2+X 3C=3A=12 P2 3+X 4D=4A=16 P+3 4+X-N 5E=5A-N-3 P+4 2+5 6B+E7A-N-11=R 2P+5=7a in its leaft (Pre. 5.) Value. Example 2. What Number is that a, which if multiplied by 25-M, and divided by 7=N, the Remainder will be 6=R? Here 25 leaves 44, 25×2 7 7 leaves 1B; therefore P1, and X÷B+N-A4, Then Example 3. What Number is that a, which if multiplied by 121=M, and divided by 49 N, the Remainder will be 17-R? Here 121X2 A, 49 121X1 leaves 23= leaves 46-B-24; therefore P-1 and X=A=23. Then Example 4. What Number is that a, which if multiplied by 23=M, and divided by 37 N, the Remainder will be 30-R? Here 23×2 leaves 94 37 23×3 27 leaves 32 B; therefore P=2, and X-B—A—23. Example 5. What Number is that a, which if multiplied by 6M, and divided by 21N, will leave 4=R? 6x4 6x5 Here leaves 3 A, leaves 6 B=2A; therefore P = 4, and 21 X=B-A=A=3• | 1|1=3 1+X 2B-6 21 1x7-N 47A=21, 21-21=0 P P-1-2 &c. It is therefore apparent that the Remainder will always be fome one of thefe Numbers 3, 6, 9, 12, 15, 18, and confequently can never be 4, as is required; therefore is no fuch Integer as the Queftion supposes. PROBLEM XII. 668. What two Integers are thofe a, e, whereof b times the former lefs, c times the latter is equal to a given Integer d: Or ba➡ce=d? Example 1. Suppofe b=23, C=37, and d=30. Then 1234-376=30 by the Question. Therefore a must be fome Number, which multiplied by 23, and divided by 37, leaves 30. Confequently a=19 in its least Value (In. 667.) accord ing to which the leaft Value of e is 23a-30 ing 37 to a 19, and 23 to e11, we shall have Anfwers to the Question ad infinitum, as follow. lale Example 2. Suppose b=39, 56, and d=20. Then 1394-56e=20 by the Question. Whence 2 394-20 56 Therefore a must be some Number, which multiplied by 39, and divided by 56, leaves 20. Confequently the least Value of a is 12, (In. 667.) according to which the leaft Value of e is 394-20 8. And by continually add56 ing 56 to a=12, and 39 to e=8, we shall have answers to the Question ad infinitum. SCHOLIUM VI. 37e+30 669. But this laft Problem will be more readily answered by seeking the leffer unknown Number first, especially when the Difference between the unknown Numbers is great; in which cafe proceed as follows. Ex. gr. If 23a -378-30, then by due Reduction a= fome Number, which multiplied into 37, 2×23—30=16: If 23a—37e=60, or a= 23 which fhews that e must be and divided by 23, will leave 37e+60 then e muft be fome 23 Number, 23 Number, which multiplied into 37, and divided by 23, will leave 3×2337e+90 then must be fome Number, 60=9: If 234-37e=90, or a== which multiplied into 37, and divided by 23, will leave 4×23-90-2: &c. And in general, whenever the Divifor N is lefs than the abfolute Number in the Dividend, it must be multiplied into the leaft Integer that makes it greater, and the faid abfolute Number fubtracted from the Product will give the Remainder R, to be proceeded with as in In. 667. But if N be greater, the Difference between it and the abfolute Number will be the Remainder 56e5 then e is fome Number which 39 , R. Ex. gr. If 39a-56e=5, or a= multiplied into 56, and divided by 39 will leave 39-5=34. If 39a-56e then e is fome Number, which multiplied into 56, and 10, or a= 56e+10 56e415 divided by 39, will leave 39-10=29. If 39a-56e=15, or a= 39 then e is fome Number, which multipled into 56, and divided by 39, will 56e+20 thene is fome Numleave 39-15=24. If 39a-56e=20, or a= ber, which multiplied into 56, and divided by 39, will leave 39—20=19, &c. PROBLEM XIII." 39 J 670. To find whether a given Integer be compofed of two or more given Integers, and if it be, to fhew how many ways... Example 1. To find what two Integers thofe are a, e, whereof 21 times the former, added to 17 times the latter, equals 2000. Or in other Terms. It is required to find how many Ways 100l. or 2000 s. may be paid by Guineas of 21 s. the Piece, and Piftols of 17 s. the Piece. Hence a must be fome Number, which multiplied into 21, and divided by 17, will leave 11; therefore the leaft Value of a is 7, (In. 667.) and confequently the greatest Value of e is 117+ 11-21×7 17 109. Then by continually adding 17 to a=7, and fubftracting 21 from 109, there will be found five other Anfwers to this Question in Integers,, as in the Table following. E Otherwife we may begin with finding the leaft Value of e, and greatest Value of a, thus; Which fhews to be fome Number, which being multiplied by 17, and divided by 21, will leave 5, therefore the leaft Value of e is 4 (In. 667.) and confequently the greatest Value of a is 95+ ble above. 5-17x4 21 92, as in the Ta Example 2. It is required to pay 611. 7s. or 1227 s. in Moidores of 11. 6s. 6 d. the Piece, Piftols of 17 s. 6d. the Piece, and Pieces of eight of 43. 62. the Piece. Ör in other Terms, It is required to find three Integers a, e, y, whereof 26.5a+17.50 +4.5y may equal 1227. According to which, the leaft Value of a is 60 by the last Example, and the greatest Value of is 693953×60 35 21 a defective Number; con fequently |