A palindrome number is that which is same even where read backwards. Example 363. A Super-palindrome number is that palindrome number whose square root is also palindrome. For example- 121 is a palindrome number and its square root is 11 which is also a palindrome number. List first 15 Super Palindrome numbers (note – Single digit numbers are not Palindromes)
📌 Challenge Details and Links
ExcelBI Excel Challenge Number: 394
Challenge Difficulty: ⭐️⭐️⭐️⭐️
📥Download Sample File
📥Link to the solutions on LinkedIn
Solving the challenge of List Super-Palindrome Numbers with Power Query
Power Query solution 1 for List Super-Palindrome Numbers, proposed by John V.:
let
T = Text.From,
R = Text.Reverse,
r = List.Generate(
() => [i = 10, v = null, n = 0],
each [n] <= 15,
each [
i = 1 + [i],
v = if (T(i) = R(T(i))) and (T(i * i) = R(T(i * i))) then i * i else null,
n = if v = null then [n] else 1 + [n]
],
each [v]
)
in
List.RemoveNulls(r)Power Query solution 2 for List Super-Palindrome Numbers, proposed by Kris Jaganah:
let
Source = Excel.CurrentWorkbook(){[Name = "Table1"]}[Content],
Ans = Table.SelectRows(
Table.TransformColumns(
Source,
{
"Number",
each
let
a = Number.Sqrt(_),
b = Text.From(a),
c = if Text.Reverse(b) = b then _ else null
in
c
}
),
each ([Number] > 0)
)
in
AnsPower Query solution 3 for List Super-Palindrome Numbers, proposed by Alejandro Simón 🇵🇦 🇪🇸:
let
Sol = List.RemoveNulls(List.Skip(List.Generate(
()=> [x = 119, y = 0, z = 0],
each [z]<=15,
each [x=[x]+1, y = if Text.From(x)= Text.Reverse(Text.From(x)) and Text.From(Number.Sqrt(x)) = Text.Reverse(Text.From(Number.Sqrt(x))) then x else null,
z = if y <> null then [z]+1 else [z]],
each [y])))
in
Sol
El resultado difiere del presentado en el reto, pero a varios participantes les pasó igual.
Show translation
Show translation of this commentPower Query solution 4 for List Super-Palindrome Numbers, proposed by Ramiro Ayala Chávez:
let
l = {11 .. 1000000},
a = List.Transform(l, Text.From),
b = List.Select(a, each Text.Reverse(_) = _),
c = List.Transform(b, Number.From),
d = List.Transform(c, each Number.Power(_, 2)),
e = List.Transform(d, Text.From),
f = List.Select(e, each Text.Reverse(_) = _),
g = List.Transform(f, Number.From),
h = Table.FromColumns({g}, {"Answer Expected"}),
Sol = Table.FirstN(h, 15)
in
SolPower Query solution 5 for List Super-Palindrome Numbers, proposed by Rafael González B.:
let
Origen = List.Generate(
() => [i = 0, l = {}, N = 11],
each [i] <= 15,
each [
N = [N] + 1,
i = if oo and pp then [i] + 1 else [i],
o = Text.From([N]),
oo = o = Text.Reverse(o),
p = Number.Power([N],2),
pp = Text.From(p) = Text.Reverse(Text.From(p)),
