THE PROBLEM OF THE COURIERS

447. Two couriers, A and B, travel on the same road in the direction from X to Y at the rates of m and n miles an hour, respectively. At a certain time, say 12 o'clock, A is at P, and B is at Q, a miles from P. Find when and where they are together.

X

P

♦

Y

SOLUTION

Suppose that time reckoned from 12 o'clock toward a later time is positive, and toward an earlier time, negative; also, that distances measured from P toward the right are positive, and toward the left, negative.

Let x represent the number of hours from 12 o'clock, and y the number of miles from P, when A and B are together. Then, they will be together y a miles from Q.

Since A travels mx miles and B travels nx miles before they are together,

When a > 0 and m>n, the numerator and denominator in (3) and also in (4) are positive; hence, x and y are positive.

That is, A overtakes B some time after 12 o'clock, somewhere at the right of P.

2. When a > 0 and m<n.

When a > 0 and m<n, both x and y are negative.

That is, at 12 o'clock B is ahead of A and gaining on him, and they were together some time before 12 o'clock and somewhere at the left of P.

3. When a > 0 and m = n.

When a > 0 and m = n, x and y are positive and infinitely great.

That is, at 12 o'clock B is ahead of A and traveling at the same rate; consequently, he will never be overtaken by A.

4. When a = 0 and m>n or m <n.

When a 0 and m>n or m< n, x = 0 and y = 0.

If m>n, x = +0 and y = +0. That is, at 12 o'clock A and B are together, and A is passing B.

If m < n, x = − O and y = gether, and B is passing A.

-0. That is, at 12 o'clock A and B are to

That is, A and B are together at 12 o'clock, and since they travel at the same rate they will be together at all times.

INDETERMINATE EQUATIONS

448. While a problem that presents more unknown literal numbers than independent equations involving them is in general indeterminate (§ 214), yet frequently by the introduction of a condition or conditions not leading to equations, the number of values of the unknown numbers may be limited and these values algebraically determined. A common condition is that the results shall be positive integers.

1. Solve the equation 5x+3y= 35 in positive integers.

SOLUTION

Since x and y are positive integers, 5 x must be equal to 5 or a multiple of 5, and 3y must be equal to 3 or a multiple of 3. Since the sum of these multiples is 35, if the multiples of 5 are subtracted from 35, one or more of the remainders will be a multiple of 3, if the problem is possible.

The only multiples of 5 that subtracted from 35 leave multiples of 3 are 5 and 4 times 5.

Or, since x must be a positive integer and by transposition 5x = 35 — 3y, the values of x must be 1, 2, 3, 4, 5, or 6, if the equation is possible. Substituting these values of x in the given equation and rejecting all those that give negative or fractional values for y, the positive integral values are found to be = 1 or 4, and y = 10 or 5.

2. Solve the equation 5x+8y = 107 in positive integers.

multiple of it is equal to an integer. Since 5 is contained in 3 times - 3 y, - 2 y times with a remainder of y, multiplying (4) by 3,

Equations (7) and (8) are called the general solution of the given equation

in integers.

To make y and x positive integers, it is evident from (7) that we must take w>0; and from (8) that we must take w<3.

Since w is an integer greater than 0 and less than 3, w = 1 or 2.

3. Determine whether the equation 10x + 15 = 53 may be satisfied by integral values of x and y.

If x and y are integers, the first member is integral.

Since the first member is equal to the fraction 33, it cannot be an integer Hence, x and y cannot be integers at the same time; that is, the equation is not satisfied by integral values of x and y.

ADY. ALG. -26

Find the least integral values of x and y in the following:

21. Separate 100 into two parts one of which is a multiple of 11, and the other a multiple of 6.

22. In what ways may a weight of 19 pounds be weighed with 5-pound and 2-pound weights?

23. A man has $300 that he wishes to expend for cows and sheep. If cows cost $45 apiece and sheep $6 apiece, how many can he buy of each?

24. If 9 apples and 5 oranges together cost 52 cents, what is the cost of one of each?

25. A grocer sold two packages of sugar for $1.25. One package contained a certain number of pounds of 7-cent sugar, the other a certain number of pounds of 5-cent sugar.

pounds were there in each package?

How many

sheep, hogs, and cows

26. A man sold 9 animals for $100. If he received $3 for a sheep, $6 for a hog, and $35 for a cow, how many of each did he sell?

27. A woman expended 93 cents for 14 yards of cloth, some at 5, some at 7, and the rest at 10 cents a yard. How many whole yards of each did she buy?

28. Divide 74 into three parts that shall give integral quotients when divided by 5, 6, and 7, respectively, the sum of which quotients shall be 12.

29. A purse contained 30 coins, consisting of half-dollars, quarters, and dimes. How many coins of each kind were there, if their aggregate value was $6.50?

30. A man bought 100 animals for $99. There were pigs, sheep, and ducks. If he paid $6 for a pig, $4 for a sheep, and 50 cents for a duck, how many of each did he buy?

31. What is the least number that will contain 25 with a remainder of 1, and 33 with a remainder of 2?

32. Find the least number that divided by 10 and by 11 will leave remainders of 3 and 6, respectively.

33. What is the least number that will contain 2, 3, 4, 5, and 6, each with a remainder of 1, and 7 without a remainder?

34. A man selling eggs to a grocer took them out of his basket 4 at a time and there was 1 egg over. The grocer put them into a box 5 at a time and there were 3 over. Both lost the count; but knowing that there were between 6 and 7 dozen eggs, the grocer paid for 6 dozen. How many eggs did he lose?

35. Four boys have a pile of marbles. A throws away 1 and takes of the remainder; B throws away 1 and takes of the remainder; C throws away 1 and takes of the remainder; D throws away 1, and each boy takes of the remainder. At least how many marbles must have been in the pile, and how many does each boy now have?