No matter how many digits the multiplier may have, the principle of multiplication is the same as that of multiplying by two-digit numbers. You have only to see that you do not mistake the order of multiplication and the rods on which to set products.
The12 rod soroban, notation as below:
Example : 37 x 432 = 15 984
Step 1: Set 37 on FG, with G as the unit rod, and set 432 on ABC.
Step 2: Multiplying the 7 in 37 by the 4 in 432, set the product 28 on HI.
Step 3: Multiplying the same 7 in 37 by the 3 in 432, set the product 21 on IJ. This makes a total of 301 on HIJ.
Step 4: Multiplying the same 7 in 37 by the 2 in 432, set the product 14 on JK, and clear G of its 7. This makes a total of 3,024 on HIJK.
Step 5: Multiplying the 3 in 37 by the 4 in 432, set the product 12 on GH. This makes a total of 15,024 on GHIJK.
Step 6: Multiplying the 3 in 37 by the 3 in 432, set the product 9 on 1. This makes a total of 15,924 on GHIJK.
Step 7: Multiplying the 3 in 37 by the 2 in 432, set the product 6 on J, and clear F of its 3. This makes, on GHIJK, a total of 15,984, which is the answer.
Example : 78 x 503 = 39 234
Step 1: Set 78 on FG, with G as the unit rod, and set 503 on ABC.
Step 2: Multiplying the 8 in 78 by the 5 in 508, set the 4 of the product 40 on H.
Step 3: Multiplying the same 8 in 78 by the 3 in 503, set the product 24 on JK and clear G of the 8. This makes a total of 4,024 on HIJK. In setting this product skip rod I as the second figure of the multiplier 503 is zero. In other words, the product must be set on JK instead of on IJ.
Step 4: Multiplying the 7 in 78 by the 5 in 503, set the product 35 on GH. This makes a total of 39 024 on GHIJK.
Step 5: Multiplying the same 7 in 78 by the 3 in 503, set the product 21 on IJ instead of HI, as the second figure of 503 is zero, and clear F of its 7. This leaves, on GHIJK, a total of 39 234, which is the answer.
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