l = if oo and pp then [l] & {p} else [l]
],
each [l]
)
in
List.Last(Origen)
🧙♂️🧙♂️🧙♂️
Power Query solution 6 for List Super-Palindrome Numbers, proposed by Glyn Willis:
let
Source = List.Sort(
List.Transform(
List.Select(
List.Combine(
List.Transform(
{1 .. 100},
(x) =>
List.Transform(
{""} & {0 .. 9},
(y) => Number.From(Text.From(x) & Text.From(y) & Text.Reverse(Text.From(x)))
)
)
),
(z) => Number.Mod(z * z, 1) = 0
),
(w) => w * w
)
),
Custom1 = List.Select(Source, each _ = Number.From(Text.Reverse(Text.From(_))))
in
Custom1Power Query solution 7 for List Super-Palindrome Numbers, proposed by Ian Segard:
let Source = {10..15000}, #"Converted to Table" = Table.FromList(Source, Splitter.SplitByNothing(), null, null, ExtraValues.Error), #"Changed Type" = Table.TransformColumnTypes(#"Converted to Table",{{"Column1", type text}}), #"Added Custom" = Table.AddColumn(#"Changed Type", "Custom", each Text.Reverse([Column1])), #"Added Conditional Column" = Table.AddColumn(#"Added Custom", "Custom.1", each if [Column1] = [Custom] then "y" else "m"), #"Filtered Rows" = Table.SelectRows(#"Added Conditional Column", each ([Custom.1] = "y")), #"Removed Columns" = Table.RemoveColumns(#"Filtered Rows",{"Custom", "Custom.1"}), #"Changed Type1" = Table.TransformColumnTypes(#"Removed Columns",{{"Column1", Int64.Type}}), #"Inserted Square" = Table.AddColumn(#"Changed Type1", "Square", each Number.Power([Column1], 2), Int64.Type), #"Changed Type2" = Table.TransformColumnTypes(#"Inserted Square",{{"Square", type text}}), #"Added Custom1" = Table.AddColumn(#"Changed Type2",
Solving the challenge of List Super-Palindrome Numbers with Excel
Excel solution 1 for List Super-Palindrome Numbers, proposed by Bo Rydobon 🇹🇭:
=LET(
p,
LAMBDA(
a,
FILTER(
a,
--BYROW(
MID(
a,
15-SEQUENCE(
,
14
),
1
),
CONCAT
)=a
)
),
p(
p(
SEQUENCE(
11111,
,
11
)
)^2
)
)Excel solution 2 for List Super-Palindrome Numbers, proposed by John V.:
=TOCOL(MAP(ROW(11:11111),LAMBDA(n,LET(p,LAMBDA(x,x=--CONCAT(MID(x,10-ROW(1:9),1))),n^2/p(n)/p(n^2)))),2)
or with Bo Rydobon 🇹🇭 idea:
✅=LET(p,LAMBDA(x,TOCOL(x/(x=--BYROW(MID(x,10-COLUMN(A:I),1),CONCAT)),2)),p(p(ROW(11:11111))^2))Excel solution 3 for List Super-Palindrome Numbers, proposed by محمد حلمي:
=LET(
s,
SEQUENCE(
11111,
,
11
),
v,
LAMBDA(
a,
a=--CONCAT(
MID(
a,
20-SEQUENCE(
19
),
1
)
)
),
e,
FILTER(
s,
MAP(
s,
v
)
)^2,
FILTER(
e,
MAP(
e,
v
)
)
)
///
=LET(
e,
LAMBDA(
x,
FILTER(
x,
MAP(
x,
LAMBDA(
a,
a=--CONCAT(
MID(
a,
20-SEQUENCE(
19
),
1
)
)
)
)
)
),
e(
e(
SEQUENCE(
11111,
,
11
)
)^2
)
)Excel solution 4 for List Super-Palindrome Numbers, proposed by Kris Jaganah:
=TOCOL(MAP(A2:A10,
LAMBDA(x,
LET(a,
