Speed Mathematics Simplified (53 page)

Numbers to which all four approaches apply without remainders or fractions are few, but numbers for which two of the short cuts work are plentiful. Exercise your understanding of the four approaches by converting each of the following numbers in at least
two
ways:

Cover the explanations with your pad until you have found two or more short cuts for each of these numbers.

Here are the possibilities:

This does not mean that the different short cuts are of equal value whenever a number can be converted in two or more different ways. The value in each case depends not only on the possibilities of that number, but also on the other number involved. It also depends on which of the ways you find most adapted to your own ease and speed. The idea is to pick the simplest method for working the problem. Sometimes it will be one of the short cuts, and other times it will be to go ahead and do it with your new streamlined arithmetic. Flexibility is the key. You do not dig a hole for a rose bush with a steam shovel, or use a garden spade for a house foundation. Equally, you do not use three two-digit factors in place of multiplying by the number you factored, because it would not save you any work.

Combining Short Cuts

The possibilities of combining two short cuts in one problem are quite extensive and rather intriguing.

Breakdown by itself, for instance, makes sense only if you can break a number down to a simple base and an adjustment that is a simple fraction of the other number or the product. If you combine methods, however, you can use any aliquot or any easily factored or any easily changed number as a base.

In the range of numbers in which breakdown alone would save you work, 25 was not a useful base because you would still have to multiply through by two digits. But with aliquots to help, you could break down 26 into the combination technique “divide by 4 (aliquot) plus the other number (breakdown).”

Here is how you would do it:

Note that since you are dividing by the easy-to-handle divider 4, your short-cut method is to jot down only the answer.

Surprisingly difficult-looking problems can sometimes be solved almost at sight when one of the numbers happens to contain an aliquot as an easy breakdown base. 1375 × 8642 becomes merely 1/8 of 86420000, plus
of the product—because 1375 can be broken down into 1250 (1/8 of 10,000) plus 125 (
of 1250). Try it and see.

Do the following problem with an aliquot-breakdown.

Cover the explanation with your pad while you give it your best.

We break the number 385 into 375 plus 10—3/8 of 1,000 plus 10 times the other number. First you divide 4782000 by 8, jotting down only the answer. Multiply the result by 3. Then add 47820—10 × 4782:

Just for comparison, here is the usual way of solving the same problem:

Try breaking down these multipliers to aliquot bases:

Some of these get a little tricky, but each of them can be broken down to an aliquot base. Here is how:

Just as you can break down a complex multiplier to an aliquot base as well as a rounded-off base, so can you break down multipliers to a factorable base. There would be no point in breaking down 37 by the breakdown method alone. But by combining it with the factor short cut, 37 becomes 6 × 6 (factors) plus the other number (breakdown).

See if you can find the surprisingly easy breakdown-factor short cut for this problem: