{"id":56,"date":"2023-06-05T15:29:38","date_gmt":"2023-06-05T15:29:38","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/tulsacc-math1473\/chapter\/the-positional-system-and-base-10\/"},"modified":"2025-08-27T18:56:54","modified_gmt":"2025-08-27T18:56:54","slug":"the-positional-system-and-base-10","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/tulsacc-math1473\/chapter\/the-positional-system-and-base-10\/","title":{"raw":"The Positional System with Other Bases","rendered":"The Positional System with Other Bases"},"content":{"raw":"<div class=\"textbox learning-objectives\">\r\n<h3>Learning Outcomes<\/h3>\r\n<ul>\r\n \t<li>Convert numbers between bases<\/li>\r\n<\/ul>\r\n<\/div>\r\n<span style=\"color: #077fab; font-size: 1.15em; font-weight: 600;\">The Development and Use of Different Number Bases<\/span>\r\n<h3>Other Bases<\/h3>\r\n<div class=\"textbox examples\">\r\n<h3>Powers of numbers other than 10<\/h3>\r\nRecall that in our base-ten number system, each place value in a number represents a power of ten. Our numbers take the form\r\n\r\n...\u00a0<span style=\"text-decoration: underline;\">thousands<\/span>\u00a0<span style=\"text-decoration: underline;\">hundreds<\/span>\u00a0<span style=\"text-decoration: underline;\">tens<\/span>\u00a0<span style=\"text-decoration: underline;\">ones<\/span>\u00a0.\r\n\r\n...\u00a0<span style=\"text-decoration: underline;\">10<\/span><sup>3<\/sup>\u00a0+ <span style=\"text-decoration: underline;\">10<\/span><sup>2<\/sup>\u00a0+\u00a0<span style=\"text-decoration: underline;\">10<\/span><sup>1<\/sup>\u00a0+ <span style=\"text-decoration: underline;\">10<\/span><sup>0\u00a0<\/sup>, where\u00a0[latex]10^{0}=1[\/latex] (In fact any number raised to the zero<sup>th\u00a0<\/sup>power equals one).\r\n\r\nWe have an intuitive understanding that [latex]10^{1}=10[\/latex] because we are using [latex]10[\/latex] as a factor [latex]1[\/latex] time. And certainly [latex]10^{2} = 10\\times 10 = 100, \\\\ 10^{3}=1000,[\/latex] and so on.\r\n\r\nThis pattern works with bases other than [latex]10[\/latex] as well. As you'll see below, using [latex]5[\/latex] as a base yields the following.\r\n\r\n[latex]5^{0}=1 \\\\ 5^{1}=5 \\\\ 5^{2}=5\\times 5 = 25 \\\\ 5^{3}=5\\times 5\\times 5 = 125,[\/latex] and so on.\r\n\r\n<\/div>\r\nFor example, let\u2019s suppose we adopt a base-five system. The only modern digits we would need for this system are 0, 1, 2, 3 and 4. What are the place values in such a system? To answer that, we start with the ones place, as most base systems do. However, if we were to count in this system, we could only get to four (4) before we had to jump up to the next place. Our base is 5, after all! What is that next place that we would jump to? It would not be tens, since we are no longer in base-ten. We\u2019re in a different numerical world. As the base-ten system progresses from 10<sup>0<\/sup> to 10<sup>1<\/sup>, so the base-five system moves from 5<sup>0<\/sup> to 5<sup>1<\/sup> = 5. Thus, we move from the ones to the fives.\r\n\r\nAfter the fives, we would move to the 5<sup>2<\/sup> place, or the twenty fives. Note that in base-ten, we would have gone from the tens to the hundreds, which is, of course, 10<sup>2<\/sup>.\r\n\r\nLet\u2019s take an example and build a table. Consider the number 30412 in base five. We will write this as 30412<sub>5<\/sub>, where the subscript 5 is not part of the number but indicates the base we\u2019re using. First off, note that this is NOT the number \u201cthirty thousand, four hundred twelve.\u201d We must be careful not to impose the base-ten system on this number. Here\u2019s what our table might look like. We will use it to convert this number to our more familiar base-ten system.