{"id":480,"date":"2024-02-06T15:23:30","date_gmt":"2024-02-06T23:23:30","guid":{"rendered":"https:\/\/www.rightlobemath.com\/blog\/?p=480"},"modified":"2020-02-06T15:23:32","modified_gmt":"2020-02-06T23:23:32","slug":"how-to-multiply-on-the-abacus-with-a-two-digit-multiplier","status":"publish","type":"post","link":"https:\/\/sumomath.com\/content\/how-to-multiply-on-the-abacus-with-a-two-digit-multiplier\/","title":{"rendered":"How to Multiply on the Abacus with a two digit multiplier?"},"content":{"rendered":"

Let’s generalize what we did with Multiplication with Single Digit Multiplier<\/u><\/a> to a multi-digit multiplier. In this case we will look at a 2 digit by 2 digit multiplication. Again we will transform multiplication problems into a series of additions using our memorized 9×9 multiplication facts as we move both multiplier digits through the multiplicand digits. The technique we learn here will enable us to multiply any size multiplicand with any size multiplier.<\/p>\n

We will determine the starting rod the same way by summing the number of multiplicand and multiplier digits. As we work through the multiplier digits our starting rod will move 1 rod to the right as we shift to the next multiplier digit. Again, this is the simple mechanism used to ensure all additions are done at the correct place value working through the problem.<\/p>\n

Starting Rod = Number of Multiplicand Digits + Number of Multiplier Digits<\/p>\n

Let’s look at the following example:<\/p>\n

 52 x 63 = 3,276<\/p>\n

We first determine the starting rod location to begin entering our first product. Since the number of multiplicand digits is 2 and the number of multiplier digits is 2, our starting rod is 4 rods to the left of the chosen unit rod. Including our chosen unit rod we count 4 rods to the left and enter the first product by multiplying the 5 of 52 by the 6 of 63, 5×6 = 30. Enter 63 on the abacus on the 4th rod.<\/p>\n

\"\"<\/p>\n

Next we multiply the 2 of 52 by the 6 of 63, 2×6=12. Since we moved one digit to the right in the multiplicand we also move one rod to the right on the abacus to the 3rd rod. Add the product of 6×2 = 12 on the 3rd rod from the unit rod on the abacus.<\/p>\n

\"\"<\/p>\n

Now that we have multiplied each multiplicand digit with the first multiplier digit we move to the second multiplier digit the 3 of 63. Since we moved to the next multiplier digit we also move our starting rod one rod to the right. So for the second multiplier digit we enter the first product on rod 3 instead of rod 4. Please recognize that the reason for adjusting the starting rod for each multiplier digit is a simple mechanic used on the abacus to ensure we always maintain the correct place value for entering each product fact. Alternatively you could now think of the multiplier as a single digit number so our starting rod for the second multiplier digit will be 2 +1 = 3 (2 multiplicand digits and 1 multiplier digit). Next we add the product of the 5 in 52 by the 3 in 63, or 5×3=15 on the 3rd rod.<\/p>\n

\"\"<\/p>\n

To complete the problem we move 1 rod right on the abacus to the 2nd rod since we moved one digit to the right in the multiplicand and add the product of the 2 of 52 by the 3 of 63, 2×3=6 for a final answer of 3276. Please remember to maintain the correct place value as we move through a calculation we must enter each product fact as a 2 digit number. Even though the product 3×2 = 6 is a single digit number we enter it on the abacus as a two digit number 06 placing 0 on the 2nd rod and 6 on the unit rod.<\/p>\n

\"\"<\/p>\n","protected":false},"excerpt":{"rendered":"

Let’s generalize what we did with Multiplication with Single Digit Multiplier to a multi-digit multiplier. In this case we will look at a 2 digit by 2 digit multiplication. Again we will transform multiplication problems into a series of additions using our memorized 9×9 multiplication facts as we move both multiplier digits through the multiplicand digits.<\/p>\n","protected":false},"author":2,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[],"tags":[],"class_list":["post-480","post","type-post","status-publish","format-standard","hentry"],"_links":{"self":[{"href":"https:\/\/sumomath.com\/content\/wp-json\/wp\/v2\/posts\/480"}],"collection":[{"href":"https:\/\/sumomath.com\/content\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/sumomath.com\/content\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/sumomath.com\/content\/wp-json\/wp\/v2\/users\/2"}],"replies":[{"embeddable":true,"href":"https:\/\/sumomath.com\/content\/wp-json\/wp\/v2\/comments?post=480"}],"version-history":[{"count":0,"href":"https:\/\/sumomath.com\/content\/wp-json\/wp\/v2\/posts\/480\/revisions"}],"wp:attachment":[{"href":"https:\/\/sumomath.com\/content\/wp-json\/wp\/v2\/media?parent=480"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/sumomath.com\/content\/wp-json\/wp\/v2\/categories?post=480"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/sumomath.com\/content\/wp-json\/wp\/v2\/tags?post=480"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}