Number Palindromes (version 5 March 2018)
A palindromic number is one that is the same when the digits are reversed, for example 18381.
All numbers turn out to be palindromes
in some number base. We look at sequences of Palindromes and find a surprisingly small number of
palindromes are needed to make a sum for any given number (in a given base).
A previous page on Number Bases is useful if you are not familiar with
binary numbers, number in bases other than 10 (decimal),
Fibonacci base or Factorial base.
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Contents of this page
The
icon means there is a
You Do the Maths... section of questions to start your own investigations.
The
calculator icon
indicates that there is a live interactive calculator in that section.
Palindromes
Every family has a palindrome in it: MUM and DAD, since a palindrome is a word that is the
same when reversed. Perhaps your name is a palindrome (BOB, ANNA, EVE)?
We can have palindromic numbers also for example 19291.
Whether a number is a palindrome or not depends on which base it is represented in but all numbers
are palindromic in some base.
All numbers are palindromes in some base
For example 21 is not a palindrome
in base 10, but it is in base 2: 21 = 10101_{2}.
We usually don't write initial 0s in a number but if we did then 10 would also be a palindrome since it is 010.
On this page we will not count such numbers as palindromes.
It is perhaps surprising at first to learn that all numbers are palindromes in some base.
There are several simple reasons why. How many can you think of?
Show some
 First choose a base that makes number N a single digit. All single digits are palindromes.
A suitable base is N+1: N = {N}_{N+1}
 Choose base 1 since N = {1,1,1, .... ,1,1}_{1} (digit 1 occurs N times in the brackets).
 Choose base N1 then we have N
= 1×(N1) + 1 = {1,1}_{N1}
 If N is not prime we can write N = P × Q so N = P (Q − 1) + P.
If P is smaller than Q−1 then choosing base Q−1
we have N = {P,P}_{Q−1}, a palindrome.
There are some numbers that are not palindromic in any of the bases from 2 to n−2.
For example,
4 = 100_{2} is not a palindrome;
6 = 110_{2} = 20_{3}
= 12_{4} none of which are palindromes
but
7 = 111_{2} is a palindrome
So the list of numbers n not palindromic in any base 2 to n−2 starts 4, 6, ... .
What are the next five numbers in this list?
Show the answer
4, 6, 11, 19, 47, 53, 79, ...
A016038
Sums of Palindromes
The palindromes in base 10 in order are 0,1,2,3,4,5,6,7,8,9,11,22,33, ..., 99,101,111,121,... A002113.
What is the 100th or 10,000th palindrome in this list?
Given a palindrome what is its position (its index number) in this list?
If every number is a sum of three triangular numbers, or four squares
(see more about this here),
can we make every number by summing palindromes?
If so, how many do we need to guarantee such a sum exists?
In which bases, if any, is this possible?
There are some surprisingly simple answers to these questions and we investigate them in this section.
The position of a palindrome
There is an amazingly simple way to find the position (index number) of any palindrome in this list, writes Hugo Pfoertner (see the OIES link below):
 If the palindrome has an even number of digits,
 put a 1 on the front half of the palindrome's digits
for example: 98766789 is at position 19876.
 If the number of digits is odd
 add 1 to the value of the first digit followed by the rest of the digits up to and including the central digit of the palindrome.
for example: 515 is the 61st palindrome,
8206028 is at position 9206 in the list,
9230329 has index number 10230.
Every number is a Sum of 3 Palindromes
If not all numbers are palindromes in base 10, then can we use palindromes to write any value by summing some of the palindromes?
How many would we need
to guarantee a sum for any given number?
It has been proved that the answer is that we can always do this and...
Every integer is a sum of at most 3 palindromes in base 10.
The number of ways to write n as sum of (up to) 3 palindromes is
1, 1, 2, 3, 4, 5, 7, 8, 10, 12, 13, 15, 16, 17, 17, 18, 17, 17, 16, 15, ... A261132
and since this is always possible, we know that all entries in this sequence are at least 1.
Recently (2016) Cilleruelo, Luca and Baxter proved that 10 is not unique in having the property that every number
is the sum of 3 palindromes in that base. They proved it applies to all bases from 5 upwards.
It is not true for base 2 because 176 needs 4 palindromes as do these numbers up to 1000:
176, 188, 208, 242, 244, 310, 524, 628, 656, 736, 754, 794, 832, 862,
866, 868, 880, 932, 944, 994, 1000, ... A261678.
Then in June 2016 Rajasekaran, Shallit and Smith of the University of Waterloo filled in the gap for bases 2, 3 and 4
and proved that we can add bases 3 and 4 to the list but sometimes four base 2 palindromes are needed but a maximum of four
applies to all binary numbers.
The reason base 2 is an exception is that palindromes in base 2 begin and end with 1.
A sum of 3 binary palindromes will therefore end in 1 and so must be odd.
Any even number will need an even number of palindromes in its sum.
All other numbers in all other bases need only 3 at most.
Use the Calculator here to try out some numbers and bases.
A set is a collection of distinct numbers, shown in increasing order.
A list is a collection of numbers which may occur more than once, shown in order.
Palindromic BASE Numbers Calculator
 Bases in this Calculator
 may be any positive number greater than 1 with a size limit of about 15 digits or Factorial base: ! or
Fibonacci base: F.
 The position of a palindrome
 is its index number in the list of that base's palindromes where the first palindrome
at position 1 is always 0.
Positions may be of any length up to hundreds of digits but larger numbers may slow down some calculations.
All numbers N are palindromic in every base beyond N+1 as well as in base 1 and base N−1
as we saw above.
We can regard these as the 'obvious' (mathematicians use the term 'trivial') cases and other answers are the 'interesting' ones when looking for
bases in which a number is palindromic.
You Do The Maths...

