II. +---------------+ | ● ● ● ● ● ● ● | | ● ● ● ● ● ● ● | +---------------+ +---------------+ | ● ● ● ● ● ● ● | | ● ● ● ● ● ● ● | +---------------+ +---------------+ | ● ● ● ● ● ● ● | | ● ● ● ● ● ● ● | +---------------+ +---------------+ | ● ● ● ● ● ● ● | | ● ● ● ● ● ● ● | +---------------+ The number of counters in all is 8 times 7, or 56. But (as in fig. I.) enclose each four rows in oblong figures, such as A and B. The number in each oblong is 4 times 7, or 28, and there are two of those oblongs; so that in the whole the number of counters is twice 28, or 28 x 2, or 7 first multiplied by 4, and that product multiplied by 2. In figure II. it is shewn that 7 multiplied by 8 is also 7 first multiplied by 2, and that product multiplied by 4. The same method may be applied to other numbers. Thus, since 80 is 8 times 10, 256 times 80 is 256 multiplied by 8, and that product multiplied by 10. If we use the signs, the foregoing assertions are made thus: 7 × 8 = 7 × 4 × 2 = 7 × 2 × 4. 256 × 80 = 256 × 8 × 10 = 256 × 10 × 8. EXERCISES. Shew that 2 × 3 × 4 × 5 = 2 × 4 × 3 × 5 = 5 × 4 × 2 × 3, &c. Shew that 18 × 100 = 18 × 57 + 18 × 43. 56. Articles (51) and (55) may be expressed in the following way, where by _ab_ we mean _a_ taken _b_ times; by _abc_, _a_ taken _b_ times, and the result taken _c_ times. _ab_ = _ba_. _abc_ = _acb_ = _bca_ = _bac_, &c. _abc_ = _a_ × (_bc_) = _b_ × (_ca_) = _c_ × (_ab_). If we would say that the same results are produced by multiplying by _b_, _c_, and _d_, one after the other, and by the product _bcd_ at once, we write the following: _a_ × _b_ × _c_ × _d_ = _a_ × _bcd_. The fact is, that if any numbers are to be multiplied together, the product of any two or more may be formed, and substituted instead of those two or more; thus, the product _abcdef_ may be formed by multiplying _ab_ _cde_ _f_ _abf_ _de_ _c_ _abc_ _def_ &c. 57. In order to multiply by 10, annex a cipher to the right hand of the multiplicand. Thus, 10 times 2356 is 23560. To shew this, write 2356 at length which is 2 thousands, 3 hundreds, 5 tens, and 6 units. Take each of these parts ten times, which, by (52), is the same as multiplying the whole number by 10, and it will then become 2 tens of thou. 3 tens of hun. 5 tens of tens, and 6 tens, which is 2 ten-thou. 3 thous. 5 hun. and 6 tens. This must be written 23560, because 6 is not to be 6 units, but 6 tens. Therefore 2356 × 10 = 23560. In the same way you may shew, that in order to multiply by 100 you must affix two ciphers to the right; to multiply by 1000 you must affix three ciphers, and so on. The rule will be best caught from the following table: