Want to share your content on R-bloggers? click here if you have a blog, or here if you don't. 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==0)&(x>=1e7)&(x<1e8)){
r1=trunc(x/1e4)-digz*y
if ((digz*y>=1e3)&(digz*y<1e4) &(r1>9)&(r1<100)){
r2=10*r1+digx-7*y
if ((7*y>=1e2)&(7*y<1e3)&(r2>=1e2)&(r2<1e3)){
r3=10*r2+digx-digz*y
if ((digz*y>=1e2)&(digz*y<1e3)&(r3>9)&(r3<1e2)){
r4=10*r3+digx
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 this digit 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
Filed under: Books, Kids, pictures, R Tagged: arithmetics, division, FiveThirtyEight, long division, The Riddler  