\r\n<table>\r\n<tbody>\r\n<tr>\r\n<td><\/td>\r\n<td>Base 5<\/td>\r\n<td>This column coverts to base-ten<\/td>\r\n<td>In Base-Ten<\/td>\r\n<\/tr>\r\n<tr>\r\n<td><\/td>\r\n<td>3 \u00d7 5<sup>4<\/sup><\/td>\r\n<td>= 3 \u00d7 625<\/td>\r\n<td>= 1875<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>+<\/td>\r\n<td>0 \u00d7 5<sup>3<\/sup><\/td>\r\n<td>= 0 \u00d7 125<\/td>\r\n<td>= 0<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>+<\/td>\r\n<td>4 \u00d7 5<sup>2<\/sup><\/td>\r\n<td>= 4 \u00d7 25<\/td>\r\n<td>= 100<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>+<\/td>\r\n<td>1 \u00d7 5<sup>1<\/sup><\/td>\r\n<td>= 1 \u00d7 5<\/td>\r\n<td>= 5<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>+<\/td>\r\n<td>2 \u00d7 5<sup>0<\/sup><\/td>\r\n<td>= 2 \u00d7 1<\/td>\r\n<td>= 2<\/td>\r\n<\/tr>\r\n<tr>\r\n<td><\/td>\r\n<td><\/td>\r\n<td>Total<\/td>\r\n<td>1982<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nAs you can see, the number 30412<sub>5<\/sub> is equivalent to 1,982 in base-ten. We will say 30412<sub>5<\/sub> = 1982<sub>10<\/sub>. All of this may seem strange to you, but that\u2019s only because you are so used to the only system that you\u2019ve probably ever seen.\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nConvert [latex]6234_{7}[\/latex]\u00a0to a base [latex]10[\/latex] number.\r\n[reveal-answer q=\"482364\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"482364\"]We first note that we are given a base-7 number that we are to convert. Thus, our places will start at the ones (7<sup>0<\/sup>), and then move up to the 7s, 49s (7<sup>2<\/sup>), etc. Here\u2019s the breakdown:\r\n<table>\r\n<tbody>\r\n<tr>\r\n<td><\/td>\r\n<td>Base 7<\/td>\r\n<td>Convert<\/td>\r\n<td>Base 10<\/td>\r\n<\/tr>\r\n<tr>\r\n<td><\/td>\r\n<td>= 6\u00a0\u00d7\u00a07<sup>3<\/sup><\/td>\r\n<td>= 6\u00a0\u00d7 343<\/td>\r\n<td>= 2058<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>+<\/td>\r\n<td>= 2\u00a0\u00d7\u00a07<sup>2<\/sup><\/td>\r\n<td>= 2\u00a0\u00d7 49<\/td>\r\n<td>= 98<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>+<\/td>\r\n<td>= 3\u00a0\u00d7\u00a07<sup>1<\/sup><\/td>\r\n<td>= 3\u00a0\u00d7 7<\/td>\r\n<td>= 21<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>+<\/td>\r\n<td>= 4\u00a0\u00d7 7<sup>0<\/sup><\/td>\r\n<td>= 4\u00a0\u00d7 1<\/td>\r\n<td>= 4<\/td>\r\n<\/tr>\r\n<tr>\r\n<td><\/td>\r\n<td><\/td>\r\n<td>Total<\/td>\r\n<td>2181<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n&nbsp;\r\n\r\nThus [latex]6234_{7}=2181_{10}[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\nConvert [latex]41065_{7}[\/latex] to a base [latex]10[\/latex] number.\r\n[reveal-answer q=\"896067\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"896067\"][latex]41065_{7} = 9994_{10}[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n&nbsp;\r\n\r\n[ohm_question]8680[\/ohm_question]\r\n\r\n<\/div>\r\nWatch this video to see more examples of converting numbers in bases other than 10 into a base 10 number.\r\n\r\nhttps:\/\/youtu.be\/TjvexIVV_gI\r\n<h3>Converting from Base 10 to Other Bases<\/h3>\r\nConverting from an unfamiliar base to the familiar decimal system is not that difficult once you get the hang of it. It\u2019s only a matter of identifying each place and then multiplying each digit by the appropriate power. However, going the other direction can be a little trickier. Suppose you have a base-ten number and you want to convert to base-five. Let\u2019s start with some simple examples before we get to a more complicated one.\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nConvert twelve to a base-five number.\r\n\r\n[reveal-answer q=\"97454\"]Show Solution[\/reveal-answer]\r\n\r\n[hidden-answer a=\"97454\"]\r\n\r\nWe can probably easily see that we can rewrite this number as follows:\r\n<p style=\"padding-left: 30px;\">[latex]12 = (2 \\cdot 5) + (2 \\cdot 1)[\/latex]<\/p>\r\nHence, we have two fives and two ones. Hence, in base-five we would write twelve as [latex]22_{5}[\/latex]. Thus, [latex]12_{10}=22_{5}[\/latex].