is found 13; and since 2 is more than 1, the equation admits of two solutions. 2. Given 9x+13y-2000, to find whether there are more than 17; (2) solutions which the equation admits of. Here df=17, and we have the equation 9h-13k(-c-17ab)=11: from this we find the least value of h to be 7, and since does not amount to 1, the number of solutions is but 17. 1 159. Let us suppose =d+ ; and cy=d+; showing cx' b-1 a a the greatest degree of possibility without there being an integral solution, since b 1 is the greatest remainder, and 1 the least. C ab-a-b Then we have, by subtraction, or cab-a which shows that if c is greater than ab ab, the equation axbyc is possible in integers. cx' b b-1 So also if =d+· and b; we shall have c = ab(d—ƒ)+ab—a—b; or putting d-f=d, c=ab(d'+1)—(a+b), which shows the greatest value of c for any number of solutions; 1 cy' a -1 and by putting z=d+, and ·=ƒ+- we have c= b ab(d'-1)+(a+b), for the least value of c for any number of solutions {ab(d'+1)-(a+b)}—{ab (d'—1)+(a+b)}=2{ab—(a+b)}, expresses the difference between the greatest and least value of c. Thus, in the equation 8x+9y=559, 559 is the greatest number for 7 solutions; and in the equation 8x+9y=449, 449 is the least for 7 solutions. 160. If there be three unknown quantities, and only two equations for determining them, as ax+by+cz, d, and ex+fy+gz =h; exterminate one of these quantities in the usual way, and find the values of the other two from the resulting equation, as before. Then if the values thus found be separately substituted in either of the given equations, the corresponding values of the remaining quantities will likewise be determined; thus, 1. Let there be given x-2y+z-5, and 2x+y-2=7, to find the values of x, y, and z. Here, by multiplying the first of these subtracting the second from the product, we 3' 3 3 3 1 p=1, 2, 3, 4, &c. we shall have y 3, 6, 9, 12, 15, &c. and 2= :6, 11, 16, 21, 26, &c. But from the first of the two given equations, 25+2y-z; whence, by substituting the above values for y and z, the results will give x= = 5, 6, 7, 8, 9, &c. There fore the firsr six values of x, y, and z are as below; -5 6 7 8 6 910) where the law by which they may be continued is sufficiently y3 9 12 15 18 obvious. 1. Given 3x+5y + 7z = 100, to exhibit all the differenvaluesof x, y, and z in integers. Here z cannot be greater than 100-3-5 7 134; and by proceeding as in Prob. 1, but by adding 3, the coefficient of x, to this value of y, and subtracting 5, the coefficient of y, from the value of x, we shall obtain another answer, and by repeating this process continually, we shall obtain all the possible values of x and y, for this value of z; and in a similar manner are the values of x and y to be found when z = 2, &c. when all the possible solutions will be found to be 41 in number, and to be as follow; =20|15|10| 5 2= 2 = 6Y = 258 2=7 { { y= x=1611 6 y = 369 X =138 3 9 y= 25 x= 94 z=11 161. To determine the number of solutions that the equation ax+by+czd will admit of, two at least of the coefficients a, b, c, being prime to each other.* * When this is not the case, the proposed equation must be transformed to another that shall have two, at least, of its coefficients prime to each other. Thus, if the equation be 12x+-15y+ 20z=100001, by transposing 20z, and dividing by 3, we have -1 4x+5y=33384—7z+212 is an integer, which call u; then 2-3u+1; whence, by substitution, the proposed equation becomes 12x+15u+20(3u+1) = 100001, which, by transposing But the number of solutions that the equation ax + by = e will admit of is expressed by the integral parts of cx' cy b a 1 x' and y' being determined from the equation ax-by' =1. Therefore, in the equation ax + by: =d-cz, if we make z=1. 2. 3, 4, &c. successively, then the number of solutions in the equations ax+by dc will be the integral parts of ax+by=d-2c. ax+by=d-3c.... (d-c)'x' (d-cy' a b (d-2c)x' (d— 2c)y' the sum of which will be the whole number of solutions that the equation admits of, that is, if we take the sum of the integral parts of the arithmetical series (d—c)x', (d—2c)x', (d—3c)x' b + b + + b (d-3c)x, (d-4c)x' (d-5c)x + +&c. b the difference of the two will be the whole number of integral solutions. Now in each of these series the first and last terms, as also the number of terms, are known; for the general terms being (d-cz)x' (d-cz)y' we shall have the extremes by taking and a the extreme limits of z, that is, z= 1, and z last value of z also expresses the number of terms in the series. If, therefore, we find the sum of the two whole series, and then the sums of the fractional parts in each, by deducting these last sums, each from the corresponding whole sum, the sum of the integral parts of each series will be obtained. In summing the fractional parts, there will be no