{"id":402,"date":"2024-02-04T18:06:36","date_gmt":"2024-02-05T02:06:36","guid":{"rendered":"https:\/\/www.rightlobemath.com\/blog\/?p=402"},"modified":"2020-02-04T18:33:46","modified_gmt":"2020-02-05T02:33:46","slug":"abacus-math-program-lesson-11-the-5-10-pair-combination-part-1","status":"publish","type":"post","link":"https:\/\/sumomath.com\/content\/abacus-math-program-lesson-11-the-5-10-pair-combination-part-1\/","title":{"rendered":"Abacus Math Program – Lesson 11 – The 5\/10 Pair Combination Part 1"},"content":{"rendered":"

So far in\u00a0Lesson 5<\/u><\/a>, 6<\/u><\/a>, 7<\/u><\/a>, and 8<\/u><\/a>\u00a0we introduced addition and subtraction with 10 pair complements when the operation requires more than 1 rod. In Lesson 9<\/u><\/a> and 10<\/u><\/a> we introduced addition and subtraction with 5 pair complements when the operation is done on a single rod, i.e. no carry or borrow is required from another rod. Now here in Lesson 11 we will combine both the 5 and 10 pair complements in a single bead movement.\u00a0In Lesson 11 Part 1\u00a0we will look at examples that\u00a0need to perform a 10 pair complement\u00a0addition but instead of subtracting the 10 pair directly we subtract in the form of a 5 pair complement. In all these cases we will need to subtract a 10 pair less than 5 but there will not be a sufficient number of lower\u00a0earthly beads touching the bar to subtract the 10 pair complement directly. So we will need to use the upper\u00a0heavenly 5 bead to complete the movement.<\/p>\n

Let’s look at the example problem 7 + 7. To add the second 7 to the first we use the 10 pair addition rule. So first add 1 to the next rod left and then\u00a0subtract 3, the 10 pair complement\u00a0of 7, from 7 on the target rod.\u00a0However to subtract 3 from 7 we need to use the 5 pair subtraction rule since\u00a0we only have 2 lower earthly beads touching the reckoning bar. Therefore to subtract the 10 pair complement 3, in the form of a 5 pair we add 2 earthly beads, 3’s five pair complement, and subtract the 5 bead leaving the\u00a0answer of 14.<\/p>\n

The possible\u00a05\/10 pair\u00a0addition combinations are:
<\/strong>
1. 5+6, 5+7, 5+8, 5+9
2. 6+6, 6+7, 6+8
3. 7+6, 7+7
4. 8+6<\/p>\n

In this first example we start by adding 17 to the abacus adding 1 to 0 on rod A and 7 to 0 on rod B. Next we add 6 to 17 by adding 6 to 7 on rod B. Notice this addition will require using the 5\/10 pair combination. First doing the 10 pair addition we add 1 to 1 on rod A and then subtract 4, the 10 pair complement of 6, from 7 on rod B. But since there are only 2 lower beads touching the bar we will need to subtract the 4 as a 5 pair. So subtract the 4 by adding 1, the 5 pair complement of 4, and subtract the 5 bead. The interim sum is now 23. Next add 2 to 23 by adding 2 to 3 on rod B using the 5 pair addition. Add 5 and subtract 2, the 5 pair complement of 3, on rod B for an interim sum of 25. Lastly subtract 8 from 25 by subtracting 8 from 5 on rod B using the 10 pair subtraction rule. First subtract 1 from 2 on rod A and then add 2, the 10 pair complement of 8, to 5 on rod B. The final answer is 17.<\/p>\n

\t\t\t\t\t\t\t\t\t\t\"\"\t\t\t\t\t\t\t\t\t\t\t<\/p>\n

In this next example we start by adding 5 to the abacus on rod B. Next we add 9 to 5 on rod B using the 10 pair addition rule. First add 1 to 0 on rod A and then subtract 1, the 10 pair complement of 9, from 5 on rod B. To subtract 1 from 5, we will need to use a 5 pair subtraction. So add 4, the 5 pair complement 1, to rod B and then subtract 5 for an interim sum of 14. Next add 8 to 14 by adding 8 to 4 on rod B. Again we will use a 10 pair addition by adding 1 to 1 on rod A and then subtract 2, the 10 pair complement of 8, from 4 on rod B for an interim sum of 22. Finally we add 3 to 22 by adding 3 to 2 using a 5 pair addition. So add 5 and then subtract 2, the 5 pair complement of 3 on rod B. The final answer is 25.<\/p>\n

\t\t\t\t\t\t\t\t\t\t\"\"\t\t\t\t\t\t\t\t\t\t\t<\/p>\n

In the last example we start by adding 15 to the abacus adding 1 to 0 on rod A and 5 to 0 on rod B. Next subtract 7 from 15 by using the 10 pair subtraction rule. First subtract 1 from 1 on rod A and then add 3, the 10 pair complement of 7, to 5 on rod B for an interim sum of 8. Next subtract 2 from 8 on rod B for an interim sum of 6. Lastly add 8 to 6 on rod B using the 10 pair addition rule. So first add 1 to 0 on rod A then subtract the 2, the 10 pair complement of 8, from 6 on rod B. To complete the subtraction we will need to use a 5 pair subtraction. So subtract 2 from 6 by adding 3, the 5 pair of 2, and then subtract\u00a05 for a final answer of 14.\u00a0<\/p>\n

\t\t\t\t\t\t\t\t\t\t\"\"\t\t\t\t\t\t\t\t\t\t\t<\/p>\n

Next up Lesson 12 the 5\/10 Pair Combination Part 2<\/u><\/a>.<\/p>\n","protected":false},"excerpt":{"rendered":"

So far in\u00a0Lesson 5, 6, 7, and 8\u00a0we introduced addition and subtraction with 10 pair complements when the operation requires more than 1 rod. In Lesson 9 and 10 we introduced addition and subtraction with 5 pair complements when the operation is done on a single rod, i.e. no carry or borrow is required from<\/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-402","post","type-post","status-publish","format-standard","hentry"],"_links":{"self":[{"href":"https:\/\/sumomath.com\/content\/wp-json\/wp\/v2\/posts\/402"}],"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=402"}],"version-history":[{"count":0,"href":"https:\/\/sumomath.com\/content\/wp-json\/wp\/v2\/posts\/402\/revisions"}],"wp:attachment":[{"href":"https:\/\/sumomath.com\/content\/wp-json\/wp\/v2\/media?parent=402"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/sumomath.com\/content\/wp-json\/wp\/v2\/categories?post=402"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/sumomath.com\/content\/wp-json\/wp\/v2\/tags?post=402"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}