\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nConvert sixty-nine to a base-four number.\r\n\r\n[reveal-answer q=\"90808\"]Show Solution[\/reveal-answer]\r\n\r\n[hidden-answer a=\"90808\"]\r\n\r\n64 is the highest power of four that is less than 69.\r\n\r\nSo we rewrite sixty-nine as follows:\r\n<p style=\"padding-left: 30px;\">[latex]69 = (1 \\cdot 64) + (0 \\cdot 16) + (1 \\cdot 4) + (1 \\cdot 1)[\/latex]<\/p>\r\nHere, we have one sixty-four, zero sixteens, one four, and one one. Hence, in base four we have [latex]1011[\/latex]. Thus, [latex]69_{10}=1011_{4}[\/latex].\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nConvert the base-seven number [latex]3261_{7}[\/latex]\u00a0to base 10.\r\n\r\n[reveal-answer q=\"67834\"]Show Solution[\/reveal-answer]\r\n\r\n[hidden-answer a=\"67834\"]\r\n\r\nThe powers of 7 are:\r\n<p style=\"padding-left: 30px;\">[latex]7^{0}= 1[\/latex]\r\n[latex]7^{1}= 7[\/latex]\r\n[latex]7^{2}= 49[\/latex]\r\n[latex]7^{3}= 343[\/latex]\r\nEtc\u2026<\/p>\r\n[latex]3261_{7}= (3 \\cdot 343) + (2 \\cdot 49) + (6 \\cdot 7) + (1 \\cdot 1) =1170_{10}[\/latex].\r\n\r\nThus [latex]3261_{7}=1170_{10}[\/latex].\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\nConvert [latex]143[\/latex] to base [latex]5[\/latex]\r\n\r\n&nbsp;\r\n\r\n[ohm_question]8681[\/ohm_question]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\nConvert the base-three number [latex]21021_{3}[\/latex]\u00a0to base [latex]10[\/latex].\r\n\r\n&nbsp;\r\n\r\n[ohm_question]8680[\/ohm_question]\r\n\r\n<\/div>\r\n<em>In general, when converting from base<\/em><em>-ten to some other base, it is often helpful to determine the highest power of the base that will divide into the given number at least once.<\/em>\r\n\r\nIn the last example, [latex]5^2= 25[\/latex] is the largest power of five that is present in 69, so that was our starting point. If we had moved to [latex]5^3 = 125[\/latex], then 125 would not divide into 69 at least once.\r\n<div class=\"textbox\">\r\n<h3>Converting from Base 10 to Base <em>b<\/em><\/h3>\r\n<ol>\r\n \t<li>Find the highest power of the base <em>b<\/em> that will divide into the given number at least once and then divide.<\/li>\r\n \t<li>Write down the whole number part, then use the remainder from division in the next step.<\/li>\r\n \t<li>Repeat steps 1 and 2, dividing by the next highest power of the base <em>b<\/em>, writing down the whole number part (including 0), and using the remainder in the next step.<\/li>\r\n \t<li>Continue until the remainder is smaller than the base. This last remainder will be in the \u201cones\u201d place.<\/li>\r\n \t<li>Collect all your whole number parts to get your number in base <em>b<\/em> notation.<\/li>\r\n<\/ol>\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nConvert the base-ten number [latex]348[\/latex] to base-five.\r\n<h4>Solution<\/h4>\r\nThe powers of five are:\r\n<p style=\"padding-left: 30px;\">5<sup>0<\/sup> = 1\r\n5<sup>1<\/sup> = 5\r\n5<sup>2<\/sup> = 25\r\n5<sup>3<\/sup> = 125\r\n5<sup>4<\/sup> = 625\r\nEtc\u2026<\/p>\r\nSince 348 is smaller than 625, but bigger than 125, we see that 5<sup>3\u00a0<\/sup>= 125 is the highest power of five present in 348. So we divide 125 into 348 to see how many of them there are:\r\n<p style=\"padding-left: 30px;\">348\u00a0\u00f7\u00a0125 = <span style=\"color: #ff0000;\">2<\/span> with remainder 98<\/p>\r\nWe write down the whole part, 2, and continue with the remainder. There are 98 left over, so we see how many 25s (the next power of five to the right of 125s) there are in the remainder:\r\n<p style=\"padding-left: 30px;\">98\u00a0\u00f7\u00a025 = <span style=\"color: #ff0000;\">3<\/span> with remainder 23<\/p>\r\nWe write down the whole part, 3, and continue with the remainder. There are 23 left over, so we look at the next place, the 5s:\r\n<p style=\"padding-left: 