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a second horizontal line; proceed with this second line as with the first, and so continue, until there are no two terms which can be divided. The continued product of the divisors and the numbers in the last horizontal line will be the least common multiple.

EXAMPLES.

1. What is the least common multiple of 28, 35, 42, 77, and 70?

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3,

11,

1

Hence, 7×5×2×2×3×11=4620, is the multiple

sought.

there are no two numbers that can be measured by the same divisor; then the continual product of all the divisors and numbers in the last line will be the least common multiple required."

The above we have copied from Mr. Adams' Arithmetic. Nearly all our Arithmetics give, in substance, the same rule. We will now show, by an example, that this rule may give very different results, depending upon the divisors used, and of course the rule is in fault.

EXAMPLE.

What is the least common multiple of 12, 16, and 24?

We will work this example in three ways, as follows:

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12×2×8=192.

2, 1, 1 1, 2, 1 8×3×2×2=96. 4×3×2×2=48.

These operations, which are wrought strictly by this rule, give 192, 96, and 48, for the least multiple of 12, 16, and 24. Hence, the rule is wrong, and cannot be depended upon. The least common multiple of 12, 16, and 24, is 48, as may be found by either of our rules.

In the preceding example, we find, by inspection, that all the given numbers, 28, 35, 42, 77, 70, are divisible by 7, giving, for the second horizontal line, the numbers 4, 5, 6, 11, 10. Now 7 times the least multiple of 4,1 5, 6, 11, 10, is the least multiple of 28, 35, 42, 77, 70, since the latter numbers are respectively 7 times the former. Again, of the numbers 4, 5, 6, 11, 10, we find that 5 and 10 are divisible by 5. Dividing, we find for the third horizontal line, the numbers 4, 1, 6, 11, 2; now, as before, 5 times the least multiple of 4, 1, 6, 11, and 2, is the least multiple of the numbers of the second line. Again, of the numbers 4, 1, 6, 11, 2, we find that 4, 6, and 2, are divisible by 2. Dividing, we obtain for the fourth horizontal line, 2, 1, 3, 11, 1; and, as before, twice the least multiple of the last numbers is the least multiple of 4, 1, 6, 11, 2, which, multiplied by 5, gives the least multiple of 4, 5, 6, 11, 10; and this result being multiplied by 7, gives the least multiple of the numbers sought. When the division is continued until there are no two terms which can be divided, the continued product of the numbers constituting the last horizontal line is the least multiple. Hence the correctness of the rule.

2. What is the least common multiple of 46, 92, 374, and 23? Ans. 22 x 23 x 187=17204. 3. What is the least common multiple of 5, 15, 36 and 72? Ans. 23 x 32 x 5=360.

4. What is the least common multiple of 11, 77, 88, and 92? Ans. 23 x7x 11 x 23=14168.

5. What is the least common multiple of 14, 51, 102, Ans. 2x3x 7 x 17 x 250=178500.

and 500?

12. Suppose we wish to find all the divisors of 36, we proceed as follows: We resolve 36 into its prime factors, and thus obtain 36=22 × 32.

Now it is obvious that any combination of 2 and 3, which does not make use of these factors in a higher power than they occur in 22 x 32, must be a divisor of 36. All such combinations can be found by multiplying 1+2+4 by 1+3+9. Performing this multiplication, we obtain

1+2+4
1+3+9

1+2+4+3+6+12+9+18+36.

Therefore, the divisors of 36 are 1, 2, 3, 4, 6, 9, 12, 18 and 36.

Hence, to find all the divisors of any number, we have this

RULE.

Resolve the number into its prime factors; form as many series of terms as there are prime factors, by making 1 the first term of any one of the series, the first power of one of the prime factors for the second term, the second power of this factor for the third term, and so on, until we reach a power as high as occurred in the decomposition. Then multiply these series together, (by Rule under Art. 4,) and the partial products thus obtained will be the divisor sought.

EXAMPLES.

1. What are the divisors of 48?

Here we find 48-2' x 3. Therefore, our series of terms will be 1+2+4+8+16 and 1+3; multiplying these together, (by rule under Art. 4,) we get

1+2+4+8+16

1+3

1+2+4+8+16+3+6+12+24+48.

Therefore, the divisors of 48 are 1, 2, 3, 4, 6, 8, 12,

16, 24, and 48.

2. What are the divisors of 360?

Ans. {1, 2, 3, 4, 5, 6, 8, 9, 10, 12, 15, 18, 20, 24,

30, 36, 40, 45, 60, 72, 90, 120, 180, 360.

3. What are the divisors of 100?

Ans. 1, 2, 4, 5, 10, 20, 25, 50, 100.

4. What are the divisors of 810?

Ans. {

1, 2, 3, 5, 6, 9, 10, 15, 18, 27, 30, 45, 54, 81, 90, 135, 162, 270, 405, 810.

5. What are the divisors of 920?

Ans. {

1, 2, 4, 5, 8, 10, 20, 23, 40, 46, 92, 115, 184, 230, 460, 920.

6. What are the divisors of 840?

Ans.

1, 2, 3, 4, 5, 6, 7, 8, 10, 12, 14, 15, 20, 21, 24, 28, 30, 35, 40, 42, 56, 60, 70, 84, 105, 120, 140, 168, 210, 280, 420, 840.

7. What are the divisors of 1000?

Ans. {

1, 2, 4, 5, 8, 10, 20, 25, 40, 50, 100, 125, 200, 250, 500, 1000.

NOTE. We may observe that all the divisors of these respective divisors are included in the same series, since they must evidently be divisors of the original number.

13. Since the series of terms which we multiplied together by the last rule, to obtain the divisors of any number commenced with 1, it follows that the number of terms in each series will be one more than the units in the exponent of the factor used.

Hence, to find the number of divisors of any number without exhibiting them, we have this

RULE.

Resolve the number into its prime factors; increase each exponent by a unit, and then take their continued product, and it will express the number of divisors.

EXAMPLES.

1. How many divisors has 4320 ?

4320=25×33 x 5. In this case, the exponents are 5, 3, and 1, each of which being increased by one, we obtain 6, 4, and 2, the continued product of which is 6x4x2=48, the number of divisors sought.

2. How many divisors has 300?
3. How many divisors has 3500?
4. How many divisors has 162000?
5. How many divisors has 824 ?
6. How many divisors has 1172?
7. How many divisors has 6336 ?
8. How many divisors has 75600 ?

Ans. 18.

Ans. 24.

Ans. 100.
Ans. 8.

Ans. 6.

Ans. 42.

Ans. 120.

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