**W**hile 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…