30px;\">23\u00a0\u00f7\u00a05 = <span style=\"color: #ff0000;\">4<\/span> with remainder <span style=\"color: #ff0000;\">3<\/span><\/p>\r\nThis leaves us with 3, which is less than our base, so this number will be in the \u201cones\u201d place. We are ready to assemble our base-five number:\r\n<p style=\"padding-left: 30px;\">348 = (<span style=\"color: #ff0000;\">2<\/span> \u00d7 5<sup>3<\/sup>) + (<span style=\"color: #ff0000;\">3<\/span> \u00d7 5<sup>2<\/sup>) + (<span style=\"color: #ff0000;\">4<\/span> \u00d7 5<sup>1<\/sup>) + (<span style=\"color: #ff0000;\">3<\/span> \u00d7 1)<\/p>\r\nHence, our base-five number is <span style=\"color: #ff0000;\">2343<\/span>. We\u2019ll say that [latex]348_{10}=2343_{5}[\/latex].\r\n\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nConvert the base-ten number [latex]4,509[\/latex] to base-seven.\r\n<h4>Solution<\/h4>\r\nThe powers of 7 are:\r\n<p style=\"padding-left: 30px;\">7<sup>0<\/sup> = 1\r\n7<sup>1<\/sup> = 7\r\n7<sup>2<\/sup> = 49\r\n7<sup>3<\/sup> = 343\r\n7<sup>4<\/sup> = 2401\r\n7<sup>5<\/sup> = 16807\r\nEtc\u2026<\/p>\r\nThe highest power of 7 that will divide into 4,509 is 7<sup>4<\/sup> = 2401.\u00a0With division, we see that it will go in 1 time with a remainder of 2108. So we have 1 in the 7<sup>4<\/sup> place.\r\n\r\nThe next power down is 7<sup>3<\/sup> = 343, which goes into 2108 six times with a new remainder of 50. So we have 6 in the 7<sup>3<\/sup> place.\r\n\r\nThe next power down is 7<sup>2<\/sup> = 49, which goes into 50 once with a new remainder of 1. So there is a 1 in the 7<sup>2<\/sup> place.\r\n\r\nThe next power down is 7<sup>1<\/sup> but there was only a remainder of 1, so that means there is a 0 in the 7s place and we still have 1 as a remainder.\r\n\r\nThat, of course, means that we have 1 in the ones place.\r\n\r\nPutting all of this together means that [latex]4,509_{10}=16101_{7}[\/latex].\r\n<div class=\"textbox\">\r\n\r\n4509 \u00f7 7<sup>4<\/sup> = <span style=\"color: #ff0000;\">1<\/span> R 2108\r\n\r\n2108 \u00f7 7<sup>3<\/sup> = <span style=\"color: #ff0000;\">6<\/span> R 50\r\n\r\n50 \u00f7 7<sup>2<\/sup> = <span style=\"color: #ff0000;\">1<\/span> R 1\r\n\r\n1 \u00f7 7<sup>1<\/sup> = <span style=\"color: #ff0000;\">1<\/span>\r\n\r\n4,509<sub>10<\/sub> = <span style=\"color: #ff0000;\">16101<\/span><sub>7<\/sub>.\r\n\r\n<\/div>\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\nConvert [latex]657_{10}[\/latex]\u00a0to a base [latex]4[\/latex] number.\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\nConvert [latex]8377_{10}[\/latex]\u00a0to a base [latex]8[\/latex] number.\r\n\r\n<\/div>","rendered":"<div class=\"textbox learning-objectives\">\n<h3>Learning Outcomes<\/h3>\n<ul>\n<li>Convert numbers between bases<\/li>\n<\/ul>\n<\/div>\n<p><span style=\"color: #077fab; font-size: 1.15em; font-weight: 600;\">The Development and Use of Different Number Bases<\/span><\/p>\n<h3>Other Bases<\/h3>\n<div class=\"textbox examples\">\n<h3>Powers of numbers other than 10<\/h3>\n<p>Recall that in our base-ten number system, each place value in a number represents a power of ten. Our numbers take the form<\/p>\n<p>&#8230;\u00a0<span style=\"text-decoration: underline;\">thousands<\/span>\u00a0<span style=\"text-decoration: underline;\">hundreds<\/span>\u00a0<span style=\"text-decoration: underline;\">tens<\/span>\u00a0<span style=\"text-decoration: underline;\">ones<\/span>\u00a0.<\/p>\n<p>&#8230;\u00a0<span style=\"text-decoration: underline;\">10<\/span><sup>3<\/sup>\u00a0+ <span style=\"text-decoration: underline;\">10<\/span><sup>2<\/sup>\u00a0+\u00a0<span style=\"text-decoration: underline;\">10<\/span><sup>1<\/sup>\u00a0+ <span style=\"text-decoration: underline;\">10<\/span><sup>0\u00a0<\/sup>, where\u00a0[latex]10^{0}=1[\/latex] (In fact any number raised to the zero<sup>th\u00a0<\/sup>power equals one).