## plusquamperfect squares

Posted in Books, Kids, R with tags , , , on April 2, 2021 by xi'an

For some perfect squares, when you remove the last digit, you get another perfect square. The first five perfect squares are 16, 49, 169, 256 and 361. What are the next three ones? Is there a more than perfect square other than 169 such that removing the last two digits returns a perfect square?

Writing an R code for plusquamperfect squares is straightforward and returns the following first 20 values

``` [1]         16         49        169        256        361       1444
[7]       3249      18496      64009     237169     364816     519841
[13]    2079364    4678569   26666896   92294449  341991049  526060096
[19]  749609641 2998438564
```

Adding the second constraint does not return a solution other than 169.

## Le Monde puzzle [#1119]

Posted in Kids, R with tags , , , , on December 8, 2019 by xi'an

A digit puzzle as Le weekly Monde current mathematical puzzle that sounds close to some earlier versions:

Perfect squares are pairs (a²,b²) with the same number of digits such that a²b² is itself a square. What is the pair providing a²b² less than 10⁶? Is there a solution with both integers enjoying ten digits?

The run of a brute force R code like

```cek<-function(a,b){
u<-trunc
if ((n<-u(log(a^2,ba=10)))==u(log(b^2,ba=10))&
(u(sqrt(a^2*10^(n+1)+b^2))^2==(a^2*10^(n+1)+b^2))) print(c(a,b))}```

provides solutions to the first question.

```[1] 2 3
[1] 4 9
[1] 12 20
[1] 15 25
[1] 18 30
[1] 49 99
[1] 126 155
[1] 154 300
[1] 159 281
[1] 177 277
[1] 228 100
[1] 252 310
[1] 285 125
```

with the (demonstrable) conclusion that the only pairs with an even number of digits are of the form (49…9²,9…9²), as for instance (49999²,99999²) with ten digits each.

## Le Monde puzzle [#1114]

Posted in Kids, R with tags , , , , , , on October 16, 2019 by xi'an

Another very low-key arithmetic problem as Le Monde current mathematical puzzle:

32761 is 181² and the difference of two cubes, which ones? And 181=9²+10², the sum of two consecutive integers. Is this a general rule, i.e. the root z of a perfect square that is the difference of two cubes is always the sum of two consecutive integers squared?

The solution proceeds by a very dumb R search of cubes, leading to

34761=105³-104³

The general rule can be failed by a single counter-example. Running

```sol=0;while(!sol){
x=sample(2:1e3,1)
y=sample(1:x,1)-1
sol=is.sqr(z<-x^3-y^3)
z=round(sqrt(z))
if (sol)
sol=(trunc(sqrt(z/2))^2+ceiling(sqrt(z/2))^2!=z)}
```

which is based on the fact that, if z is the sum of two consecutive integers squared, a² and (a+1)² then

2 a²<z<2 (a+1)²

Running the R code produces

x=14, y=7

as a counter-example. (Note that, however, if the difference of cubes of two consecutive integers is a square, then this square can be written as the sum of the squares of two different integers.) Reading the solution in the following issue led me to realise I had missed the consecutive in the statement of the puzzle!

## Le Monde puzzle [#1045]

Posted in Books, Kids with tags , , , , , , on May 13, 2018 by xi'an

An minor arithmetic Le Monde mathematical puzzle:

Take a sequence of 16  integers with 4 digits each, separated by 2,  such that it contains a perfect square and its sum is a perfect cube. What are the possible squares and cubes?

The question is dead easy to code in R

```for (x in as.integer(1e3:(1e4-16))){
if (max(round(sqrt(x+2*(0:15)))^2==x+2*(0:15))==1) {
b=sqrt((x+2*(0:15))[round(sqrt(x+2*(0:15)))^2==x+2*(0:15)])
if ((round((2*x+30)^(1/3)))^3==(2*x+30))
print(c(x,b,(16*(x+15))^(1/3)))}}
```

and return the following solutions:

```[1] 1357   37   28
[1] 5309   73   44
```

Nothing that exciting…!

## Le Monde puzzle [#944]

Posted in Books, Kids, pictures, Statistics, Travel, University life with tags , , , , on January 20, 2016 by xi'an

A completely dull Le Monde mathematical puzzle:

Find all integer pairs (a,b) less than 10⁶ such that a-b=2015 and ab is a perfect square.

If  I write the only condition as the function

```perfect=function(c){
return(c==trunc(sqrt(c))^2)}
```

and brute-force checked for all possible solutions

```for (A in 2015:1e6)
if (perfect(A*(A-2015))) print(A)
trunc(y/10)*11+(y-10*trunc(y/10))-(y==10*trunc(y/10))}
```

which produced

```[1] 2015
[1] 2304
[1] 2420
[1] 2511
[1] 3627
[1] 4212
[1] 7056
[1] 7595
[1] 16640
[1] 33759
[1] 41616
[1] 79092
[1] 204020
```

as the 13 possible answers. If one checks between 10⁶ and 5 10⁶, the only remaining solution is 1,016,064. Which is the highest possible one according to Le Monde.