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9.10: Lines of Numbers Extra for Experts

Difficulty Level: At Grade Created by: CK-12

Extras for Experts: Lines of Numbers - Identify Patterns and Reason Proportionally

Solutions

1.a. The 50th number is 1: 50÷6 is 8 with 2 left over and 8×6=48. So the 48th number is 3, the last number in the set. The 50th number is 2, the second number in the set.b. The 100th number is 3: 100÷6=16 with 4 left over and 6×16=96. So the 96th number is 3. Then the 100th number is 3.c. The sum of the first 100 numbers is 232: The sum of one set of 1, 2, 2, 3, 3, and 3 is 14. From the answer to question b,\ we know that 16 sets of six numbers ends with 96. The sum of the first 96 numbers is 16×14, or 224. The four left over numbers are 1, 2, 2, and 3, and their sum is 8. So the sum of the first 100 numbers in the pattern is 224+8, or 232.2.a. The 70th number is 4:70÷4 is 17 with two left over and 4×17=68. So the 68th number is 6, the last numberon the set. The 70th number is 4, the second number in the set.b. The 125th number is 3:125÷4=31 with one left over and 4×31=124. So the 124th number is 6.The 125th number will be 3.c. The sum of the first 125 numbers is 561:The sum of one set of 3, 4, 5, and 6 is 18. From the answer to question b,\ we know that 31sets of three numbers ends with 124. The sum of the first 124 numbers is 18×31,or 558. The left over numbers is 3. So the sum of the first 125 numbers in the pattern is558+3, or 561.\begin{align*}1. \!\\ {\;} \quad \text{a}.\ \text{The}\ 50^{\text{th}}\ \text{number is}\ 1: \!\\ {\;} \qquad \ 50 \div 6\ \text{is}\ 8\ \text{with} \ 2 \ \text{left over and}\ 8 \times 6 = 48.\ \text{So the}\ 48^{\text{th}}\ \text{number is}\ 3,\ \text{the last number in} \!\\ {\;} \qquad \ \text{the set. The}\ 50^{\text{th}}\ \text{number is}\ 2,\ \text{the second number in the set}. \!\\ {\;} \quad \text{b}.\ \text{The}\ 100^{\text{th}}\ \text{number is}\ 3: \!\\ {\;} \qquad \ 100 \div 6 = 16\ \text{with} \ 4 \ \text{left over and}\ 6 \times 16 = 96.\ \text{So the}\ 96^{\text{th}}\ \text{number is}\ 3.\!\\ {\;} \qquad \ \text{Then the}\ 100^{\text{th}}\ \text{number is}\ 3.\!\\ {\;} \quad \text{c}.\ \text{The sum of the first}\ 100\ \text{numbers is}\ 232: \!\\ {\;} \qquad \ \text{The sum of one set of}\ 1,\ 2,\ 2,\ 3,\ 3,\ \text{and}\ 3\ \text{is}\ 14.\ \text{From the answer to question b,\ we know that} \!\\ {\;} \qquad \ 16\ \text{sets of six numbers ends with}\ 96.\ \text{The sum of the first 96 numbers is} \!\\ {\;} \qquad \ 16 \times 14,\ \text{or}\ 224.\ \text{The four left over numbers are}\ 1,\ 2,\ 2,\ \text{and}\ 3,\ \text{and their sum is}\ 8.\ \text{So the} \!\\ {\;} \qquad \ \text{sum of the first}\ 100\ \text{numbers in the pattern is}\ 224 + 8,\ \text{or}\ 232.\!\\ 2. \!\\ {\;} \quad \text{a}.\ \text{The}\ 70^{\text{th}}\ \text{number is}\ 4: \!\\ {\;} \qquad 70 \div 4\ \text{is}\ 17\ \text{with two left over and}\ 4 \times 17 = 68.\ \text{So the}\ 68^{\text{th}}\ \text{number is}\ 6,\ \text{the last number} \!\\ {\;} \qquad \text{on the set. The}\ 70^{\text{th}}\ \text{number is}\ 4,\ \text{the second number in the set}. \!\\ {\;} \quad \text{b}.\ \text{The}\ 125^{\text{th}}\ \text{number is}\ 3: \!\\ {\;} \qquad 125 \div 4 = 31\ \text{with one left over and}\ 4 \times 31 = 124.\ \text{So the}\ 124^{\text{th}}\ \text{number is}\ 6. \!\\ {\;} \qquad \text{The}\ 125^{\text{th}}\ \text{number will be}\ 3.\!\\ {\;} \quad \text{c}.\ \text{The sum of the first}\ 125\ \text{numbers is}\ 561:\!\\ {\;} \qquad \text{The sum of one set of}\ 3,\ 4,\ 5,\ \text{and}\ 6\ \text{is}\ 18.\ \text{From the answer to question b,\ we know that}\ 31 \!\\ {\;} \qquad \text{sets of three numbers ends with}\ 124.\ \text{The sum of the first}\ 124\ \text{numbers is}\ 18 \times 31,\!\\ {\;} \qquad \text{or}\ 558.\ \text{The left over numbers is}\ 3. \ \text{So the sum of the first}\ 125\ \text{numbers in the pattern is}\!\\ {\;} \qquad 558 + 3,\ \text{or}\ 561.\end{align*}