<\/p>\n<p>We have an intuitive understanding that [latex]10^{1}=10[\/latex] because we are using [latex]10[\/latex] as a factor [latex]1[\/latex] time. And certainly [latex]10^{2} = 10\\times 10 = 100, \\\\ 10^{3}=1000,[\/latex] and so on.<\/p>\n<p>This pattern works with bases other than [latex]10[\/latex] as well. As you&#8217;ll see below, using [latex]5[\/latex] as a base yields the following.<\/p>\n<p>[latex]5^{0}=1 \\\\ 5^{1}=5 \\\\ 5^{2}=5\\times 5 = 25 \\\\ 5^{3}=5\\times 5\\times 5 = 125,[\/latex] and so on.<\/p>\n<\/div>\n<p>For example, let\u2019s suppose we adopt a base-five system. The only modern digits we would need for this system are 0, 1, 2, 3 and 4. What are the place values in such a system? To answer that, we start with the ones place, as most base systems do. However, if we were to count in this system, we could only get to four (4) before we had to jump up to the next place. Our base is 5, after all! What is that next place that we would jump to? It would not be tens, since we are no longer in base-ten. We\u2019re in a different numerical world. As the base-ten system progresses from 10<sup>0<\/sup> to 10<sup>1<\/sup>, so the base-five system moves from 5<sup>0<\/sup> to 5<sup>1<\/sup> = 5. Thus, we move from the ones to the fives.<\/p>\n<p>After the fives, we would move to the 5<sup>2<\/sup> place, or the twenty fives. Note that in base-ten, we would have gone from the tens to the hundreds, which is, of course, 10<sup>2<\/sup>.<\/p>\n<p>Let\u2019s take an example and build a table. Consider the number 30412 in base five. We will write this as 30412<sub>5<\/sub>, where the subscript 5 is not part of the number but indicates the base we\u2019re using. First off, note that this is NOT the number \u201cthirty thousand, four hundred twelve.\u201d We must be careful not to impose the base-ten system on this number. Here\u2019s what our table might look like. We will use it to convert this number to our more familiar base-ten system.<\/p>\n<table>\n<tbody>\n<tr>\n<td><\/td>\n<td>Base 5<\/td>\n<td>This column coverts to base-ten<\/td>\n<td>In Base-Ten<\/td>\n<\/tr>\n<tr>\n<td><\/td>\n<td>3 \u00d7 5<sup>4<\/sup><\/td>\n<td>= 3 \u00d7 625<\/td>\n<td>= 1875<\/td>\n<\/tr>\n<tr>\n<td>+<\/td>\n<td>0 \u00d7 5<sup>3<\/sup><\/td>\n<td>= 0 \u00d7 125<\/td>\n<td>= 0<\/td>\n<\/tr>\n<tr>\n<td>+<\/td>\n<td>4 \u00d7 5<sup>2<\/sup><\/td>\n<td>= 4 \u00d7 25<\/td>\n<td>= 100<\/td>\n<\/tr>\n<tr>\n<td>+<\/td>\n<td>1 \u00d7 5<sup>1<\/sup><\/td>\n<td>= 1 \u00d7 5<\/td>\n<td>= 5<\/td>\n<\/tr>\n<tr>\n<td>+<\/td>\n<td>2 \u00d7 5<sup>0<\/sup><\/td>\n<td>= 2 \u00d7 1<\/td>\n<td>= 2<\/td>\n<\/tr>\n<tr>\n<td><\/td>\n<td><\/td>\n<td>Total<\/td>\n<td>1982<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>As you can see, the number 30412<sub>5<\/sub> is equivalent to 1,982 in base-ten. We will say 30412<sub>5<\/sub> = 1982<sub>10<\/sub>. All of this may seem strange to you, but that\u2019s only because you are so used to the only system that you\u2019ve probably ever seen.<\/p>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>Convert [latex]6234_{7}[\/latex]\u00a0to a base [latex]10[\/latex] number.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q482364\">Show Solution<\/span><\/p>\n<div id=\"q482364\" class=\"hidden-answer\" style=\"display: none\">We first note that we are given a base-7 number that we are to convert. Thus, our places will start at the ones (7<sup>0<\/sup>), and then move up to the 7s, 49s (7<sup>2<\/sup>), etc. Here\u2019s the breakdown:<\/p>\n<table>\n<tbody>\n<tr>\n<td><\/td>\n<td>Base 7<\/td>\n<td>Convert<\/td>\n<td>Base 10<\/td>\n<\/tr>\n<tr>\n<td><\/td>\n<td>= 6\u00a0\u00d7\u00a07<sup>3<\/sup><\/td>\n<td>= 6\u00a0\u00d7 343<\/td>\n<td>= 2058<\/td>\n<\/tr>\n<tr>\n<td>+<\/td>\n<td>= 2\u00a0\u00d7\u00a07<sup>2<\/sup><\/td>\n<td>= 2\u00a0\u00d7 49<\/td>\n<td>= 98<\/td>\n<\/tr>\n<tr>\n<td>+<\/td>\n<td>= 3\u00a0\u00d7\u00a07<sup>1<\/sup><\/td>\n<td>= 3\u00a0\u00d7 7<\/td>\n<td>= 21<\/td>\n<\/tr>\n<tr>\n<td>+<\/td>\n<td>= 4\u00a0\u00d7 7<sup>0<\/sup><\/td>\n<td>= 4\u00a0\u00d7 1<\/td>\n<td>= 4<\/td>\n<\/tr>\n<tr>\n<td><\/td>\n<td><\/td>\n<td>Total<\/td>\n<td>2181<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>&nbsp;<\/p>\n<p>Thus [latex]6234_{7}=2181_{10}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p>Convert [latex]41065_{7}[\/latex] to a base [latex]10[\/latex] number.