 Base 2 Palindromes: What can you say about numbers which are binary palindromes and have an even number of bits?
 Base 3 Palindromes: What can you say about numbers which are palindromes in base 3 and have an even number of 'digits'?
 Base 4 Palindromes: What can you say about numbers which are palindromes in base 4 and have an even number of 'digits'?
 Can you make a conjecture about the numbers that are palindromes in base β and have an even number of digits?
 Base 3: What can you say about the odd numbers in the list of base 3 palindromes?
 All base 5 palindromes of lengths 4 and 6 are multiple of 6. What about length 8? length 10?
 In base Fibonacci (Zeckendorf representations) which numbers are palindromes?
Check your answer with A094202
Palindromes and Sequences
First we look at some sequences that are palindromes and then at palindromic numbers in some other sequences
such as squares that are paindromes and in which bases. And what about the primes or the Fibonacci numbers?
Palindromic Compositions
Here a composition is a mathematical name for a sequence with a given sum. But does the order of the numbers matter in
the sequence or can numbers be repeated? For example with a sum of 4 we have the sequences:
4 = 1+1+1+1 = 1+1+2 = 1+3 = 2+2
There are 5 ways to sum palindromic numbers to make a total of 4
if the order is irrelevant (only the frequency of occurrence of each separate number matters)
If the order matters, there are 3 extra sums:
1+2+1, 2+1+1 and 3+1
making 8 in all.
If a different order of the elements still counts as the same solution, we have a partition of n.
If different orders are different solutions, we have a composition of n.
If we restrict the elements of the compositions to be the numbers 1 and 2 only, we have the following:
n  Compositions from {1,2}  Count 
1  1 
2  1,1; 2 
3  1,1,1; 1,2; 2,1 
4  1,1,1,1; 1,1,2; 1,2,1; 2,1,1; 2,2 
5  1,1,1,1,1; 1,1,1,2; 1,1,2,1; 1,2,1,1; 2,1,1,1; 1,2,2; 2,1,2; 2,2,1; 
The counts are the FIbonacci numbers and we can prove this quite simply.
Show how
Let's use C(n) for the number of compositions on n using only 1 and 2.
Either the composition of n begins with 1 or else it begins with 2:
If it begins with 1 then the rest is a composition of n−1 and there are C(n1) of them
If it begins with 2 then the rest is a composition of n−2 and there are C(n2) of them
So we have
C(n) = C(n − 1) + C(n − 2)
There is 1 composition of 1: C(1)=1.
There are 2 compositions of 2: C(2)=2.
This is the recursive definition of the Fibonacci Numbers but with a different starting point:
C(n) = Fib(n + 1)
Square numbers
11^{2} = 121 is a palindrome in base 10.
12^{2} = 144 is not a palindrome in base 10 but it is in base 11:
144 = 12×12 = 11×12 + 12 = 11(11 + 1) + 11 + 1 =
1×11^{2} + 2×11 + 1 = 121_{11}
Can you see why all squares are palindromes in some base?
Show why
n^{2} = 1×(n1)^{2} + 2(n1) + 1 = 121_{n1} if n>2.
This algebraic equation fails in we want binary squares as 2 is not a binary digit.
 The squares in base 2
 are 1 and 3^{2}=1001_{2} and then there is a long gap before the next
palindromic binary square number:
4523^{2} = 20457529 = 1001110000010100000111001_{2}.
They are quite rare and there are only 34 numbers with up to 100 bits whose squares are binary palindromes (see
A003166).
There may well be an infinite number in base 2 also, but no one has yet found a proof or an infinite series of square binary palindromes.
 in base 3:
 the squares of 1, 2, 4, 10, 11, 20, 22, 28, ... A029984
 in base 4:
 the squares of 1, 5, 17, 21, ... A029986 and so on.
 in base 10:
 the squares of 1, 2, 3, 11, 22, 26, 101, 111, 121, 202, 212, 264, 307, 836, 1001, ... A002778
For bases bigger than 2 we can find an infinite series of numbers that are not only palindromic
but their squares are too. Hint: look at the example at the start of this section.
Show why
For example, 11^{2} = 121, 101^{2} = 10201, 1001^{2} = 1002001, 10001^{2}= 100020001, ...
This works in any base which allows the "digit" 2, that is any base bigger than 2.
There are many other similar infinite sequences. How many can you find?
Many palindromic squares are palindromic because the number squared is a "simple" palindrome,
for example 1001^{2} = 1002001.
But there are other palindromes that are not so simple and whose squares are palindromes,
for example 26^{2} = 676, 264^{2} = 69696, 307^{2} = 94249,
836^{2} = 698896.
You will not be able to find any base 10 palindromic square number with 2, 4, 8 or 10 digits and there are other lengths too
that are missing in the lengths of such squares. Why?
The Calculator below helps you investigate these and many other sequences.

