# पृष्ठम्:गणितसारसङ्ग्रहः॒रङ्गाचार्येणानूदितः॒१९१२.djvu/३६३

एतत् पृष्ठम् परिष्कृतम् अस्ति
167
CHAPTER VI--MIXED PROBLEMS.

divided by ${\tfrac {3}{5}}$ , (then) halved, and then reduced to its square root, happens to be the number 5 ?

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 circumferential 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 tho quiver? O mathematician, give this out if you have taken pains in relation to the process of calculation known as vicitra-kuṭṭīkāra.

Thus ends vicitra-kuṭṭīkāra in the chapter on mixed problems.

288.^  The formula here given to find out the total number of arrows is ${\tfrac {(n+3)^{2}+3}{12}}$ where n is the number of circumferential arrows. This formula can be arrived at from the following considerations. It can be proved geometrically that only six circles can be described round another circle, 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 circles, the second 12, the third 18, and so on. So that the number of circles in any round, say p, is equal to 6 p.

Now the total number of circles in the given number of rounds p, calculated from the central circle, is $1+1\times 6+2\times 6+3\times 6+\cdots +p\times 6$ $=1+6(1+2+3+\cdots +p)$ $=1+6{\tfrac {p(p+1)}{2}}$ $=1+3p(p+1)$ . If the value of 6p is given, say, as n, the total number of circles in $1+3\times {\tfrac {n}{6}}({\tfrac {n}{6}}+1)$ . which is easily reducible to the formula given at the beginning of this note, 