Proceeding in the same way, we see that 3m+1 is integral. The reasoning here given, taken backwards, shews that, if The following is a somewhat different solution of the same problem. Suppose 3+1 divisible by 10; then 3" must have 9 for its last digit. Now 3* = 81: hence 3* has 1 for its last digit. Hence 3" x 3" has 9 for its last digit, and therefore 1 + 3′′ × 3a is divisible by 10. Suppose 3+1 divisible by 10: then 3+ has 9 for its last digit: therefore 3" must have 9 for its last digit; for otherwise 3" x 3" would not have 9 for its last digit. Hence 3+1 is divisible by 19. 5. Find the value of tana or tanß from the equations tan (a + B) = tana cotẞ + cota tanß, tan (a–B) = tana cotß - cota tanß. or 1-tan'a tan'ß tanẞ(1+tan'ẞ) = 1 - tan'a tan B... By symmetry, tana (1+tan3a) = 1 – tan3ß tan3a.......... .(1). ..(2). From (1) and (2), tana - tanß + tan3a — tan3ß = 0, and therefore, since the second factor cannot be zero, 6. If A + B + C = 90°, shew that the least value of tan A+ tan'B + tan2C is 1. 0 = cot (A+B+C) = 1-tan A tan B-tan B tan C-tan Ctan A tan A+ tan B+ tan C – tan A tan B tan C' = 1. therefore tan A tan B+ tan B tan C + tan C tan A But since tan'A + tan3B = (tan A − tan B)2 + 2 tan A tan B, tan* B+tan C = tan C+tan A = (tan B- tan C) + 2 tan B tan C, (tan C-tan 4)2 + 2 tan C tan A; therefore tan A+ tan B+tan C1+ {(tan A-tan B) + (tan B - tan C)2 + (tan C-tan A)*}; therefore tan A+ tan B+tan' C is not < 1. 7. Lines, drawn through Y, Z, at right angles to the major axis of an ellipse, cut the circles, of which SP, HP, are diameters, in I, J, respectively. Prove that IS, JH, BC, produced indefinitely, intersect each other in a single point. Let IY, JZ, fig. (4), produced if necessary, intersect the major axis in Y', Z', respectively: then whence the triangle formed by producing IS, HJ, is isosceles, and therefore, CS, CH, being equal, the vertex of the triangle must lie in BC produced. Since the angles SIY', HJZ', are equal respectively to the angles SPY, HPZ, they can never be zero, and therefore SI, HJ, can never be perpendicular to the major axis. Thus the point of intersection of IS, JH, BC, can never move off to an infinite distance from C. 8. From any point T, (fig. 5), two tangents are drawn to a given ellipse, the points of contact being Q, Q': CQ, CQ', QQ', CT, are joined; V is the intersection of QQ, CT. Prove that the area of the rectilinear triangle QCQ' varies inversely as 9. A piece of uniform wire is bent into three sides of a square ABCD, of which the side AD is wanting; shew that, if it be hung up by the two points A and B successively, the angle between the two positions of BC is tan 18. -1 Let EF, fig. (6), be drawn parallel to BA, through E the middle point of BC. Then, if G be the centre of gravity of the piece of wire, EG equals two-thirds of BE. Draw HG parallel to BC, and join AG, BG. When the wire is hung up by A, AG will be vertical, and when hung up by B, BG will be vertical; therefore the inclinations of BC to the vertical will be equal to the angles which BC makes with AG and BG. Therefore the angle between the two positions of BC, (supposing it to be kept in the same plane,) will be the angle between AG and BG. Now tan AGB = tan (AGH+HGB) therefore the angle between the two positions of BC is tan ̄118. 10. A weight of given magnitude moves along the circumference of a circle, in which are fixed also two other weights: prove that the locus of the centre of gravity of the three weights is a circle. If the immoveable weights be varied in magnitude, their sum being constant, prove that the corresponding circular loci intercept equal portions of the chord joining the two immoveable weights. Let R, fig. (7), be the moveable weight, P and Q the stationary ones. Let G be the centre of gravity of P and Q, H that of P, Q, R. R. GR ∞ GR. But the locus of R is a circle; hence that of H is a circle, G being a similar point in the two circles, and GR, GH, similar lines. Hence, if HP', HQ', be drawn parallel to RP, RQ, P' and Q' will be points in the locus of H. and therefore, GH: GR being constant, and PQ being constant, P'Q' is constant. 11. A ball of elasticity e is projected from a point in an inclined plane, and, after once impinging upon the inclined plane, rebounds to its point of projection: prove that, a being the inclination of the inclined plane to the horizon, and ẞ that of the direction of projection to the inclined plane, cota. cotß = 1 + e. Let V be the velocity of projection. This is equivalent to V sinẞ and V cosẞ respectively perpendicular and parallel to the plane. Also the force of gravity is equivalent to g cosa and g sina, perpendicular and parallel to the plane. Consider the motion perpendicular to the plane. The time of flight = twice the time in which the velocity V sinẞ can be generated by the force g cosa after rebounding, the velocity perpendicular to the plane is eV sin ß, therefore time of returning to the point of projection Again, the motion parallel to the plane is not affected by the impact, therefore |