 459 = 495 again, so 495 repeats.
- 4.Use a three-digit start with at least two different digits. The three-digit and four-digit procedures use different numbers of positions and have different repeating constants.
Compare the largest and smallest digit arrangements
Digit arrangement also helps compare sums and differences. With nonzero digits, descending order gives the largest number and ascending order the smallest. The digits 4, 7, 3, 2 give 7432 and 2347. Their difference is 5085 and their sum is 9779. Changing the chosen digits changes both results, so a comparison task must check the sum or difference actually requested.
Problem
Choose four digits so that the largest-minus-smallest difference is above 5085, then choose another set making it below 5085. Compare the sums as well.
- 1.Choose 7, 4, 3, 1. Their extreme arrangements are 7431 and 1347. The difference is 7431 − 1347 = 6084, above 5085; their sum is 8778, below 9779.
- 2.Choose 7, 4, 3, 3. Their extreme arrangements are 7433 and 3347. The difference is 4086, below 5085; their sum is 10,780, above 9779.
- 3.These two choices satisfy all four comparison requests. A larger difference does not force a larger sum: the first pair has the larger difference but the smaller sum.
When asked for the smallest ordinary four-digit number using digits that include zero, its first digit must be the smallest nonzero digit; put zero next. A Kaprekar routine instead allows an ascending arrangement such as 0558 because it retains four positions. Read the rule before choosing how to handle zero.
Find valid patterns in clocks and dates
Digit patterns also occur in times you read every day. On a twelve-hour display, 4:44 repeats a digit, 10:10 repeats a pair, and 12:21 mirrors the two sides. To test a pattern, say how the time is written. For mirrored two-digit hour and minute pairs, use a consistent HH:MM display, such as 05:50, and ignore the colon when comparing digits. A digit pattern must still be a valid time: minutes cannot be 60 or more.
| Pattern | Examples | Constraint to check |
|---|---|---|
| One digit repeated | 1:11, 2:22, 3:33, 4:44, 5:55 | 6:66 fails because 66 is not a minute reading. |
| The hour pair repeats | 01:01, 09:09, 10:10, 11:11, 12:12 | Use hour values from 01 to 12. |
| Mirrored pairs | 01:10, 02:20, 03:30, 04:40, 05:50, 10:01, 11:11, 12:21 | 06:60 fails; reverse the hour digits and check the minutes. |
Problem
On an HH:MM twelve-hour display, the time is 10:01. How long until the next two palindromic times?
- 1.Within hour 10, the mirrored minute pair is 01, which has already appeared. The next eligible hour is 11, whose mirrored minutes are 11.
- 2.From 10:01 to 11:01 is 60 minutes, then to 11:11 is 10 more: a wait of 70 minutes.
- 3.The following palindromic time is 12:21, another 70 minutes after 11:11. It is 140 minutes after the original 10:01.
For dates, fix the format as DD/MM/YYYY and keep the zeros. Removing the slashes from 20/12/2012 gives 20122012, two copies of 2012. Likewise, 20/04/2004 gives two copies of 2004. A different pattern is a palindrome: 11/02/2011 gives 11022011, and 02/02/2020 gives 02022020, both unchanged on reversal. Every proposed answer must also represent a real date; a symmetric string with an impossible month does not qualify.
Problem
Find their sum and their difference.
- 1.A five-digit number begins with a nonzero digit. The smallest mirrored choice uses 1 at both ends and zeros inside: 10,001.
- 2.The largest uses 9 in every place: 99,999.
- 3.Their sum is 10,001 + 99,999 = 110,000. Their difference is 99,999 − 10,001 = 89,998.
When can a yearly calendar be reused?
A matching yearly calendar needs more than matching date digits. The year must begin on the same weekday and have the same length: both years must be ordinary years or both leap years. Then the same sequence of month lengths places every date on the same weekday. A leap year includes 29 February, so its layout cannot match an ordinary year completely.
An ordinary year has 365 days, which is 52 full weeks plus one day. The following 1 January therefore moves forward one weekday. A leap year has 366 days, or 52 weeks plus two days, so it moves forward two weekdays. To investigate repeated calendars, track these shifts over several years and check both the starting weekday and the presence of 29 February. Do not assume a single fixed waiting period works for every starting year.
Choose two printed yearly calendars. Compare 1 January, whether 29 February exists, and the weekday of a date later in the year. If they do not match, continue through later calendars. Record the evidence that establishes a complete match, rather than checking only one convenient date.
Quiz
Which is a palindrome?
How many three-digit palindromes use only 1, 2, and 3, with repetition allowed?
What is the first reverse-and-add result for 48?
What is the correct ascending arrangement for a Kaprekar round starting with 5085?
Why is 7777 excluded as a starting number for the four-digit Kaprekar investigation?
Which HH:MM reading is both valid and palindromic?
Which conditions are enough for two complete yearly calendars to match?
Practice Problems
- List all three-digit palindromes using only 2 and 5, with repetition allowed.
- Apply reverse-and-add to 76 and show the stopping check.
- Explain why the five-digit clue gives only 12,421.
- Use Kaprekar’s procedure on 1000, preserving four positions until 6174 appears.
- Repeat one Kaprekar round starting with 6174 and one starting with 495 in the three-digit version.
- List all mirrored HH:MM twelve-hour readings and explain the missing hours.
- Test 12/02/2021 for a palindrome in DD/MM/YYYY format. Construct a valid date made by repeating a four-digit block.
- Explain why the weekday of 1 January alone cannot establish that two calendars match.
222, 252, 525, 555. There are two choices for each of the outside and middle digits.
Key Takeaways
• Palindromes have matching digit positions at equal distances from their ends. • Follow each reverse-and-add or Kaprekar instruction in the stated order. • Keep leading zeros when a Kaprekar routine requires a fixed number of digit positions. • Clock and date patterns require a consistent display format and valid times or dates. • Repeated examples support an investigation; they do not automatically prove every case. • Matching yearly calendars need the same starting weekday and the same leap-year status.