Le Monde puzzle [#1053]

An easy arithmetic Le Monde mathematical puzzle again:

1. If coins come in units of 1, x, and y, what is the optimal value of (x,y) that minimises the number of coins representing an arbitrary price between 1 and 149?
2.  If the number of units is now four, what is the optimal choice?
   

The first question is fairly easy to code

coinz <- function(x,y){
  z=(1:149)
  if (y<x){xx=x;x=y;y=xx}
  ny=z%/%y
  nx=(z%%y)%/%x
  no=z-ny*y-nx*x
  return(max(no+nx+ny))
}

and returns M=12 as the maximal number of coins, corresponding to x=4 and y=22. And a price tag of 129.  For the second question, one unit is necessarily 1 (!) and there is just an extra loop to the above, which returns M=8, with other units taking several possible values:

[1] 40 11  3
[1] 41 11  3
[1] 55 15  4
[1] 56 15  4

A quick search revealed that this problem (or a variant) is solved in many places, from stackexchange (for an average—why average?, as it does not make sense when looking at real prices—number of coins, rather than maximal), to a paper by Shalit calling for the 18¢ coin, to Freakonomics, to Wikipedia, although this is about finding the minimum number of coins summing up to a given value, using fixed currency denominations (a knapsack problem). This Wikipedia page made me realise that my solution is not necessarily optimal, as I use the remainders from the larger denominations in my code, while there may be more efficient divisions. For instance, running the following dynamic programming code

coz=function(x,y){
  minco=1:149
  if (x<y){ xx=x;x=y;y=xx}
  for (i in 2:149){
    if (i%%x==0)
      minco[i]=i%/%x
    if (i%%y==0)
      minco[i]=min(minco[i],i%/%y)
    for (j in 1:max(1,trunc((i+1)/2)))
          minco[i]=min(minco[i],minco[j]+minco[i-j])
      }
  return(max(minco))}

returns the lower value of M=11 (with x=7,y=23) in the first case and M=7 in the second one.

Le Monde puzzle [#1018]

An arithmetic Le Monde mathematical puzzle (that first did not seem to involve R programming because of the large number of digits in the quantity involved):

An integer x with less than 100 digits is such that adding the digit 1 on both sides of x produces the integer 99x.  What are the last nine digits of x? And what are the possible numbers of digits of x?

The integer x satisfies the identity

where ω is the number of digits of x. This amounts to

10….01 = 89 x,

where there are ω zeros. Working with long integers in R could bring an immediate solution, but I went for a pedestrian version, handling each digit at a time and starting from the final one which is necessarily 9:

#multiply by 9
rap=0;row=NULL
for (i in length(x):1){
prud=rap+x[i]*9
row=c(prud%%10,row)
rap=prud%/%10}
row=c(rap,row)
#multiply by 80
rep=raw=0
for (i in length(x):1){
prud=rep+x[i]*8
raw=c(prud%%10,raw)
rep=prud%/%10}
#find next digit
y=(row[1]+raw[1]+(length(x)>1))%%10

returning

7 9 7 7 5 2 8 0 9

as the (only) last digits of x. The same code can be exploited to check that the complete multiplication produces a number of the form 10….01, hence to deduce that the length of x is either 21 or 65, with solutions

[1] 1 1 2 3 5 9 5 5 0 5 6 1 7 9 7 7 5 2 8 0 9
[1] 1 1 2 3 5 9 5 5 0 5 6 1 7 9 7 7 5 2 8 0 8 9 8 8 7 6 4 0 4 4 9 4 3 8 2 0 2 2
[39] 4 7 1 9 1 0 1 1 2 3 5 9 5 5 0 5 6 1 7 9 7 7 5 2 8 0 9

The maths question behind is to figure out the powers k of 10 such that

For instance, 10²≡11 mod (89) and 11¹¹≡88 mod (89) leads to the first solution ω=21. And then, since 10⁴⁴≡1 mod (89), ω=21+44=65 is another solution…

continental divide

While the Riddler puzzle this week was anticlimactic,  as it meant filling all digits in the above division towards a null remainder, it came as an interesting illustration of how different division is taught in the US versus France: when I saw the picture above, I had to go and check an American primary school on-line introduction to division, since the way I was taught in France is something like that

with the solution being that 12128316 = 124 x 97809… Solved by a dumb R exploration of all constraints:

for (y in 111:143)
for (z4 in 8:9)
for (oz in 0:999){
  z=oz+7e3+z4*1e4
  x=y*z
  digx=digits(x)
  digz=digits(z)
  if ((digz[2]==0)&(x>=1e7)&(x<1e8)){ 
   r1=trunc(x/1e4)-digz[5]*y 
   if ((digz[5]*y>=1e3)&(digz[4]*y<1e4) &(r1>9)&(r1<100)){ 
    r2=10*r1+digx[4]-7*y 
    if ((7*y>=1e2)&(7*y<1e3)&(r2>=1e2)&(r2<1e3)){     
     r3=10*r2+digx[3]-digz[3]*y 
     if ((digz[3]*y>=1e2)&(digz[3]*y<1e3)&(r3>9)&(r3<1e2)){
       r4=10*r3+digx[2]
       if (r4<y) solz=rbind(solz,c(y,z,x))
  }}}}

Looking for a computer-free resolution, the constraints on z exhibited by the picture are that (a) the second digit is 0 and the fourth digit is 7.  Moreover, the first and fifth digits are larger than 7 since y times these digits is a four-digit number. Better, since the second subtraction from a three-digit number by 7y returns a three-digit number and the third subtraction from a four-digit number by ny returns a two-digit number, n is larger than 7 but less than the first and fifth digits. Ergo, z is necessarily 97809! Furthermore, 8y<10³ and 9y≥10³, which means 111<y<125. Plus the constraint that 1000-8y≤99 implies y≥112. Nothing gained there! This leaves 12 values of y to study, unless there is another restriction I missed…