TIL QUEST. 42. Solved by Robert Adrain.

Let r=10, and a, b, c, &c. the digits of any number in our notation: then the number itself will be ar+b+c, from which taking the sum of its digits, a+b+c, there remains axr2-1+bxr-1=(axr+1 +b)xr-1=(axr+1+b)x9: since, therefore, the difference between any number and the sum of its digits is divisible by 9, it evidently follows that the number and the sum of its digits, divided by 9, leave either none or equal remainders. Further, since r-1 in any other system would not be equal to 9, this is not a property of the number nine, but merely an effect of our present notation.

Scholium by the Editors.

Our young mathematical friends will easily demonstrate, that the difference of the like powers of any two numbers is divisible by the difference of the same two numbers; or that xy" is divisible by x-yi and consequently that 1 is divisible by r-1; and hence they will easily perceive that Mr. Adrain's elegant demonstration is not here limited to any particular number of digits.

n

The ingenious proposer, Thomas Bulmer, of Sunderland, in England, after answering this question nearly in the same manner as above, adds, "What ought we to think of some of our most eminent English mathematicians of the present day, who have gravely informed us, that this property of the digit 9 was unknown to the ancients? The ancients could not possibly be acquainted with an effect before its cause existed; and hence our profound sages might just as well have told us that old Pythagoras was never on board an American gun-boat.”

IV. QUEST. 43. Solved by James McCormic, Carlisle, Pennsylvania.

1, 3, 5, 7, 9, &c. continued to n terms, is evidently a series in arithmetical progression, the common difference being 2, and the last term 1+n-1×2 or 2n-1; hence, by a well known theorem, the sum of 2n-1+1 this series is= Xn=n2, as was to be proved. 2

V. QUEST. 44. Solved by the Proposer, Diarius Yankee,

VI. QUEST. 45. Solved by Robert Adrain.

Let d=AD the given side of the trapezium, a= the given area, m=sine, and n=cosine of the given angle A, and p=sine, and q=cosine of the other given angle D: the radius being here considered=1. Also put the length of the required side AB⇒CD= x; and let BE and CF be perpendicular G

to AD: then BE=mx, AE=nx, CF=
px, and DF-qx; and, consequently, B
AF=d-qx, and DE-d-nx. By men-
suration BE.AF+CF.DE=twice the

given area; that is, mxxd-qx+p¤× AF F D d-nx=dm+dpxx-mq+npxx2=2a, a

quadratic equation from which x is easily found.

"

Scholium by G. Baron.

Some negative Mathematicians will probably say that our learned friend ought here to have remarked, that the cosine of an obtuse angle is negative. Pray, scientific gentlemen, can any of you give a correct definition of a negative quantity?* If such a quantity cannot be defined, it must be either a creature of your own imaginations, or something which you do not rightly understand; and if you do not know what is meant by a negative quantity, you can have no definite idea of a negative cosine. Let us dive into the ocean of negative mysteries, when we have sounded its depths with the line of philosophy. If you contemplate the above solution, you will easily perceive that the perpendiculars BE and CF must always fall either on AD or AD produced; and that the theorem 'for finding the double area of the trapezium is universally true. Now, if to each of the different cases which the problem affords, you apply the same reasoning here used by Mr. Adrain, you will be enabled, in a rational manner, to account for all the changes of the signs which will occur in certain terms of the final equation; and hence you will see that this solution is universally true, and not limited to the case of acute angles. Remember that a solution and its final equation are two things which differ essentially.

E

B

VII. QUEST. 46. Solved by Henry Smith, New-York. · Let ABC be a plane rectilineal triangle, B its vertical angle, and C an angle adjacent to the base AC. From C, at right angles to AB produced when necessary, draw the straight line CE; and from B to D, the middle point of the base, draw the straight line

A

* Jared Mansfield, in his Mathematical Essays, has attempted to define a negative quantity. Jared's discourse on this subject reminds me of a Connecticut carle, who said that the moon was exactly the size of a johnny-cake.

BD. Then by prop. 23 and 22, book 2, Emerson's Geometry, AB+BC2=AC2+2.AB.BE=4.DC2+ 2.AB.BE. But by prop. 28, book 2, Emerson's Geometry, AB+BC=2.DC2+2.BD'; consequently 4.DC2+2.AB.BE=2.DC2+2.BD2, or DC+ AB.BE=BD': whence it is evident that AB.BE= BD DC, as was to be proved.

Corollary. From the above demonstration it is plain that BD is greater or less than DC, according as the angle B is acute or obtuse.

VIII. QUEST. 47. Solved by the Proposer, James North, Philadelphia.

Put c=100 gallons, the content of the vessel, s=10 seconds of time, and x=the quantity of wine remaining in the vessel at the end of any variable time t; then from the nature of the question s:t :: 1 the quantity of the mixture discharged in the infinitely small particle of time i; and c:x:::: whence whose fluent corrected gives hyp. log.

,

t

8

*=hyp. log. c——. Now, assume t=3600 the num

CS

nt

ber of seconds in one hour, and put n=43429448; and we have com. log.x=com.log.c =-4365399; consequently x=2.73237 gallons, the quantity of wine required.

CS

Corollary. The wine would never be entirely exhausted in the vessel; for in the above general equa tion it is easy to perceive that when x=o, t is infinite.

IX. QUEST. 48. Solved by Ben. Cheetham, Billiard-Hall.

Let ABCD represent the two plantations separated by the crooked fence EFG, and let BC and AD be straight lines. By a well known problem a straight Tine G/ may be drawn, which will separate these enclosures, without varying their respective areas. Then, if BC be parallel to AD, a straight line passing through the middle of Gl, and meeting BC and AD at right angles, will evidently be the required fence. But when BC and AD are not parallel, they may be produced to meet in some point P. In this case find PI-a mean proportional between Pl and PG; make PK=PI, and join KI, which will be the fence required. For in the triangles P/G and PIK, the angle P is common, and by construction PI.PG==PI2: PI.PK, or P/: PI:: PK: PG; therefore, by prop. 15, book 6, Euclid, the triangles P/G and PIK have equal areas; and consequently KI separates the two plantations, without changing their respective areas. Also, since PI-PK, or the angle PIK-PKI; by the lemma to prob. 68, Appendix to Simpson's Algebra, the straight line KI is a minimum, as was required.

=

Corollary. In every case of this problem the angle BKI is equal to the angle AIK.