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The foregoing method may be used with small numbers also, as,
L. C. M.2 × 3 × 7 × 25 × 12, or 12600.
We have shown that each of the numbers is contained in it a whole number of times. Therefore, 12600 must be a common multiple of all three numbers.
It must be the least common multiple, because 300, 252, and 175 have no common factor.
Some arithmetics give problems which involve the finding of the Least Common Multiple or Greatest Common Divisor of two or more numbers. Most of them are problems that have no practical application. They are deemed a waste of time here.
A good practical knowledge of how to find quickly the Least Common Multiple or Greatest Common Divisor is what is needed. This can be better accomplished by dealing with mere numbers, than in spending time on problems that do not occur in real life often enough to give compensation for the energy expended.
There is enough practical work to be done in arithmetic to give all mind training that might be secured by the use of impractical problems.
PRESENT IN THIS WAY:
Divide 36 by 18. What is the quotient? (Child) 2. Divide 9 times 4 by 9 times 2. What is the quotient? (Child) 2.
How often is 3 times 6 contained in 9 times 6? (Child) 3 times.
Divide 95 by 3× 5. (Child) The quotient is 3. Suppose we cut out or cancel the factor 5, because it is found both in the divisor and dividend, would the quotient be the same? (Child) Yes.
If the dividend and divisor both have common factors, the work can be made much easier by cancellation.
In order to cancel, write the dividend above the divisor, with a line between.
Cancelling like factors from both dividend and divisor does not change the quotient, but makes the work easier.
Cancelling the factor 3 from both 18 and 9, we
6 + 3.
6+3 = 2. The quotient is not changed.
Cancelling the factor 3 from both 18 and 9, we have
NOTE. In cancellation we use the last form.
To the Teacher :
Here is a good place to test the pupil on the effect of a 0 as a factor. Ask him what 8×2×0 × 4 will equal., Show him that 16 and 4 are incorrect results. Show him that whenever 0 is used as a factor, the product is always 0.
Multiply 620 by 18 and divide the product by 20.
13 into 65, 5 times.
Cancel 13 and 65 and write 5 below
85. 5 into 15, 3 times. Cancel 5 and 15 and write 3 above 15.
Dividing any one of a series of factors by a number divides their product by that number.
Take the series 9 × 12 × 15. Suppose we wish to divide this series of factors by 3.
The pupil when asked how to do it may say, "Divide the 9, 12, and 15, each by 3." Show him that he is wrong.
9 × 12 × 15 is really a number broken up into factors, and if you divide even one factor, it is as if you divide the whole number by that divisor. 9 × 12 × 15 = 1620. 1620 + 3 = 540.
Suppose we divide each of the factors 9, 12, and 15 by 3. We have left 3×4 × 5, or 60.
But, suppose we divide one factor by 3, say the first. We have then 3× 12 × 15, or 540. This is dividing the whole number by 3. 16203540, which is correct.
A farmer took to town 12 bu. of potatoes, which he sold at 35 per bu. In exchange he bought sugar at 4 per pound. How many pounds of sugar did he get!
His potatoes were worth 12 × 35.
He received as many pounds of sugar as 4 is contained times in 12 x 35, or 105.