x^0.5,
x/(--CONCAT(
MID(
a,
SEQUENCE(
LEN(
a
),
,
LEN(
a
),
-1
),
1
)
)=a)))),
3)Excel solution 5 for List Super-Palindrome Numbers, proposed by Sunny Baggu:
=LET(
_a,
SEQUENCE(
ROUNDUP(
7000000 / 100,
0
),
100,
100
),
_b,
TOCOL(
IF(
SQRT(
_a
) = INT(
SQRT(
_a
)
),
_a,
x
),
3
),
TAKE(
FILTER(
_b,
MAP(
_b,
LAMBDA(
x,
LET(
_e1,
LAMBDA(
x,
x = CONCAT(
MID(
x,
LEN(
x
) + 1 - SEQUENCE(
LEN(
x
)
),
1
)
) + 0
),
_e1(
x
)
)
)
)
),
15
)
)Excel solution 6 for List Super-Palindrome Numbers, proposed by 🇵🇪 Ned Navarrete C.:
=LET(e,
LAMBDA(m,
TOCOL(MAP(m,
LAMBDA(r,
LET(i,
--CONCAT(
MID(
r,
10-SEQUENCE(
9
),
1
)
),
r/(i=r)))),
3)),
TAKE(
e(
e(
SEQUENCE(
10^5,
,
11
)
)^2
),
15
))Excel solution 7 for List Super-Palindrome Numbers, proposed by Charles Roldan:
=LET(M,
LAMBDA(
g,
g(
g
)
),
K,
LAMBDA(
f,
LAMBDA(
n,
[m],
f(
n
)
)
),
Succ,
LAMBDA(
f,
M(
LAMBDA(
g,
LAMBDA(
n,
IF(
f(
n + 1
),
n + 1,
g(
g
)(
n + 1
)
)
)
)
)
),
_isPal,
M(LAMBDA(g,
LAMBDA(n,
IF(LEN(
n
) < 2,
TRUE,
IF(LEFT(
n
) = RIGHT(
n
),
g(
g
)(MID(
n,
2,
LEN(
n
) - 2
))))))),
_isSubPal,
LAMBDA(
n,
AND(
MAP(
n ^ {1,
2},
_isPal
)
)
),
SCAN(
10,
SEQUENCE(
15
),
K(
Succ(
_isSubPal
)
)
) ^ 2
)Excel solution 8 for List Super-Palindrome Numbers, proposed by Tyler Cameron:
=LET(
u,
SEQUENCE(
11101,
,
11
)^2,
FILTER(
u,
MAP(
u,
LAMBDA(
x,
LET(
a,
SQRT(
x
),
b,
LEN(
a
),
d,
ROUNDDOWN(
b/2,
0
),
e,
LEN(
x
),
f,
ROUNDDOWN(
e/2,
0
),
AND(
CONCAT(
MID(
a,
SEQUENCE(
d
),
1
)
)=CONCAT(
MID(
a,
SEQUENCE(
d,
,
b,
-1
),
1
)
),
CONCAT(
MID(
x,
SEQUENCE(
f
),
1
)
)=CONCAT(
MID(
x,
SEQUENCE(
f,
,
e,
-1
),
1
)
)
)
)
)
)
)
)Solving the challenge of List Super-Palindrome Numbers with Python in Excel
Python in Excel solution 1 for List Super-Palindrome Numbers, proposed by Abdallah Ally:
import pandas as pd
df = pd.read_excel(file_path, usecols='A')
def is_palindrome(n):
return str(n) == str(n)[::-1]
def super_palindrome(n):
numbers = []
number = 10
while len(numbers) < n:
sqrt_num = number ** 0.5
if (is_palindrome(number) and is_palindrome(int(sqrt_num))
and sqrt_num == int(sqrt_num)):
numbers.append(number)
number += 1
return numbers
df['My Answer'] = pd.Series(super_palindrome(15))
print(df)
Solving the challenge of List Super-Palindrome Numbers with R
R solution 1 for List Super-Palindrome Numbers, proposed by Konrad Gryczan, PhD:
library(tidyverse)
library(readxl)
is_palindrome = function(x) {
x = as.character(x)
x == str_c(rev(str_split(x, "")[[1]]), collapse = "")
}
result = tibble(palindrome = keep(10:1e5, is_palindrome)) %>%
mutate(square = palindrome^2,
is_palindrome = map_lgl(sq&uare, is_palindrome)) %>%
filter(is_palindrome) %>%
slice(1:15) %>%
select(square)
&&