Palindromic squares for various number system bases I Korec
Mathematica Slovaca, Vol. 41, 1991, pages 261276. PDF
He calls base B numbers "badic" and explores squares that are palindromes by looking at polynomials.
 Feng Yuan's list of all the base 10 palindromic squares up to 45 digits long (in 2002).
PDF (2002)
.
He finds none of length
2, 4, 8, 10, 14, 18, 20, 24, 30, 38 and 40. Other even numbers with very few palindromic squares (the count is
in brackets) are
6(1), 12(1), 16(1), 22(1), 26(1), 28(2), 32(1), 34(1), 36(3), 42(1) and 44(1).
Fibonacci Numbers
On the Number Bases page we saw that all numbers are the sum of a set of Fibonacci numbers.
"Set" means there that no Fibonacci number is duplicated in the sum.
There are usually many ways to write a number as such a set but
There is a problem when we look at palindromes in this representation:
n_{Fibonacci} palindromes  0  1  11  101  111  1001  1111  10001  10101  11011  11111 
n  0  1  3  4  6  6  11  9  12  16  19 
We see that there are two palindromes in the Fibonacci base for the same number 6 and that the ordering
is wrong in that 9 has five 'digits' (9 = 8 + 1)
but comes after 11 which has just four (11 = 5 + 3 + 2 + 1)!
One way that we can resolve these problems is if we exclude consecutive Fibonacci numbers from the sum, or,
in terms of base Fibonacci, if we exclude any base Fibonacci representation that contains
two consecutive 1s. We call this the Zeckendorf representation. For any number
n the Zeckendorf collection of Fibonacci numbers that sum to n has the fewest.
Its alternative name is the Minimal Fibonacci representation.
Using only the Zeckendorf representations would mean that we can exclude 111 for 6
and 1111 for 11 from the table of Fibonacci palindromes but we also lose
3 and 16 too.
n_{F}  0  1  101  1001  10001  10101  100001  1000001  1001001  1010101  10000001  10100101 
n  0  1  4  6  9  12  14  22  27  33  35  51 
Palindromic sequences Calculator
Enter an expression using the letter n as the variable. The Calculator calculates the value
of the expression for
each value of n in your given range and shows which are palindromes either in the BASE given
or for each n it will find the bases in which the expression's value is a palindrome.
Use brackets ( and ) as usual with the builtin operations and functions as follows:
 + − / *
 always include the * operator for multiplication: write 2*n for
2n
 use ^ to raise a number to a power: x^y means
x^{y}
 fib(n) for Fibonacci numbers 0,1,1,2,3,5,8,... fib(0)=0, fib(1)=1, ...
 luc(n) for Lucas numbers 2,1,3,4,7,11,18,... luc(0)=2, luc(1)=1, ...
 gfib(a,b,n) for the general Fibonacci function
starting from a, b, a+b, .... for example,
fib(n) is gfib(0,1,n),
luc(n) is gfib(2,1,n),
gfib(a,b,0)=a, gfib(a,b,1)=b
 factorial(n) is 1*2*...*(n1)*n
 isprime(n) will find those n that are prime and a BASE palindrome.
isrect(n) will find those n that are composite (not a prime) and a BASE palindrome.
 polyg(s,r) for the polygonal numbers with polygons of s sides and outsides of length r:
polyg(3,n) is the nth triangular number n*(n+1)/2 and
polyg(4,n) is the nth square number n*n and so on.
polyg(n,5) is all the polygon shapes for the given range of n with sides of length 5
Bases are
 any whole number, positive or negtive except 1, 0 and 1
 F or Fmin for Fibonacci (Zeckendorf) with no consecutive 1s
 Fmax for Fibonacci representatives with maximum number of 1s and no consecutive 0s
 ! for Factorial base
 Binomial bases are not relevant here because all representations are a list of descending 'digits' so can never be palindromes (except 0).
Values for n can be of arbitrary length.
You Do the Maths...
 Base 10 Palindromic Squares
 Find the three numbers whose squares are palindromes with three digits in base 10.
Hint: Use f(n) = n^2 and BASE 10. What range of n should you use?
 Find all the palindromic squares of up to 7 digits (there are none with 8 digits).
 Without using the Calculator, write down eight (simple) 5digit palindromes whose squares are palindromes of 9 digits.
Check that there are a total of eleven palindromic squares with 9 digits.
 Without using the Calculator write down the five 6digit palindromes which when squared are palindromes with 11 digits.
Hint: Three contain just 0s and 1s, one contains two 2s and 0s. What is the other?
 How many of the twentyfive 9 digit palindromes which have 17 digit palindromic squares can you write down in 5 minutes?
 Why does the list of base 10 square palindromes never end?
Show why
The palindromic squares are
1, 4, 9, 121, 484, 676, 10201, 12321, 14641, 40804, ... A002779
they are the squares of
1, 2, 3, 11, 22, 26, 101, 111, 121, 202, ... A002778
In these lists is the sequence
2^{2} =4, 22^{2} = 484, 202^{2} = 40804, 2002^{2} = 4008004, ... which goes on for ever.
There are other such sequences too but
1^{2} =1, 11^{2} = 121, 111^{2} = 12321, 1111^{2} = 1234321, ... is
not
one of them: consider
1111111111^{2}.
On the other hand,
1^{2} =1, 11^{2} = 121, 101^{2} = 10201, 1001^{2} = 1002001, ... is a valid answer
and you should be able to find several more like this.
 Look at the palindromic squares in base 3 (use BASE 3
and f(n) = n^2 for n = 1 to 300).
Can you find
 a sequence which shows there is always a base 3 square with odd length ≥3?
 a sequence which shows there is always a base 3 square with even length ≥6?
Show the answer
 11_{3}^{2} = 121_{3},
101_{3}^{2} = 10201_{3},
1001_{3}^{2} = 1002001_{3}, ...

