NEW REBUSES, CHARADES, and QUERIES. I. REBUS, by Aminicus. Unto three-eighths of Saul's beloved fon, II. REBUS, by Mr. William Wardley. If to a cloyfter'd maid you join What all must do whene'er they dine; To thefe, reverfed, what we fay, III. REBUS, by Mrs. Hallilay. The royal title greets your lift'ning ear; The daughter too, by joyous fportsmen nam'd: IV. REBUS, by Eliza S The ancient feat of every art refin'd, A goddefs that o'er wisdom bears the fway, of the Dale. I. CHARADE, by Mrs. Richardfon. My first rides aloft on the loud roaring waves, My next ftands as Centinel; fervant to flaves, As my first and my fecond accomplish my whole, I II. CHA U. CHARADE, by Mr. Philip Rufher. To Job from midft the whirlwind once was told, When errant heroes fought, and ladies nurs'd. III. CHARADE, by Mr. John Savage, of Coventry. To many-yet to fome their kindeft friend. IV. CHARADE, by Mifs M. Fitzgerald, of Old Mofs. My first to the face of the fick is confin'd; My next is a weight of great moment you'll find; Or your credit, your fortune, and health you'll impair. I. QUERY, by Aramont. Whether can a handfome widow or a virtuous old maid beftow the more agreeable charms on a husband. II. QUERY, by Mifs Eliz. Baylie, of Chefbam, Bucks. Is it a good way for a young woman to get a busband, to entertain many lovers at once. HI. QUERY, by Mr. John Merriland, of Stony Stratford. Is it harder to gain a woman's love, or to keep it when gained. IV. QUERY, by Mr. Ralph Dutton, of Hull. Required the origin of the popular ftory of Dr. Fauftus. V. ASTRONOMICAL QUERY, by a Lunarian. Is the outward arc of the enlightened part of the moon's apparent difk, at any time apparently less than a femi-circle. ANSWERS to the MATHEMATICAL QUESTIONS. I-QUESTION 924 anfwered by Mr Jof. Garnett, from Mr Rodham's Academy, Richmond, Yorkshire. By the Queftion x, y, and z, are to be whole numbers, and from the 3d equation no one can be above 20, and fome one must be more than 14. Now from the ift equation the three roots must evidently be fome of the divifors of 160, which are 1, 2, 4, 5, 8, 10, 16, and 20, among which there are only two above 14, viz. 16 and 20; therefore the numbers are either 16 { 10 and 1, 5 or 20 and I, 4 and 2; of which 20, 8, and I are the only ones that will anfwer the conditions, and the word is HAT. The fame answered by Mr J. Holt, of Manchester. Because no word can be formed without a vowel, the value of one of the unknown quantities must answer to a vowel, and be fuch that the product of the other two in the first equation be a compofite number; but I and 5 only have these properties. Now if 5 be fubftituted for x in the 2d equation, then y2+z2=440; but no two perfect squares whatever will make this number; therefore x1, confequently 2+z2464, and y≈ 160; from this latter equation ≈ — which value of ≈ substituted in the former, it becomes y2+ - 2 160 25600 32 Hence the 464, or 4. -464y2 — — 25600, the two roots of which quadratic equation are 8 and 20, which are the values of y and z. required word is HAT. The fame by Mr Da. Kinnebrook, Jun. of Norwich. 2r; Let x+y+, and xy+xx+y=r; then will x+y2+ alfo x3 +3 +3 — 3xyz(480) = s2 = 2rxsrs, and x3 y3+z3 = 53 — 3 rs +480, whence come these two equations 22r=465, and s3 — 3rs + 480 =8513; from the first r= · 1395s — — 16066, where s +x. Now by the Ift and 2d original equations xy = 2—465 — x2, therefore x+y=√465-≈2+ - 160, the roots of which equation are 1, 8, 20; which, as the C. unknown quantities are alike concerned, are the values of x, y, z anfwering to HAT. Nearly according to one or other of these methods was the solution given by Meffieurs James Adams, Amicus, Job Ayres, Geo. Barnes, A Bengal Officer, Geo. Befwick, T. Bournley, A. Buchanan, T. Bulmer, Wm. Burden, John Campbell, John Cavill, Peter Charlton Fof. Cowing, Chrif. Cox, John Craggs, John Cullyer, James Cunliffe, James