ALGEBRA

1851. (A). Prove the rule for finding the greatest common measure of two quantities.

(B). Shew that the greatest common measure of the two numbers is equal to the greatest common measure of any divisor made use of in the process and the corresponding dividend.

Let the usual process for finding G.C. M. of a and b, be pursued, C1, C2 С„..... being the successive remainders, P1, P2, P37 the successive quotients. One step in the process will be

27 3)

+1

from which it follows that C contains the whole system of factors common to C, and C, and that C, contains all those common to C, and C

Hence, G. C. M. of

7+1°

C, and CG.C. M. of C, and C

1850.

(by the same reasoning), G. C. M. of C and C =G.C.M. of C and C-99

&c.

G.C. M. of a and b.

(A). Shew how to find the least whole number which is accurately divisible by cach of two given whole num

bers.

(B). Find the least number of ounces of standard gold that can be coined into an exact number of halfsovereigns; standard gold being coined at the rate of £3 17s. 10 d. to an ounce.

In (4), if a, b be the two given numbers, d their greatest common divisor, the number required, i.e. their least common

In (B), if we find the number of shillings in the value of an ounce of gold and in half-a-sovereign, the least common multiple of these numbers divided by the number of shillings in the value of an ounce will give the least number of ounces that can be coined into an exact number of half-sovereigns.

and we have to find the least common multiple of 10 and 77·875. By (4), the least common multiple of 10000 and 77875 is

Hence the least common multiple of 10 and 77.875 is 6230, and

1848. (A). Prove the rules for finding the greatest common measure and least common multiple of two integers.

(B). Find the least number of pounds which can be paid in either half-crowns or guineas.

Here (B) is an example of finding the least common multiple of three numbers.

Reducing pounds, half-crowns, and guineas to sixpences, we have to find the least common multiple of 40, 5, and 42.

Now L. C. M. of 5 and 42 : 5 × 42 = 210, and L. C. M. of 40, 5, and 42 = L. C. M. of 40 and 210

1850.

(4). If a quantity vary directly as a when b is invariable, and inversely as b when a is invariable; prove

α

it will vary as

b

when both a and b are variable.

prove that

(B). If 5 men and 7 boys can reap a field of corn of 125 acres in 15 days, in how many days will 10 men and 3 boys reap a field of corn of 75 acres, each boy's work being of a man's?

Since each boy's work is of a man's, therefore 5 men and 7 boys are equivalent to (5+) men, or 22 men; and 10 men and 3 boys are equivalent to (10+ 1) men, or 11 men.

Again, since a given number of men will do more work in proportion as the time is increased; and a given piece of work will require more men to do it in proportion as the time allowed them is diminished; therefore

But the question gives that when the days 15 and acres 125, the men = 22; and we have to find the days when 75 and men = 11.

acres

1848.

(4). If Ax B when C is constant, and A ∞ C when B is constant; prove that A BC, when B and C both vary.

(B). Given that the area of an ellipse varies as either axis when the other is constant, and that the area of a circle of radius unity 3·14..., find the area of the ellipse whose axes are 3 and 5.

Let a, b be the semi-axes of the ellipse. Then, if A be the area of the ellipse,

and

therefore, by (A),

A a, b constant,

Ab, a constant,

A ∞ ab, a, b, both variable,

= p.ab, suppose.

1; then A becomes the area of a circle whose

.. 3·14... p ;

.. A = (3·14...) ab.

Now

Therefore

1851.

(A). Find the sum of a series of quantities in arithmetical progression.

(B). The square of the arithmetic mean of two quantities is equal to the arithmetic mean of the arithmetic and geometric means of the squares of the same two quantities.

We find from (4), if a be first term of the series, c the last, and n number of terms,

sum In (a + c).

Let there be three terms a, b, c of the progression. Then a + b + c = 3(a + c),

1850.

arithmetic mean between

a2 + c2

, Vac, i.e. between the

2

arithmetic and geometric means of a2, c2.

(A). Find the sum of a series of quantities in geometrical progression.

(B). Apply the result to find a common fraction equivalent to a recurring decimal fraction.

(C). If a be the first and the last of a series of n quantities in geometrical progression, prove that the continued product of the terms of the series is (al)1”.

(B). Let ·PPP... be the recurring decimal, where P, the recurring part, contains p digits. The decimal may be written

Now we have from (A), if a be the first term, and r the common ratio of the series,