202_{3}^{2} = 112211_{3},
2002_{3}^{2} = 11022011_{3},
20002_{3}^{2} = 1100220011_{3}, ...
 The triangular numbers are 1, 3, 6, 10,... and are the sum of the first n whole numbers 1 + 2 + 3 + ... + n = n(n+1)/2.
Using the Calculator generate these by putting either
n*(n+1)/2 or polyg(3,n)
into the f(n) input box.
Which triangular numbers below 1000 are palindromic in base 10?
(more about the Triangular numbers)
 Find all the palindromic prime numbers up to 100, in base 10.
Check your answer with A002385.
 Why are there no base 10 palindromic primes with an even number of digits except 11?
Hint: what is a simple test on the digits of a number to see if it is divisible by 11?.
 Which prime numbers are palindromes in binary?
Check your answer with A016041.
Will this list include any of the numbers of the form 2^{n} ± 1?
 Which Fibonacci numbers are palindromes in base 10?
Check your answer with A045504
Palindromes in several bases
We know that every number n is palindromic in base (n−1), if we ignore base 1 and bases bigger than the number.
But for some numbers this base is the only one
in which it is a palindrome whereas others are palindromes in many bases.
Smaller examples of those palindromic in just one base are 1, 2, 3, 4, 6, 11, 19, ... A016038.
Are there patterns here?
Multiple Base Palindromic Numbers Calculator
The bases in this Calculator are only numeric and bigger than 2 but the range limits can be arbitrarily large.
You Do the Maths...
 Show that any composite (nonprime) number N>2
is palindromic in some (nontrivial) base except for two nonprime values of N.
What are they and why?
Show the answer
If N>2 is not prime number then we can factor it: N = A×B.
There are three cases:
 If A is less than B−1:
 A is a valid "digit" in base B−1: N = A(B−1)+A = {A,A}_{B−1}
 If A = B−1:
 N = AB = (B−1)B.
One of B or B−1 is even so N is even: N = 2 M.
N = 2(M − 1) + 2 = 22_{M−1}
Exception:
The only exception is when M=3 when 2 is not a valid digit in base M−1.
This is when N=6.
 If A = B, N is a square number:
 N = A×A = (A−1)^{2} + 2(A−1) + 1
N = 121_{A−1}
Exception
The only problem is when A=2 because A−1 = 2 is not a valid Base (A−1) digit.
This is when N=4
The two exceptions are 2 and 4. All other composite N are covered by the above cases.
 Find the numbers from 2 up to 100 that are palindromes in only the trivial cases.
From the previous question these must be prime apart from the two composite exceptions.
Show the answer
2, 3, 4, 6, 11, 19, 47, 53, 79, 103, ...
A016038

