Therefore, the greyhound will make 72 leaps to overtake the fox, 3 and during this time the fox will make 72X or 108. Verification. 2 The 72 leaps of the greyhound are equivalent to leaps of the fox. And 60+108=168, the leaps which the fox made from the beginning. 8. A father who had three children, ordered in his will, that his property should be divided amongst them in the following manner: the first to have a sum a, plus the nth part of what remained after subtracting a from the whole estate; the second, a sum 2a plus the nth part of what remained after subtracting from it the first part and 2a; the third, to have a sum 3a plus the nth part of what remained after subtracting from it the first two parts and 3a. In this manner his property was entirely divided; required the amount of it. Let a denote the property of the father. If by means of this quantity, algebraic expressions can be formed for the three parts, we may subtract their sum from the whole property x, and the remainder placed equal to zero, will give the equation of the problem. We will then endeavour to determine successively these three parts. Since x denotes the property of the father, x-a is what remains after having subtracted a from it; therefore x-a is the first remainder, and the part which the first child is to have, is a+- 9 or reducing to a common denominator, n In order to form the 2d part, this first part and 2a must be subtract (an+x-a) ed from this gives x- -2a or reducing to a com n mon denominator and subtracting, nx-3an-x+a 2d. remainder. n remainder; therefore, it is 2a+ Now, the second part is composed of 2a, plus the nth part of this a common denominator, 2an2+nx-3an−x+a =2d. part. Subtracting the two first parts plus 3a, from x, we have (an+x-a) X- -За n (2an2+nx-3an-x+a) n2 Or, reducing to a common denominator, and performing the operations indicated, But from the enunciation, the estate of the father is found to be entirely divided. Hence, the difference between x, and the sum of the three parts should be equal to zero. This gives the equation by making the denominators disappear, and performing the opera. tions indicated, we have n3x-6an3 — 3n3x+10an2+3nx-5an−x+a=0. 6an3-10an2+5an-a__a(6n3-10n2+5n—1) A more simple equation and result may be obtained, by observ. ing, that the part which goes to the third child is composed of 3a, plus the nth part of what remains, and that the estate is then entirely exhausted; that is, the third child has only the sum 3a, and the remainder just mentioned is nothing. Now the expression for this remainder has been found to be n2x-6an2-2nx+4an+x-a n2 Placing this equal to zero, and making the denominator disappear, we have n2x-6an2-2nx+4an+x-a=0. 6an2-4an+a_a(6n2-4n+1) Whence X= n2-2n+1 Verification. n2-2n+1 Το prove the numerical identity of this expression with the preceding, it is only necessary to show that the second can be deduced from the first, by suppressing a factor common to its numerator and denominator. Now if we apply the rule for finding the greatest common divisor (Art. 70.), to the two polynomials a(6n3—10n2+5n-1) and n3-3n2+3n—1, it will be seen that n-1 is a common factor, and by dividing the numerator and denominator of the first expression by this factor, the result will be the second. This problem shows the beginner how important it is to seize upon every circumstance in the enunciation of a question, which may facilitate the formation of the equation, otherwise he runs the risk of arriving at results more complicated than the nature of the the question requires. The conditions which have served to form successively the ex pressions for the three parts, are the explicit conditions of the problem; and the condition which has served to determine the most simple equation of the problem, is an implicit condition, which a little attention has sufficed to show, was comprehended in the enunciation. To obtain the values of the three parts, it is only necessary to substitute for x its value in the three expressions obtained for these parts. x== 10000(6X25-4x5+1) 10000×131 1310000 25-10+1 =81875. To verify the enunciation in this case: 81875-10000 The first child should have, 10000+ 5 or 24375. There remains then 81875-24375, or 57500, to divide between the other two children. The second should have, 20000+ 57500-20000 or 27500. Then there remains 57500-27500, or 30000, for the third child. Now 30000 is triple of 10000; hence the problem is verified. We can give a more simple and elegant solution to this problem, but it is less direct. It also depends upon the remark, that after having subtracted 3a and the two first parts from the whole estate, nothing remains. Denote the three remainders mentioned in the enunciation by r, r', r". The algebraic expressions for the three parts will be Now, 1st. From the enunciation, it is evident that r"=0. Therefore the third part is 3a. 2d. What remains after giving to the second child 2a+ can be represented by r'- or n (n-1)r' n n Moreover, this remainder also forms the third part. Therefore Зап За or convert n-1 n-1' 2an+a Then the second part is 2a+; ·÷n=2a+; ing the whole number into a fraction, and reducing, 3d. The remainder, after giving to the first a+ n-1 can be ex. n Now this remainder should form Or, by reducing the whole number and fractions to a common denominator, |