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Difference between revisions of "Euclid 2020/Problem 4"

(Created page with "(a) The positive integers <math>a</math> and <math>b</math> have no common divisor larger than <math>1</math>. If the difference between b and a is <math>15</math> and<math>5/...")
 
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==Problem==
 
(a) The positive integers <math>a</math> and <math>b</math> have no common divisor larger than <math>1</math>. If the
 
(a) The positive integers <math>a</math> and <math>b</math> have no common divisor larger than <math>1</math>. If the
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difference between b and a is <math>15</math> and<math>5/9 < a/b <4/7</math>, what is the value of <math>a/b</math>?
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difference between b and a is <math>15</math> and <math>\frac59<\frac{a}b<\frac47</math>, what is the value of <math>\frac{a}b</math>?
  
 
(b) A geometric sequence has  first term <math>10</math> and common ratio <math>1/2</math> .
 
(b) A geometric sequence has  first term <math>10</math> and common ratio <math>1/2</math> .
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from the previous term by multiplying it by a non-zero constant, called the
 
from the previous term by multiplying it by a non-zero constant, called the
 
common ratio. For example, <math>3, 6, 12</math> is a geometric sequence with three terms.)
 
common ratio. For example, <math>3, 6, 12</math> is a geometric sequence with three terms.)
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==Solution==
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(a) Because both <math>\frac59</math> and <math>\frac47</math> are less than one, we have <math>a<b</math>. Because the difference between <math>a</math> and <math>b</math> is 15, we have <math>a=b-15</math>, so we have <math>\frac59<\frac{b-15}b<\frac47</math>. Using the left side of the equation, we have <math>\frac59<\frac{b-15}b</math> so <math>5b<9b-135</math>, so <math>135<4b</math>, so <math>b>\frac{135}4</math>. Using the right side of the equation, we have <math>\frac{b-15}b<\frac47</math>, so

Revision as of 18:47, 12 October 2025

Problem

(a) The positive integers $a$ and $b$ have no common divisor larger than $1$. If the difference between b and a is $15$ and $\frac59<\frac{a}b<\frac47$, what is the value of $\frac{a}b$?

(b) A geometric sequence has  first term $10$ and common ratio $1/2$ . An arithmetic sequence has  first term $10$ and common difference $d$. The ratio of the $6th$ term in the geometric sequence to the $4th$ term in the geometric sequence equals the ratio of the $6th$ term in the arithmetic sequence to the $4th$ term in the arithmetic sequence. Determine all possible values of $d$. (An arithmetic sequence is a sequence in which each term after the  first is obtained from the previous term by adding a constant, called the common difference. For example, $3; 5; 7; 9$ are the  first four terms of an arithmetic sequence. A geometric sequence is a sequence in which each term after the  first is obtained from the previous term by multiplying it by a non-zero constant, called the common ratio. For example, $3, 6, 12$ is a geometric sequence with three terms.)

Solution

(a) Because both $\frac59$ and $\frac47$ are less than one, we have $a<b$. Because the difference between $a$ and $b$ is 15, we have $a=b-15$, so we have $\frac59<\frac{b-15}b<\frac47$. Using the left side of the equation, we have $\frac59<\frac{b-15}b$ so $5b<9b-135$, so $135<4b$, so $b>\frac{135}4$. Using the right side of the equation, we have $\frac{b-15}b<\frac47$, so

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