 Can a palindrome in base β ever be a multiple of β?
 What can you say about the remainder when divided by 6 if a number is a palindrome in both base 2 and base 3?
Show the answer
 A multiple of β in base β must end in 0 and we are excluding such numbers as palindromes.
If we included such then 10 becomes a palindrome in base 10 as do 20, 30, ..., 90, 100, 110 and so on.

A palindrome in base 2 must therefore be ODD and a palindrome in base 3 is not a multiple of 3.
So a number which is a palindrome in both when divided by 6 cannot have a remainder
of 0, 2 or 4 (such numbers are even), nor a remainder of 3 (it would be a multiple of 3)
so must have a remainder of 1 or 5.
Palindromes in bases 2 and 3 must be either one more or one less than a multiple of 6.
 33=100001_{2},
99=1100011_{2}
and 313=100111001_{2}
are examples of numbers that are palindromes in both bases 2 and 10.
 What is the complete list up to 1000?
 Another much larger number is 7475703079870789703075747.
Can you find a larger one?
Hint: Find the index number of this palindrome in base 2. Search using 'positions' from this position upwards.
Check your answers with A007632.
See the References below for more on this problem.

 Ignoring bases where a number is a single digit,
5 is smallest number palindromic in two bases and 10 the smallest palindromic in three bases.
What is the smallest number palindromic in four bases? in five bases?
 Continue the list up to thirteen bases.
Check your answer in A037183

 What is 123456_{10} in base 100?
What is 10110100_{2} in base 4?
What is the relationship between the digits of any number in base β and its digits in base β^{2}?
 What is 26^{2} in bases 5 and 25?
What is 50^{2} in bases 7 and 49?
What is the pattern here?
 In how many bases is 64^{2} a palindrome?
Palindromic Twosums and Products
Palindromic Twosums
A sum of two numbers gives a palindrome and the reversal of both the numbers summed gives the same sum:
23 + 98 = 121 = 89 + 32
Apart from single digit numbers, are there other pairs or twosomes (twosums?) like this? We can exclude any sum which does
not involve a carry as those are again "trivial".
Can you find examples in other bases?
Palindromic Products
If we take the product 134 × 201 = 26934 we can reverse the whole equation
and it still remains true:
134 × 201 = 26934
43962 = 102 × 431
Are there any more? What do they look like?
Product: Reversed: 
12 × 102 = 1224
21 × 201 = 4221

