CH. XII] 



MISCELLANEOUS PROBLEMS 



251 



that every step in the working consists of two lines each of 

 which contains an equal number of digits. The problem is to 

 restore the whole working of the sum. The solution is unique 

 and gives a divisor of 125473 and a quotient of 58781. 



7. ) • -7 ( . .7.. 



7 . . 



(iv) The second problem is similar and requires the restora- 

 tion of the digits in the following division sum, where the position 

 of four "4's" is given, 



. . . ) 4( . 4. . 



4. 



4 . 



To this problem there are four solutions, the divisors being 846, 

 848, 943, 949 ; and the respective quotients 1419, 1418, 1418, 



1416. 



If we propound the problem (using five "4's") thus : 



••)• 



. . 4( . 4 



4. 



