CH. XIIl] CALCULATING PRODIGIES 267 



how many hairs one inch long would be required to fill the 

 same space — the dimensions of a grain and a hair being given. 

 These problems involve high numbers, but are not intrinsically 

 difficult, though they could not be solved mentally unless the 

 calculator had a phenomenally good memory. In each case he 

 gave the correct answer, though only after considerable effort. 

 In 1753 he was asked to give the dimensions of a cubical 

 cornbin, which holds exactly one quarter of malt. He recog- 

 nized that to answer this required a process equivalent to the 

 extraction of a cube root, which was a novel idea to him, 

 but in an hour he said that the edge of the cube would be 

 between 25£ and 26 inches, which is correct : it has been sug- 

 gested that he got this answer by trying various numbers. 



Accounts of his performances were published, and his repu- 

 tation reached London, which he visited in 1754. During his 

 stay there he was examined by various members of the Royal 

 Society, who were satisfied as to the genuineness of his perform- 

 ances. Some of his acquaintances took him to Drury Lane 

 Theatre to see Garrick, being curious to see how a play would 

 impress his imagination. He was entirely unaffected by the 

 scene, but on coming out informed his hosts of the exact 

 number of words uttered by the various actors, and of the 

 number of steps taken by others in their dances. 



It was only in rare cases that he was able to explain his 

 methods of work, but enough is known of them to enable us to 

 say that they were clumsy. He described the process by which 

 he arrived at the product of 456 and 378 : shortly it was as 

 follows :— If we denote the former of these numbers by a, he 

 proceeded first to find 5a = (say) b; then to find 206 = (say) c; 

 and then to find 3c = (say) d. He next formed 156 = (say) e, 

 which he added to d. Lastly he formed 3a which, added to 

 the sum last obtained, gave the result. This is equivalent 

 to saying that he used the multiplier 378 in the form 

 (5 x 20 X 3) + (5 X 15) + 3. Mitchell suggests that this may 

 mean that Buxton counted by multiples of 60 and of 15, and 

 thus reduced the multiplication to addition. It may be so, for 

 it is difficult to suppose that he did not realize that successive 



