continental divide

May 18, 2017

(This article was originally published at R – Xi'an's Og, and syndicated at StatsBlogs.)

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){
  if ((digz[2]==0)&(x>=1e7)&(x<1e8)){ 
   if ((digz[5]*y>=1e3)&(digz[4]*y<1e4) &(r1>9)&(r1<100)){ 
    if ((7*y>=1e2)&(7*y<1e3)&(r2>=1e2)&(r2<1e3)){     
     if ((digz[3]*y>=1e2)&(digz[3]*y<1e3)&(r3>9)&(r3<1e2)){
       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…

Filed under: Books, Kids, pictures, R Tagged: arithmetics, division, FiveThirtyEight, long division, The Riddler

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