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q896067\">Show Solution<\/span><\/p>\n<div id=\"q896067\" class=\"hidden-answer\" style=\"display: none\">[latex]41065_{7} = 9994_{10}[\/latex]<\/p>\n<\/div>\n<\/div>\n<p>&nbsp;<\/p>\n<p><iframe loading=\"lazy\" id=\"ohm8680\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=8680&theme=oea&iframe_resize_id=ohm8680&show_question_numbers\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<p>Watch this video to see more examples of converting numbers in bases other than 10 into a base 10 number.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-1\" title=\"Convert Numbers in Base Ten to Different Bases:  Remainder Method\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/TjvexIVV_gI?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<h3>Converting from Base 10 to Other Bases<\/h3>\n<p>Converting from an unfamiliar base to the familiar decimal system is not that difficult once you get the hang of it. It\u2019s only a matter of identifying each place and then multiplying each digit by the appropriate power. However, going the other direction can be a little trickier. Suppose you have a base-ten number and you want to convert to base-five. Let\u2019s start with some simple examples before we get to a more complicated one.<\/p>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>Convert twelve to a base-five number.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q97454\">Show Solution<\/span><\/p>\n<div id=\"q97454\" class=\"hidden-answer\" style=\"display: none\">\n<p>We can probably easily see that we can rewrite this number as follows:<\/p>\n<p style=\"padding-left: 30px;\">[latex]12 = (2 \\cdot 5) + (2 \\cdot 1)[\/latex]<\/p>\n<p>Hence, we have two fives and two ones. Hence, in base-five we would write twelve as [latex]22_{5}[\/latex]. Thus, [latex]12_{10}=22_{5}[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>Convert sixty-nine to a base-four number.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q90808\">Show Solution<\/span><\/p>\n<div id=\"q90808\" class=\"hidden-answer\" style=\"display: none\">\n<p>64 is the highest power of four that is less than 69.<\/p>\n<p>So we rewrite sixty-nine as follows:<\/p>\n<p style=\"padding-left: 30px;\">[latex]69 = (1 \\cdot 64) + (0 \\cdot 16) + (1 \\cdot 4) + (1 \\cdot 1)[\/latex]<\/p>\n<p>Here, we have one sixty-four, zero sixteens, one four, and one one. Hence, in base four we have [latex]1011[\/latex]. Thus, [latex]69_{10}=1011_{4}[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>Convert the base-seven number [latex]3261_{7}[\/latex]\u00a0to base 10.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q67834\">Show Solution<\/span><\/p>\n<div id=\"q67834\" class=\"hidden-answer\" style=\"display: none\">\n<p>The powers of 7 are:<\/p>\n<p style=\"padding-left: 30px;\">[latex]7^{0}= 1[\/latex]<br \/>\n[latex]7^{1}= 7[\/latex]<br \/>\n[latex]7^{2}= 49[\/latex]<br \/>\n[latex]7^{3}= 343[\/latex]<br \/>\nEtc\u2026<\/p>\n<p>[latex]3261_{7}= (3 \\cdot 343) + (2 \\cdot 49) + (6 \\cdot 7) + (1 \\cdot 1) =1170_{10}[\/latex].<\/p>\n<p>Thus [latex]3261_{7}=1170_{10}[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p>Convert [latex]143[\/latex] to base [latex]5[\/latex]<\/p>\n<p>&nbsp;<\/p>\n<p><iframe loading=\"lazy\" id=\"ohm8681\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=8681&theme=oea&iframe_resize_id=ohm8681&show_question_numbers\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p>Convert the base-three number [latex]21021_{3}[\/latex]\u00a0to base [latex]10[\/latex].