Endeterminate Analysis. INDETERMINATE ANALYSIS is the resolution of a species of equations where the number of the unknown quantities is greater than the number of equations, and where the answers are restricted to whole and positive numbers. AXIOMS. 1. If a whole number be subtracted from a whole number, the remainder is a whole number. 2. Any multiple of a whole number is a whole number. PROBLEM. Given the indeterminate equation ax± by =c to find all the possible values of x and y in integer numbers, suppose the numbers a, b, c, prime to each other. Find the value of one of the unknown quantities in terms of the other. Thus, if the equation be by+c ax-by=c, then x= a by-c Or, ax+by=c, then x=- a Increase the values of y successively by the coefficient of x, and diminish the values of x successively by the coefficient of y, and all the values of x and y will be obtained. RULE. 1. Of the two quantities (viz. the divisor and dividend), divide that which has the greatest coefficient by the least, the divisor by the remainder, and thus continue the process until a remainder appear in which the coefficient of y is unity, observing in each step to take the quotient figure such, that the coefficient of y in the multiple of the divisor may be the nearest possible to the coefficient of y in the dividend, whether greater or less than the dividend, and to subtract the less from the greater, whether the least be the multiple or the divisor. 2. Divide the absolute number in the remainder by the coefficient of x, and if the sign of the dividend be negative the remainder is the value of y; but if it be affirmative subtract the remainder from the divisor, and the last remainder will be the value of y.* EXAMPLES. (1) Given 19x-14y=11, to find the least affirmative values of x and y. 2 ay cient m of x the nearest possible to the coefficient a of ; then since ny-d a Let a-m=n, then less than either or a, but is less than either a or b; therefore n is much less than either a or b. It is evident by continuing this process, the operation will be similar to that of finding a common measure, and will be performed without the use of the denominator a, and that as the coefficient of y is continually diminishing, it may be reduced to unity. must also be a whole number, and ʼn will be And because the sign of the dividend 44 is negative, the remainder 6 is the least value of y; Whence the least value of y is 6, and the least value of x is 5. Note. If all the values of x and y were required in this example it would be impossible to give them, as the number of answers would be infinite. For by increasing the value of x by the coefficient of y, and increasing the value of y by the coefficient of x, We have { } x=5, 19, 33, 47, 61, 75, 89, &c. (2) Given 17x+29y=573, to find all the positive values of X and y in whole numbers. Then because the sign of 4011 in the dividend is affirmative, subtract the remainder 16 from 17, and the remainder is 1, which is the least value of y. Whence x= 29y-573 -32. Then diminishing x by the coefficient of y, in the original equation, and increasing y by the coefficient of x, We have x=32, 3 And y= 1, 18. (3) Given 21x+17y=2000, to find all the positive values of x and y in whole numbers. У Then diminishing x by the coefficient of y, and increasing y by the coefficient of x, We have x=92, 75, 58, 41, 24, And 7 y= 4, 25, 46, 67, 88, 109. (4) Given 3x+5y+7z=55, to find all the values of x, y, and z, in whole numbers. It is evident that 6 is the greatest value of z, because, if we put z=7, we have 72=7x7=49, then 49+3+5=57, which is too great by 2. Now in order to find the values of x and y we may proceed as before, thus, Hence it may be seen that the first values of y may be found by dividing the numbers 48, 41, 34, &c. successively by 3, and subtracting the remainders (when positive) will give the least values of y. In the two operations above the remainders are 0, and 2, whence 30=3=y, and 3-2=1=y. Observe. The other values may be found in a similar manner. (5) A company of men and women spent 1000 pence at a tavern; the men paid each 19 pence, and each woman How many men and women were there? 13 pence. Let x=the number of men, And y=the number of women. Then we shall have 192+13y=1000 And 17 is the least value of y; then diminishing x by the coefficient of y, and increasing y by the coefficient of x, we have x=41, 28, 15, 2 Now, when x=41, y=17, so that the men spent 779 pence, and the women 221. When x=28, y=36, then the men spent 532 pence, and the women 468. When x= =15, y=55, the men spent 285 pence, women 715. And, when x=2, y=74, the men spent 38 pence, women 962. (6) A company of men and women club together for the payment of a reckoning; each man pays 251. and each woman 161.; and it is found that all the women together have paid 17. more than the men. How many men and women were there? |