12 × 103 = 1236
21 × 301 = 6321

12 × 104 = 1248
21 × 401 = 8421

Squares 
13 × 13 = 169
31 × 31 = 961

112 × 112 = 12544
211 × 211 = 44521


Let's use reverse(N) for the number with the (base 10) digits of N in reverse order,
for example reverse(1213) is 3121 and let's call a pair of numbers n and m for which reverse(n)×reverse(m)=reverse(m n)
a reverse product pair.
Are all pairs of palindromes reverse product pairs?
If they were then since the two numbers are their own reversals (they are palindromes) then
their product must also be a palindrome. So we are asking if the product of two palindromes is always another palindrome.
Yes, there are some:
11 × 22 = 242; 77 × 88 = 6776; 77 × 858 = 66066
but this does not apply to every pair of palindromes:
11 × 55 = 605; 121 × 858 = 103818
More Palindrome Prime Patterns
Patrick De Geest has a wonderful World Of Numbers page which includes some nice
facts about the palindrome 323323. Here are some with suggestions about how to extend the patterns.
Here all nine digits are each raised to a power and the powers also use all nine digits once.
The sum is a palindrome.
Can you find other palindromes with this property?
What about in other bases?

1^{7} + 2^{2} + 3^{8} + 4^{9} + 5^{5} + 6^{6} + 7^{1} + 8^{4} + 9^{3} = 323323

These palindromes are a product of consecutive primes. Are there any more?

7 × 11 × 13 = 1001
5 × 7 × 11 × 13 = 5005
7 × 11 × 13 × 17 × 19 = 323323, 17 ×19 = 323

If we dropped the condition that the primes must be consecutive then we can find many more:
Can you find some patterns in these?

3 × 7 × 37 = 777
11 × 17 × 41 = 7667
41 × 43 × 53 = 93439
3 × 7 × 11 × 13 × 37 = 111111

Can you continue this pattern in the primes or find some similar ones?

2 × 3 × 11 = 66
2 × 3 × 11 × 13 = 858
2 × 3 × 7 × 11 × 13 = 6006
2 × 3 × 7 × 11 × 13 × 37 = 222222
or 2 × 3 × 7 × 11 × 13 × 47 = 282282
2 × 3 × 7 × 11 × 13 × 37 × 101 = 22444422
or 2 × 3 × 7 × 11 × 13 × 37 × 197 = 43777734

Here is another extension of the patterns above:

2 × 3 × 7 × 11 × 13 × 37 = 222222
2 × 3 × 7 × 11 × 13 × 47 = 282282
2 × 3 × 7 × 11 × 13 × 79 = 474474
2 × 3 × 7 × 11 × 13 × 101 = 606606
2 × 3 × 7 × 11 × 13 × 3467 = 20822802
2 × 3 × 7 × 11 × 13 × 4007 = 24066042
2 × 3 × 7 × 11 × 13 × 4057 = 24366342
2 × 3 × 7 × 11 × 13 × 4157 = 24966942
2 × 3 × 7 × 11 × 13 × 6869 = 41255214
2 × 3 × 7 × 11 × 13 × 7559 = 45399354
2 × 3 × 7 × 11 × 13 × 10151 = 60966906
2 × 3 × 7 × 11 × 13 × 13553 = 81399318