Dale, John Dalton, Rd. Dening, Rob. Dowden, John Elliott, Rev. L. Evans, J. Ferraby, A. Glendenning, G. Greaves, J. Griffith T.H, J. Hartley, J. Haycock, Jef. Heflop, F: Howard, Wm. Laws, Tho. Leybourn, Wm. Lawton, John Liddell, Nancy Mafon, R. Mountjoy, James Muleafter, Rd Nichelson, Jacob Park, W. Pearfon, Alex. Rowe, Rd. Rowe, John Ryley, John Sampson, Wm. Sanderfon, Ra. Simpfon, Collan Skewes, Geo. Stevenfon, W. Tarmior, Henry Taylor, Wm. Tomlinfon, John Tweddle, Wm. Virgo, Wm. Wardley, J. Whitcombe, Tho White, A. Whiteboufe, Rob. Wilkinson, and S. Woolcott. II QUESTION 925 anfwered by Mr M. Mooney, Dublin. Put 3y, and 4x=x; then by fubftitution, &c. 5% = √ 7%2; hence ; confequently x == 125 of x and y. 28 21 and y = 125 125 , the leaft values Or, univerfally: Put mxx, and nzy; then, by fubftitution, 71 3 m2 12.22. Put mana2, and cz2, or a 3x3 = m22; then the equations become az cx, and hence === (by restoring the values of a and c) 2 z3 m2 + n2) n, where m and n may be taken any numbers at pleasure, greater than n and the fum of their squares a square. If m4, and The fame by Mr John Dalton, of South-Cave. Since x and y, in the given equation ✔x2+y2 = 3/ x2 — y2, are to be rational numbers, it is plain they must denote the two legs of a right-angled triangle whofe three fides are rational Numbers. And because 32+42=52, fay as x:y :: 4:3, or y=x; this being fubftituted for y in the given equation, by reduction it becomes 125x3 This 28 =28x2, or x== ; and confequently y=x= 125 21 125 question admits of as many answers as there are right-angled triangles whofe three fides are rational numbers. 3 2 او The fame by a Bengal Officer, 2 By raifing each fide of the given equation to the 3d power, it becomes x2+2)3 = x2-2)2. Let x=ny, and it will become y 2 • n 2 + 1 ) 3 — n 2 — 12, and y22+1. Here must be greater than unity to be affirmative, and n2+1 a fquare" number to be rational. This can only happen when is taken equal the bafe of a right-angled triangle divided by the perpendicular, as 315, 24, &c. 4 12 21 28 125 125 2 2 Had the given equation been 3x2 + y2= √x2-y, then would n the hypothenufe divided by either leg. Mr. Geo. Barns fays, Your correspondent Dublinenfis. is either deceived himself, or has been very ingenious in impofing this question upon you as new: though fo much changed in form, it is the fame as Queft. 536, proposed in the Diary of 1765, and anfwered in the next year's Diary; or pa. 216, 220 of Hutton's Diarian Miscellany, vol. 3. Other ingenious anfwers were given by Meffieurs Amicus, Barnes, Bournley, Buchanan, Bulmer, Cavill, Cox, Craggs, Cullyer, Cunliffe, Dalton (of Kendal), Ferrahy, Griffith, T. H, Hartley, Haycock, Holt, Howard, Kinnebrook, Lawton, Leybourn, Muleafter, Park, Pearfon, Rowe, Ryley, Sampfon, Simpfon, Skewes, Stevenfon, Taylor, Tomlinfon, White, Wilkinson, and Woolcott. III QUESTION 926, anfwered. N. B. The number of acres was printed wrong in this question, one of the 2's being left out, for the number fhould have been 48002226, and not 4800226. Most of the correfpondents have taken notice of the inconfiftency in the numbers, and fome have corrected them by taking the area feen to be 10 times that given, and others by taking the difference of the paths of the top and bottom of the mountain to be of that which is given, viz ⚫6 inftead of 6 miles, both which alterations have the fame effect, and bring out the answers true very nearly. The Solution by Mr Geo. Barnes, of Wigton. There appears to be fome error in the numbers in this question, for in the prefent form the question is impoffible. Let therefore the number of acres of the earth's furface feen from the top of the mountain be increafed tenfold, viz. let it be 48012260 acres 75019 fquare miles. Now the radius being 3982, the circumference is 25019 miles, hence (by Dr Hutton's Menfuration 2d edit. pa. 197) 75019250193 miles T PIH R |