<\/p>\n<p>&nbsp;<\/p>\n<p><iframe loading=\"lazy\" id=\"ohm8680\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=8680&theme=oea&iframe_resize_id=ohm8680&show_question_numbers\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<p><em>In general, when converting from base<\/em><em>-ten to some other base, it is often helpful to determine the highest power of the base that will divide into the given number at least once.<\/em><\/p>\n<p>In the last example, [latex]5^2= 25[\/latex] is the largest power of five that is present in 69, so that was our starting point. If we had moved to [latex]5^3 = 125[\/latex], then 125 would not divide into 69 at least once.<\/p>\n<div class=\"textbox\">\n<h3>Converting from Base 10 to Base <em>b<\/em><\/h3>\n<ol>\n<li>Find the highest power of the base <em>b<\/em> that will divide into the given number at least once and then divide.<\/li>\n<li>Write down the whole number part, then use the remainder from division in the next step.<\/li>\n<li>Repeat steps 1 and 2, dividing by the next highest power of the base <em>b<\/em>, writing down the whole number part (including 0), and using the remainder in the next step.<\/li>\n<li>Continue until the remainder is smaller than the base. This last remainder will be in the \u201cones\u201d place.<\/li>\n<li>Collect all your whole number parts to get your number in base <em>b<\/em> notation.<\/li>\n<\/ol>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>Convert the base-ten number [latex]348[\/latex] to base-five.<\/p>\n<h4>Solution<\/h4>\n<p>The powers of five are:<\/p>\n<p style=\"padding-left: 30px;\">5<sup>0<\/sup> = 1<br \/>\n5<sup>1<\/sup> = 5<br \/>\n5<sup>2<\/sup> = 25<br \/>\n5<sup>3<\/sup> = 125<br \/>\n5<sup>4<\/sup> = 625<br \/>\nEtc\u2026<\/p>\n<p>Since 348 is smaller than 625, but bigger than 125, we see that 5<sup>3\u00a0<\/sup>= 125 is the highest power of five present in 348. So we divide 125 into 348 to see how many of them there are:<\/p>\n<p style=\"padding-left: 30px;\">348\u00a0\u00f7\u00a0125 = <span style=\"color: #ff0000;\">2<\/span> with remainder 98<\/p>\n<p>We write down the whole part, 2, and continue with the remainder. There are 98 left over, so we see how many 25s (the next power of five to the right of 125s) there are in the remainder:<\/p>\n<p style=\"padding-left: 30px;\">98\u00a0\u00f7\u00a025 = <span style=\"color: #ff0000;\">3<\/span> with remainder 23<\/p>\n<p>We write down the whole part, 3, and continue with the remainder. There are 23 left over, so we look at the next place, the 5s:<\/p>\n<p style=\"padding-left: 30px;\">23\u00a0\u00f7\u00a05 = <span style=\"color: #ff0000;\">4<\/span> with remainder <span style=\"color: #ff0000;\">3<\/span><\/p>\n<p>This leaves us with 3, which is less than our base, so this number will be in the \u201cones\u201d place. We are ready to assemble our base-five number:<\/p>\n<p style=\"padding-left: 30px;\">348 = (<span style=\"color: #ff0000;\">2<\/span> \u00d7 5<sup>3<\/sup>) + (<span style=\"color: #ff0000;\">3<\/span> \u00d7 5<sup>2<\/sup>) + (<span style=\"color: #ff0000;\">4<\/span> \u00d7 5<sup>1<\/sup>) + (<span style=\"color: #ff0000;\">3<\/span> \u00d7 1)<\/p>\n<p>Hence, our base-five number is <span style=\"color: #ff0000;\">2343<\/span>. We\u2019ll say that [latex]348_{10}=2343_{5}[\/latex].<\/p>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>Convert the base-ten number [latex]4,509[\/latex] to base-seven.<\/p>\n<h4>Solution<\/h4>\n<p>The powers of 7 are:<\/p>\n<p style=\"padding-left: 30px;\">7<sup>0<\/sup> = 1<br \/>\n7<sup>1<\/sup> = 7<br \/>\n7<sup>2<\/sup> = 49<br \/>\n7<sup>3<\/sup> = 343<br \/>\n7<sup>4<\/sup> = 2401<br \/>\n7<sup>5<\/sup> = 16807<br \/>\nEtc\u2026<\/p>\n<p>The highest power of 7 that will divide into 4,509 is 7<sup>4<\/sup> = 2401.