## Le Monde puzzle [#1107]

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

A light birthday problem as Le Monde mathematical puzzle:

Each member of a group of 35 persons writes down the number of those who share the same birth-month and the number of those who share the same birth-date [with them]. It happens that these 70 numbers include all integers from 0 to 10. Show that at least two people share a birth-day. What is the maximal number of people for this property to hold?

Which needs no R code since the result follows from the remark that the number of individuals sharing a birth-month with just one other, n¹, is a multiple of 2, the number of individuals sharing a birth-month with just two others, n², a multiple of 3, and so on. Hence, if no people share a birth-day, n¹,n²,…,n¹⁰>0 and

n¹+n²+…+n¹⁰ ≥ 2+3+…+11 = 6·11-1=65

which means that it is impossible that the 10 digits n¹,…,n¹⁰ are all positive. All the way up to 65 people. As an aside, no correction of the wrong solution to puzzle #1105 was published in the subsequent editions.

## Le Monde puzzle [#1105]

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

Another token game as Le Monde mathematical puzzle:

Archibald and Beatrix play with a pile of n>100 tokens, sequentially picking m tokens from the pile with m being a prime number [including m=1] or a multiple of 6, the winner taking the last tokens. If Beatrix knows n and proposes to Archibald to start, what is the value of n?

Which cannot be solved in a few lines of R code:

k<-function(n)n<4||all(n%%2:ceiling(sqrt(n))!=0)||!n%%6
g=(1:3)
n=c(4,i<-4)
while(max(n)<101){
if(k(i)) g=c(g,i) else{
while(i%in%g)i=i+1;j=4;o=!j
while(!o&(j<i)){
o=(j%in%n)&k(i-j);j=j+1}
if(o) g=c(g,i) else n=c(n,i)}
i=i+1}


since it returned no unsuccessful value above 100! With 4, 8, 85, 95, and 99 as predecessors. A rather surprising outcome and a big gap that most certainly has a straightforward explanation! Or a lack of understanding from yours truly: this post appears after the solution was published in Le Monde and I am more bemused than ever since the losing numbers in the journal are given as 4, 8, 85, … 89, and 129. With the slight hiccup that 89 is a prime number…. The other argument in the solution that there can only be five such losers is well-taken since there are only five possible non-zero remainders in the division by 6.

## Le Monde puzzle [#1106]

Posted in Books, Kids with tags , on July 5, 2019 by xi'an

After a translation of the original puzzle,a straight linear equation as Le Monde mathematical puzzle:

Find the ten integers x¹,x²,…. such that Ax=3x-1 where

$A=\left[\begin{matrix} 0& 1& 1& 1& 0& 0& 0& 0& 0& 0\\ 1& 0& 0& 0& 1& 1& 0& 0& 0& 0\\ 1& 0& 0& 1& 0& 0& 1& 0& 0& 0\\ 1& 0& 1& 0& 1& 0& 0& 0& 0& 0\\ 0& 1& 0& 1& 0& 1& 0& 0& 0& 0\\ 0& 1& 0& 0& 1& 0& 0& 0& 1& 1\\ 0& 0& 1& 0& 0& 0& 0& 1& 0& 0\\ 0& 0& 0& 0& 0& 0& 1& 0& 1& 0\\ 0& 0& 0& 0& 0& 1& 0& 1& 0& 1\\ 0& 0& 0& 0& 0& 1& 0& 0& 1& 0\\ \end{matrix}\right]$

Which returns 38 44 31 38 44 49 16 16 31 27 as its solution and is definitely of limited appeal…

## Le Monde puzzle [#1104]

Posted in Kids, R with tags , , , , on June 18, 2019 by xi'an

A palindromic Le Monde mathematical puzzle:

In a monetary system where all palindromic amounts between 1 and 10⁸ have a coin, find the numbers less than 10³ that cannot be paid with less than three coins. Find if 20,191,104 can be paid with two coins. Similarly, find if 11,042,019 can be paid with two or three coins.

Which can be solved in a few lines of R code:

coin=sort(c(1:9,(1:9)*11,outer(1:9*101,(0:9)*10,"+")))
amounz=sort(unique(c(coin,as.vector(outer(coin,coin,"+")))))
amounz=amounz[amounz<1e3]


and produces 9 amounts that cannot be paid with one or two coins.

21 32 43 54 65 76 87 98 201

It is also easy to check that three coins are enough to cover all amounts below 10³. For the second question, starting with n¹=20,188,102,  a simple downward search of palindromic pairs (n¹,n²) such that n¹+n²=20,188,102 led to n¹=16,755,761 and n²=3,435,343. And starting with 11,033,011, the same search does not produce any solution, while there are three coins such that n¹+n²+n³=11,042,019, for instance n¹=11,022,011, n²=20,002, and n³=6.

## easy Riddler

Posted in Kids, R with tags , , , on May 10, 2019 by xi'an

The riddle of the week is rather standard probability calculus

If N points are generated at random places on the perimeter of a circle, what is the probability that you can pick a diameter such that all of those points are on only one side of the newly halved circle?

Since it is equivalent to finding the range of N Uniform variates less than ½. And since the range of N Uniform variates is distributed as a Be(N-1,2) random variate. The resulting probability, which happens to be exactly $N/2^{N-1}$, is decreasing exponentially, as shown below…

## Le Monde puzzle [#1099]

Posted in Books, Kids, R with tags , , , , , on April 28, 2019 by xi'an

A simple 2×2 Le Monde mathematical puzzle:

Arielle and Brandwein play a game out of two distinct even integers between 1500 and 2500,  and y. Providing one another with either the pair (x/2,y+x/2) or the pair (x+y/2,y/2) until they run out of even possibilities or exceed 6 rounds. When x=2304, what is the value of y that makes Brandwein win?

Which I solved by a recursive function (under the constraint of a maximum of 11 levels of recursion):

nezt=function(x,y,i=1){
if ((i>11)||((is.odd(x)&is.odd(y)))){ return(-1)
}else{
z=-1
if (is.even(x)) z=-nezt(x/2,y+x/2,i+1)
if (is.even(y)) z=max(z,-nezt(y/2,x+y/2,i+1))
return(z)}}


and checking all values of y between 1500 and 2500 when x=2304, which produces y=1792 as the only value when Arielle loses. The reason behind (?) is that both 2304 and 1792 are divisible by 2⁸, which means no strategy avoids reaching stalemate after 8 steps, when it is Arielle’s turn to play.

## Le Monde puzzle [#1094]

Posted in Books, Kids, R with tags , , , , , , on April 23, 2019 by xi'an

A rather blah number Le Monde mathematical puzzle:

Find all integer multiples of 11111 with exactly one occurrence of each decimal digit..

Which I solved by brute force, by looking at the possible range of multiples (and  borrowing stringr:str_count from Robin!)

> combien=0
> for (i in 90001:900008){
j=i*11111
combien=combien+(min(stringr::str_count(j,paste(0:9)))==1)}
> combien
[1] 3456


And a bonus one:

Find all integers y that can write both as x³ and (10z)³+a with 1≤a≤999.

which does not offer much in terms of solutions since x³-v³=(x-v)(x²+xv+v²)=a shows that x² is less than 2a/3, meaning x is at most 25. Among such numbers only x=11,12 lead to a solution as x³=1331,1728.