necessity to go through the whole series, for, as the denominator in each is constant, these fractions will necessarily recur in periods, and the number in each period can never exceed the denominator; it will therefore only be nethe 20, becomes divisible by 3, and we then have 4x+5y+20u= 33327; in which equation x and y have, of course, the same va lues as in the one proposed, and therefore the number of solutions must be the same; but in this last one value of u may be 0, because x=3u+1. See Simpson's Alg. p. 199, Prob. 15. *This will appear from considering the above series; for, if in the first series d'and c be prime to each other, and neither of them cessary to find the sum of the fractions in one period, and to multiply this sum by the number of periods, in order to get the sum of all the fractions, observing, however, that when there are not an exact number of periods, the overplus fractions must be summed by themselves, which may be readily done, since they will be the same as the leading terms of the first period; it must also be b remembered that is to be considered as a fraction in the first se ries, as in Prob. 2. 1. It is required to determine the number of integral solutions that the equation 3x+5y+7z=100 will admit of. Ans. 41. 3x+5y=100-7z if we make z=1, 2, 3, 4........13 in succession, then the number of solutions in the equation 3x+5y=a admits of is expressed by the integral parts of ap_aq = 3 5 , p and q being determined from the equation 3p-5q 1, (where it is evident p=2, q-1,) .. in the equation 3x+5y=100-7z, if we take z=1, 2, 3, to 13, which is the limit to the value of z, then the number of solutions in the equation 3x+5y=9, and 3x+5y=16, 9p 9q 16p 9q 16p 16q &c. will be contained in the integral parts of 5 3 5 3 and the whole number of solutions will be expressed by the difference of the sums of the arithmetical series, viz. 23p 30p 37P &c., +++ + 9p, 16p 5 5 2.9 2.16 5 9q 16q 23q, 30q, 37g 2.23 2.30 2.37 or, + + + + + 5 5 5 5 5 + 2.93 5 1.93 3 and consequently the sum of these series are 265 and 221, respectively. Now as the fractions which remain amount to something, these must be deducted from the above sums. and as the number of terms is 13, the sum of the first period of fractions is (2×3+1})=7}, and of the second 1 ×4+0: prime to b, each term will be wholly integral, that is, the fractions will all be 0. If b be prime to d, and not to c, the fractions will be all equal. If b be prime to c, but not to d, then the fractions will recur after the first integral term, which can never lie beyond the bth term; and finally, if a, b, c, be all prime to each other, the series of fractions will always recur after the bth term. Similar observations evidently apply to the second series. = 41, Ans. the whole number Hence we have 2651—7—217 of solutions, as required. 221-4217 41Ans. 2. Given the equation 5x+7y+11z=224, to find the number of solutions which it admits of in integers. Here the greatest limit of z 224-5-7 is 19; also in the equation 5x-7y'-1, we have x'=3, and y'-2, also a 5, and 7; therefore the two series, of which the sums are required, b: = beginning with the least terms 3.15 2.26 3.37 3.48 be + + + --+- + 7 .... (d—19c)x' (d-19c)y' and b a 3.59 will 3.213, 7 7 7 7 7 2.48 2.59 and + + +5 = and the number of terms in each 19. Now the sum of the first series is 9284, and the sum of the second ..866; also the first period of fractions, in the first series, is +÷+8+3+3+3+4=4, and the first period in the second series, is 0++ +3=2, 7 being considered as a fraction in the first period, but not in the second. Hence the number of terms in each series being 19, we have two periods and five terms of the first series 2X4+ the first five fractions = 104, for the sum of all the fractions, and therefore 928-10% = 918 sum of the integral terms of the first series; also in the second we have three periods and four terms 3x2+1=73, and therefore 866-72-859 — sum of the integral terms of the second series; whence 918-859-59 is the whole number of integral solutions. In a similar manner may the number of solutions be obtained, when there are four or more unknown quantities. 3. Required to determine the number of integral solutions that the equation 7x+9y+23z-9999 will admit of. Ans. 34365. 4. Required to determine the number of integral solutions that the equation 5x+7y+9z=93256 will admit of. Ans. 13801148. 5. Required to determine the num. of integral solutions that the equation 5x+7y+9z6665 will admit of. Ans. 70 6. Required to determine the num. of integral solution that the equation 5x+21y+27z 20000 will admit of. 162. To find such a whole number as, being divided by other given numbers a, b, c, shall leave given remainders f, g, h, RULE I. Call the number that is to be determined, z, the num Ans. 70734 |