3.a. The 67th number is 3: 67÷5 is 13 with two left over and 5×13=65. So the 65th number is 9, the last number in the set. The 66th number is 1, and the 67th is 3.b. The 129th number is 7: 129÷5=25 with 4 left over and 5×25=125. So the 125th number is 7. The 129th is 7, the fourth number in the set.c. The sum of the first 129 numbers is 641:The sum of one set of 1, 3, 5, 7, and 9 is 25. From the answer to question b, we know that25 sets of four numbers ends with 125. The sum of the first 125 numbers is25×25, or 625. The four left over numbers are 1, 3, 5, and 7, and their sum is 16. So thesum of the first 129 numbers in the pattern is 625+16, or 641.\begin{align*}3. \!\\ {\;} \quad \text{a}.\ \text{The}\ 67^{\text{th}}\ \text{number is}\ 3: \!\\ {\;} \qquad \ 67 \div 5 \ \text{is}\ 13\ \text{with two left over and}\ 5 \times 13 = 65.\ \text{So the}\ 65^{\text{th}}\ \text{number is}\ 9,\ \text{the last number}\!\\ {\;} \qquad \ \text{in the set. The}\ 66^{\text{th}}\ \text{number is}\ 1,\ \text{and the}\ 67^{\text{th}}\ \text{is}\ 3. \!\\ {\;} \quad \text{b}.\ \text{The}\ 129^{\text{th}}\ \text{number is}\ 7: \!\\ {\;} \qquad \ 129 \div 5 = 25 \ \text{with} \ 4 \ \text{left over and}\ 5 \times 25 = 125.\ \text{So the}\ 125^{\text{th}}\ \text{number is}\ 7. \!\\ {\;} \qquad \ \text{The}\ 129^{\text{th}}\ \text{is}\ 7,\ \text{the fourth number in the set.}\!\\ {\;} \quad \text{c}.\ \text{The sum of the first}\ 129 \ \text{numbers is}\ 641: \!\\ {\;} \qquad \text{The sum of one set of}\ 1,\ 3,\ 5,\ 7,\ \text{and}\ 9\ \text{is}\ 25.\ \text{From the answer to question b, we know that}\!\\ {\;} \qquad 25\ \text{sets of four numbers ends with}\ 125.\ \text{The sum of the first}\ 125\ \text{numbers is}\!\\ {\;} \qquad 25 \times 25,\ \text{or}\ 625.\ \text{The four left over numbers are}\ 1,\ 3,\ 5, \ \text{and}\ 7,\ \text{and their sum is}\ 16.\ \text{So the} \!\\ {\;} \qquad \text{sum of the first}\ 129\ \text{numbers in the pattern is}\ 625 + 16,\ \text{or}\ 641.\end{align*}

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Date Created:
Feb 23, 2012