167 divided by , (then) halved, and then reduced to its square root, happens to be the number 5 ? CHAPTER VI-MIXED PROBLEMS. The rule for arriving at (the number of arrows in a bundle with the aid of the even number of) arrows constituting the common circumferential layer (of the bundle).- 288. Add three to the number of arrows forming the circum- ferential layer; then square this (resulting sum) and add again three (to this square quantity). If this be- further divided by 12, the quotient becomes the number of arrows to be found in the bundle. An example in illustration thereof. 289. The circumferential arrows are 18 in number. How many (in all) are the arrows to be found (in the bundle) within the quiver? O mathematician, give this out if you have taken pains in relation to the process of calculation known as vicitra- kuttikära. Thus ends vicitra-kuttikära in the chapter on mixed problems 288 The formula here given to find out the total number of ariows is (n+3)² + 3 where is the number of circumferential arrows. This formula can 12 be alived at from the following considerations It can be proved geometrically that only six circles can be described round another cucle, all of them being equal and each of them touching its two neighbouring circles as well as the central circle, that, round these circles again, only twelve circles of the same dimension can be described similarly, and that round these again, only 18 such circles are possible, and so on Thus, the first round has 6 cucles, the second 12, the third 18, and so on So that the number of cncles in any round, say p, is equal to 6 p. Now, the total number of circles in the given number of 1ounds p, calculated from the central circle, 18 1 + 1 x 6 + 2 x 6 + 3 x 6 + + px 6=1 + 6 (1+ 2+ 3+ If the + p) = 1 + 6 p (p+¹) == 1 + 3 p (p+1) 2 n 6 value of 6p is given, say, as n, the total number of circles is 1+ 3 x which is easily reducible to the formula given at the beginning of this note. + 1),
