CH. XII] MISCELLANEOUS PROBLEMS 249 



Denote the two missing digits, from right to left, by x and y. 

 Applying the theorem for 101, noting that each of the unknowns 

 cannot exceed 9, and for convenience putting y = 10 — z, this 

 equation gives z = 1, x = 7, y = 9. Hence the number is 49,187. 



(iv) The four-digit number . 8 . . is divisible by 1287. Find 

 the missing digits*. ^/ 



Denote these digits, from right to left, by x, y, z. We have 

 1287 = 9 x 11 x 13. Applying the suitable propositions, and 

 noting that each of the unknowns cannot exceed 9, we get x = l, 

 y = 6,z = S. Hence the number is 3861. 



(v) As a slightly harder example of this type suppose we 

 know that 6 . 80 . 8 . . 51 is exactly divisible by 73 and 137. 

 Find the missing digits f. The data suffice to^determine the 

 number, which is 6,780,187,951. ^ 



Glass B. Another and more difficult class of restoration 

 problems is illustrated by the following examples. Their solu- 

 tions involve analytical skill which cannot be reduced to rules. 



(i) I begin with an easy instance, said to be of Hindoo 

 origin, in which the problem is to restore the missing digits in 

 the annexed division sum where a certain six-digit number when 

 divided by a three-digit number gives a three-digit result %. 



• ■ • ) ( • • • 



. 0. . 



50 



\y 



4 . 



The solution involves no difficulty. The answer is that the 

 divisor is 215, and the quotient 573; the solution is unique. 



(ii) As a more difficult specimen I give the following prob- 

 lem, proposed in 1921 by Prof. Schub of Delft. A certain seven- 

 digit integer when divided by a certain six-digit integer gives 

 a result whose integral part is a two-digit number and whose 

 fractional part is a ten-digit expression of which the last nine 



* P. Delens, Problemes d'Arithmitique Amusante, Paris, 1914, p. 57. 



t Ibid. p. 60. 



J American Mathematical Monthly, 1921, vol. xxviii, p. 37. 



