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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 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<y<125. Plus the constraint that 1000-8y≤99 implies y≥112. 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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