248 MISCELLANEOUS PROBLEMS [CH. XII 



numbers starting with 100,000. We have 11,111 = 6 x 1851 + 5. 

 Hence we want the 5th digit in 101,851, and this is 5. 



Arithmetical Restorations. I take next a class of prob- 

 lems dealing with the reconstruction of arithmetical sums from 

 which various digits have been erased. Some of these questions 

 are easy, some difficult. This kind of exercise has attracted a 

 good deal of attention in recent years. I give examples of three 

 kinds of restoration. 



Class A. The solutions of one group of these restoration 

 questions depend on the well-known propositions that every 

 number 



a + 106 + 10 2 c + 10 s d+... 



is equal to any of certain expressions such as, 

 M(9) +a + b+c + d+... 

 M(ll) + a-b + c-d+... 

 M(33) +(a + l0b) + (c + lOd) + (e + 10f) + ... 

 M (101) + (a+ 106) - (c + lOd) + (e + 10/) - . . . 

 M (m) + (a + 106 + 1 2 c) + (d + We + 10 2 /) + . . . 

 M (n) + (a + 106 + 10 2 c) - (d + lOe + 10*/) + . . . 



where, in the penultimate line, m = 27, or 37, or 111, and in the 

 last line, n = 7, or 13, or 77, or 91, or 143. 



Examples, depending on such propositions, are not uncommon. 

 Here are four easy instances of this class of questions. 



(i) The product of 417 and . 1 ... is 9 ... 057. Find the missing 

 digits, each of which is represented by a dot. If the digits in 

 the undetermined multiplier are denoted in order by a, b, c, d, 

 and we take the steps of the multiplication in their reverse order, 

 we obtain successively d=l, c = 2, 6=9. Also the product has 

 7 digits, therefore a= 2. Hence the product is 9.1 B1.057. 



(ii) The seven-digit number 70 . . 34 . is exactly divisible 

 by 792. Find the missing digits, each of which is represented 

 by a dot. Since 792 is 8 x 9 x 11 we can easily show that the 

 number is 7,054,344. \J 



(iii) The five-digit number 4 . 18 . is divisible by 101. Find 

 the missing digits*. 



* P. Delens, ProbUmes d'Arithmgtique Amusante, Paris, 1914, p. 55. 



