Various social instincts can be utilized in matches after the pattern of the spelling match, contests between rows, certain number games, and the like. The scoring of both the play and the work of the classroom is a useful field for control by the teacher of arithmetic.
Hunt ['12] has noted the more important games which have some considerable amount of arithmetical training as a by-product and which are more or less suitable for class use. Flynn ['12] has described games, most of them for home use, which give very definite arithmetical drill, though in many cases the drills are rather behind the needs of children old enough to understand and like the game itself.
It is possible to utilize the interests in mystery, tricks, and puzzles so as to arouse a certain form of respect for arithmetic and also to get computational work done. I quote one simple case from Miss Selkin's admirable collection ['12, p. 69 f.]:--
"We must admit that there is nothing particularly interesting in a long column of numbers to be added. Let the teacher, however, suggest that he can write the answer at sight, and the task will assume a totally different aspect.
"A very simple number trick of this kind can be performed by making use of the principle of complementary addition. The arithmetical complement of a number with respect to a larger number is the difference between these two numbers. Most interesting results can be obtained by using complements with respect to 9.
"The children may be called upon to suggest several numbers of two, three, or more digits. Below these write an equal number of addends and immediately announce the answer. The children, impressed by this apparently rapid addition, will set to work enthusiastically to test the results of this lightning calculation.
"Example:-- 357 } 999 682 } A × 3 793 } ---- 2997
"Explanation:--The addends in group A are written down at random or suggested by the class. Those in group B are their complements. To write the first number in group B we look at the first number in group A and, starting at the left write 6, the complement of 3 with respect to 9; 4, the complement of 5; 2, the complement of 7. The second and third addends in group B are derived in the same way. Since we have three addends in each group, the problem reduces itself to multiplying 999 by 3, or to taking 3000 - 3. Any number of addends may be used and each addend may consist of any number of digits."