252 MISCELLANEOUS PROBLEMS [CH. XII 



there is only one solution, and some will think this is a better 

 form in which to enunciate it. 



(v) In the last of these Berwick examples, it is required to 

 restore the working of the following division sum where all the 

 digits, except five " 5's," have been erased. 



. . . . ). 55 . . 5 .( . 5 . 

 . . 5 



u/ 



To this problem there is only one solution, the divisor being 

 3926 and the quotient 652. 



Class G. A third class of digit problems depends on finding 

 the values of certain symbols which represent specified numbers. 

 Two examples will suffice. 



(i) Here is a very simple illustrative specimen. The result 

 of multiplying be by be is abc, where the letters stand for certain 

 numbers. What are the numbers ? A brief examination shows 

 that be stands for 25, and therefore a stands for 6. 



(ii) Here is another example. The object is to find the 

 digits represented by letters in the following sum*: 



ab)cd e e b(bfb 

 c eb 



goh 

 c eb 

 c eb 



A solution may be obtained thus : Since the product of b by b 

 is a number which ends in b, b must be 1, 5, or 6. Since the 

 product of ab by & is a number of three digits, b cannot be 1. 

 The result of the subtraction of h from e is e, hence h = 0, and 

 therefore if b = 5 we have / even, and if b = 6 we have /= 5. 

 Also the result of the subtraction of c from g is c, hence g = 2c, 

 and therefore c cannot be greater than 4 : from which it follows 



* Strand Magazine, September, October, 1921. 