Palindromic Multigrades
My favourite maths book of all time is A H Beiler's "Recreations in the Theory of Numbers"
and in it he has this remarkable pattern involving palindromes
1221^{ } + 5445^{ } + 6996^{ }  =  2112^{ } + 3663^{ } + 7887^{ } 
1221^{2} + 5445^{2} + 6996^{2}  =  2112^{2} + 3663^{2} + 7887^{2} 
A sum which is correct both when we add the numbers and when we add their powers up to a given maximum power is
called a multigrade.
Some simpler nonpalindromic multigrades are
1^{1} + 5^{1} + 8^{1} + 12^{1}  =  2^{1} + 3^{1} + 10^{1} + 11^{1}  =  26 
1^{2} + 5^{2} + 8^{2} + 12^{2}  =  2^{2} + 3^{2} + 10^{2} + 11^{2}  =  676 
1^{3} + 5^{3} + 8^{3} + 12^{3}  =  2^{3} + 3^{3} + 10^{3} + 11^{3}  =  17576 
He also has the two lists of numbers, each number is the reverse of one in the other list:
132, 223, 241, 243, 312, 314, 332, 423  let's call this list A
231, 322, 142, 342, 213, 413, 233, 324  let's call this list revA
have the remarkable property that
 the sum of the numbers in each list is the same: 2220
 the sums of the squares of each number in list A is the same as the sum of the squares of each number in list revA: 669376
 the sums of the cubes of each number in list A is the same as the sum of the cubes of each number in list revA: 215347770
Beiler uses the notation
132, 223, 241, 243, 312, 314, 332, 423  _{3} 
231, 322, 142, 342, 213, 413, 233, 324 
^{=} 
A
multigrade denoted
means
list L's numbers with each raised to the power P
has the same sum as
list R's numbers with each raised to the same power
and
this is true for all powers P from 1 to k.
When P is 1, we have two lists with the same sum S and these are called partitions of S.
When P is 2 we have sums of squares and the sum S is a sum of squares in two ways (L and R).
But for two lists L and R where both of these hold, we have a multigrade with k=2.
He also gives another list with this property:
1234, 2455, 2565, 3346, 4541, 5322, 5432, 6653 and the list of the reversals of each number
Here is how he derives the first palindromic multigrade at the head of this section.
He takes the two singledigit multigrades
1, 5, 6  _{2} 
2, 3, 7 
and 
2, 4, 9  _{2} 
1, 6, 8 
^{=} 
^{=} 
and multiplies the numbers in one by 1001 and the other by 110 and adds them in pairs to get
1221,
5445,
6996  _{2} 
2112,
3663,
7887 
^{=} 
He even goes a stage further and includes a third multigrade:
2, 2, 5  _{2} 
1, 4, 4 
^{=} 
to get
122221,
542245,
695596  _{2} 
211112,
364463,
784487 
^{=} 
However, the same principles give rise to:
1, 5, 6  2 
2, 3, 7 
^{=} 
11, 55, 66  2 
22, 33, 77 
^{=} 
111, 555, 666  2 
222, 333, 777 
^{=} 
... 

1, 5, 6  2 
2, 3, 7 
^{=} 
121, 535, 676  2 
212, 353, 767 
^{=} 
1221, 5335, 6776  2 
2112, 3553, 7667 
^{=} 
... 

What principles or rules are at work here?
Can you find more multigrades with this palindromic property?
What if we restrict the powers to just (1) and 2?
Can you find two such lists of palindromes which work for powers (1), 2, 3 and 4?
Can we find a multigrade with two palindromic lists as long as we like?
How much of this carries over to other bases?
For some answers and many results and theorems see the references at the foot of this page.
You Do the Maths...
 Products of palindromes::
 Are there an infinite number of pairs of palindromes whose product is a palindrome?
Hint: try to find pattern that continues for ever.
One pattern is
11 × 11 = 121, 101 × 101 = 10201, 1001 × 1001 = 1002001, ...
 Are there an infinite number of pairs of palindromes whose product is not a palindrome?
11 × 55 = 605, 111 × 55 = 6105, 1111 × 55 = 611105, 11111 × 55 = 6111105, ...
all begin with 6 but end with 5
Show the answers
 Reverse product pairs that include 12:
How many more numbers N can you find where reverse( 12 × N ) = 21 × reverse(N)?
The list begins with N = 1, 2, 3 and 4: 12×1=12, 21=21×1; 12×2=24, 42=21×2; ...
What comes next?
How many N are there less than 100?
n can be 1, 2, 3, 4, 10, 11, 12, 13, 14, 20, 21, 22, 23, 24, 30, 31, 32, 33, 40, 41, 100
 Reverse product pairs of squares:
Which numbers N when squared and reversed are the squares of their reverses, or in other words,
find numbers N where reverse( N^{2} ) = ( reverse(N) )^{2}
for example N=13 and N=112 in the examples above.
 Does the pattern
13^{2 }  =  169   31^{2 }  =  961 
113^{2 }  =  12769  311^{2 }  =  96721 
1113^{2}  =  1238769  3111^{2}  =  9678321 
continue for ever?
If not, can you find a series of squares which does;
if it does, can you find another similar series?
Show the answers
No, it stops after one more row:
11113^{2} = 123498769  31111^{2} = 967894321 