\u00a0With division, we see that it will go in 1 time with a remainder of 2108. So we have 1 in the 7<sup>4<\/sup> place.<\/p>\n<p>The next power down is 7<sup>3<\/sup> = 343, which goes into 2108 six times with a new remainder of 50. So we have 6 in the 7<sup>3<\/sup> place.<\/p>\n<p>The next power down is 7<sup>2<\/sup> = 49, which goes into 50 once with a new remainder of 1. So there is a 1 in the 7<sup>2<\/sup> place.<\/p>\n<p>The next power down is 7<sup>1<\/sup> but there was only a remainder of 1, so that means there is a 0 in the 7s place and we still have 1 as a remainder.<\/p>\n<p>That, of course, means that we have 1 in the ones place.<\/p>\n<p>Putting all of this together means that [latex]4,509_{10}=16101_{7}[\/latex].<\/p>\n<div class=\"textbox\">\n<p>4509 \u00f7 7<sup>4<\/sup> = <span style=\"color: #ff0000;\">1<\/span> R 2108<\/p>\n<p>2108 \u00f7 7<sup>3<\/sup> = <span style=\"color: #ff0000;\">6<\/span> R 50<\/p>\n<p>50 \u00f7 7<sup>2<\/sup> = <span style=\"color: #ff0000;\">1<\/span> R 1<\/p>\n<p>1 \u00f7 7<sup>1<\/sup> = <span style=\"color: #ff0000;\">1<\/span><\/p>\n<p>4,509<sub>10<\/sub> = <span style=\"color: #ff0000;\">16101<\/span><sub>7<\/sub>.<\/p>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p>Convert [latex]657_{10}[\/latex]\u00a0to a base [latex]4[\/latex] number.<\/p>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p>Convert [latex]8377_{10}[\/latex]\u00a0to a base [latex]8[\/latex] number.<\/p>\n<\/div>\n\n\t\t\t <section class=\"citations-section\" role=\"contentinfo\">\n\t\t\t <h3>Candela Citations<\/h3>\n\t\t\t\t\t <div>\n\t\t\t\t\t\t <div id=\"citation-list-56\">\n\t\t\t\t\t\t\t <div class=\"licensing\"><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Original<\/div><ul class=\"citation-list\"><li>Revision and Adaptation. <strong>Authored by<\/strong>: Lumen Learning. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><\/ul><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Shared previously<\/div><ul class=\"citation-list\"><li>Math in Society. <strong>Authored by<\/strong>: Lippman, David. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"http:\/\/www.opentextbookstore.com\/mathinsociety\/\">http:\/\/www.opentextbookstore.com\/mathinsociety\/<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><li>Convert Numbers in Base Ten to Different Bases:  Remainder Method External link. <strong>Authored by<\/strong>: James Sousa (Mathispower4u.com). <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/youtu.be\/TjvexIVV_gI\">https:\/\/youtu.be\/TjvexIVV_gI<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><li>Question ID 8680. <strong>Authored by<\/strong>: Lippman, David. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em>. <strong>License Terms<\/strong>: IMathAS Community License CC-BY + GPL<\/li><\/ul><\/div>\n\t\t\t\t\t\t <\/div>\n\t\t\t\t\t <\/div>\n\t\t\t <\/section>","protected":false},"author":503070,"menu_order":6,"template":"","meta":{"_candela_citation":"[{\"type\":\"cc\",\"description\":\"Math in Society\",\"author\":\"Lippman, David\",\"organization\":\"\",\"url\":\"http:\/\/www.opentextbookstore.com\/mathinsociety\/\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"cc\",\"description\":\"Convert Numbers in Base Ten to Different Bases:  Remainder Method External link\",\"author\":\"James Sousa (Mathispower4u.com)\",\"organization\":\"\",\"url\":\"https:\/\/youtu.be\/TjvexIVV_gI\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"cc\",\"description\":\"Question ID 8680\",\"author\":\"Lippman, David\",\"organization\":\"\",\"url\":\"\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"IMathAS Community License CC-BY + GPL\"},{\"type\":\"original\",\"description\":\"Revision 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