111113^{2} = 12346098769  311111^{2} = 96790054321 
Two other infinite patterns of palindromic numbers whose squares are palindromic are:
11^{2}  =  121   101 101^{2}  =  10221412201 
101^{2}  =  10201  101 0 101^{2}  =  1020304030201 
1001^{2}  =  1002001  101 00 101^{2}  =  102012040210201 
 ...    ... 
 Some numbers such as 2448 have two different reverse products:
12 × 204 = 24 × 102 = 2448
21 × 402 = 42 × 201 = 8442
Can you find others?
 All the examples up to now have only used the digits 0, 1, 2, 3, 4 in both numbers to be multiplied.
Can you find an example which has a larger digit?
 Henry Bottomley says that
Bottomley's Rule
If the sum of the squares of the base β digits of n is less than β
and n is not divisible by β
then the product of n and the base β reversal of n is a base β palindrome.
This refers to "small digit" numbers so that when we multiply any by its reversal we get a palindrome, as in
211 × 122 = 23632
121111 × 111121 = 13457975431
But Bottomleys' result applies to all bases not just to base 10 (β=10).
Although 113 × 311 = 35143
is not a palindrome
in base 10, the number 113 satisfies the conditions of Bottomley's Rule in base 12 and we do indeed find
{1,1,3}_{12} × {3,1,1}_{12} = {3,4,11,4,3}_{12}
What do the two conditions do?
Can you find a proof?
See World Of Numbers website.
 Palindromic Dates and Times
 When was the last palindromic year and when is the next?
 A digital clock shows hours and minutes as H:M where H is the one or twodigit hour from 1 to 12
and M is the twodigit minute from 00 to 59.
For example 1:08 for 8 minutes past 1.
If we ignore the colon separator, for how many minutes each day will the clock be showing a palindromic time?
1:01, 1:11, 1:21, 1:31, 1:41, 1:51,
2:02, 2:12, 2:22, 2:32, 2:42, 2:52
3:03, 3:13, 3:23, 3:33, 3:43, 3:53,
...
9:09, 9:19, 9:29, 9:39, 9:49, 9:59,
10:01, 11:11, 12:21
Total: 6 for each of 9 hours= 54; 3 for the 10, 11 and 12 hours: 57 minutes.
 For how many minutes would the digital clock be showing a palindrome if the digital clock's hours go from 0 to 23?
0:00, 0:10, 0:20, 0:30, 0:40, 0:50,
then 6 times each hour up to 9:59,
10:01, 11:11, 12:21, 13:31, 14:41, 15:51,
20:02, 21:12, 22:22, 23:32
10 hours with 6 minutes and 10 hours with a single minute = 70 minutes
 An analogue clock (with continuously moving hour and minute hands)
is next to the digital 24 hour clock.
Apart from 0:00 when the analogue clock's hands are overlapping,
for which one of the digital clock's palindromic times will the analogue clock's hands
be closest together?
9:49 the hands are separated by 0.5°
Show answers
References
 Every natural number is the sum of fortynine palindromes W D Banks
arXiv:1508.04721v1 (Aug 2015). The link takes you to the Abstract where there is a link to the PDF version.
This paper was soon superceded by...
 Every positive integer is a sum of three palindromes J Cilleruelo, F Luca, L Baxter (2016)
arViv:1602.06208v2
This paper proves that for all bases g≥5, all positive integers can be written as a sum of (up to) three base g palindromes.
 Sums of Palindromes: an Approach via Automata A Rajasekaran, J Shallit, T Smith (Sept 2017)
arXiv:1706.10206v3
 Patrick De Geest's World Of Number page
has lots of fascinating facts on large palindromic numbers in special bases and
palindromic in both binary and decimal. The comments contain some very interesting approaches to programming a search for
large numbers palindromic in two bases.
 The alternating sum test for divisibility by 11 (in base 10) is explained
on Divisibility by Eleven at the
Math Fun Facts website by F E Su et al.
 Patrick De Geest's World Of Numbers has a wealth of facts and figures
on palindromes.
 Equal Sums of Like Powers
by Chen Shuwen (2001)
has lots of useful information on multigrades and a useful page of Theorems
on how to extend and combine them to make others and to extend the range of powers.
 History of the Theory of Numbers:
Vol II Diophantine Analysis by L E Dickson (Dover edition 2005) Chapter XIV (pages 705  716)
is on 'Sets of Integers with Equal Sums of Like Powers'.
A reprint of the original edition of 1952 without updates since then, but
still a useful summary and
historical survey of results and works still not fully available on the web.
© 20182020 Dr Ron Knott
Dr Knott's Maths Fibonacci and HOME page
created: 20 August 2017