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Difference between revisions of "1980 AHSME Problems/Problem 25"

(Created page with "== Problem == In the non-decreasing sequence of odd integers <math>\{a_1,a_2,a_3,\ldots \}=\{1,3,3,3,5,5,5,5,5,\ldots \}</math> each odd positive integer <math>k</math> appears ...")
 
 
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\text{(E)} \ 4    </math>  
 
\text{(E)} \ 4    </math>  
  
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== Solution ==
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== Solution 1 ==
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<math>\fbox{}</math>
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Define the sequence <math>\{x_n\} = \lfloor\sqrt{n}\rfloor</math> for all non-negative integers <math>n</math>.  Then for any non-negative integer <math>m</math>, <math>x_n = m</math> if and only if <math>m \leq \sqrt{n} < m+1</math>, that is, <math>m^2 \leq n < (m+1)^2</math>.  So each non-negative integer <math>m</math> appears <math>(m+1)^2 - m^2 = 2m+1</math> times.  So <math>\{x_0, x_1, x_2, x_3, ...\} = \{0, 1, 1, 1, 2, 2, 2, 2, 2, ...\}</math>. 
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In the sequence <math>a_n</math>, each positive odd integer <math>2m+1</math> appears <math>2m+1</math> times.  Notice that <math>\{2x_0 + 1, 2x_1 + 1, 2x_2 + 1, 2x_3 + 1, ...\} = \{1, 3, 3, 3, 5, 5, 5, 5, 5, ...\} = \{a_1, a_2, a_3, a_4, ...\}</math>. 
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Therefore, <math>a_n = 2x_{n-1} + 1 = 2\lfloor{\sqrt{n-1}}\rfloor + 1</math>, so <math>b = 2</math>, <math>c = -1</math>, and <math>d = 1</math>, and <math>b + c + d = \boxed{(\textbf{C})\ 2}</math>.
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-j314andrews
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== Solution 2 ==
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Because the set consists of odd numbers, and since <math>\lfloor{}\sqrt{n+c}\rfloor{}</math> is an integer and can be odd or even, <math>b = 2</math> and <math>|a| = 1</math>. However, given that <math>\lfloor{}\sqrt{n+c}\rfloor{}</math> can be <math>0</math>, <math>a = 1</math>. Then, <math>a_1 = 1 = 2\lfloor{}\sqrt{1+c}\rfloor{}+1</math>, and <math>\lfloor{}\sqrt{1+c}\rfloor{}</math> = 0, and <math>c = -1</math> because <math>c</math> is an integer. <math>b+c+d = 2+(-1)+1 = \boxed{\text{(C)}\  2}</math>
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-e_power_pi_times_i
  
 
== See also ==
 
== See also ==

Latest revision as of 21:43, 16 August 2025

Problem

In the non-decreasing sequence of odd integers $\{a_1,a_2,a_3,\ldots \}=\{1,3,3,3,5,5,5,5,5,\ldots \}$ each odd positive integer $k$ appears $k$ times. It is a fact that there are integers $b, c$, and $d$ such that for all positive integers $n$, $a_n=b\lfloor \sqrt{n+c} \rfloor +d$, where $\lfloor x \rfloor$ denotes the largest integer not exceeding $x$. The sum $b+c+d$ equals

$\text{(A)} \ 0 \qquad \text{(B)} \ 1 \qquad \text{(C)} \ 2 \qquad \text{(D)} \ 3 \qquad \text{(E)} \ 4$

Solution 1

Define the sequence $\{x_n\} = \lfloor\sqrt{n}\rfloor$ for all non-negative integers $n$. Then for any non-negative integer $m$, $x_n = m$ if and only if $m \leq \sqrt{n} < m+1$, that is, $m^2 \leq n < (m+1)^2$. So each non-negative integer $m$ appears $(m+1)^2 - m^2 = 2m+1$ times. So $\{x_0, x_1, x_2, x_3, ...\} = \{0, 1, 1, 1, 2, 2, 2, 2, 2, ...\}$.

In the sequence $a_n$, each positive odd integer $2m+1$ appears $2m+1$ times. Notice that $\{2x_0 + 1, 2x_1 + 1, 2x_2 + 1, 2x_3 + 1, ...\} = \{1, 3, 3, 3, 5, 5, 5, 5, 5, ...\} = \{a_1, a_2, a_3, a_4, ...\}$.

Therefore, $a_n = 2x_{n-1} + 1 = 2\lfloor{\sqrt{n-1}}\rfloor + 1$, so $b = 2$, $c = -1$, and $d = 1$, and $b + c + d = \boxed{(\textbf{C})\ 2}$.

-j314andrews

Solution 2

Because the set consists of odd numbers, and since $\lfloor{}\sqrt{n+c}\rfloor{}$ is an integer and can be odd or even, $b = 2$ and $|a| = 1$. However, given that $\lfloor{}\sqrt{n+c}\rfloor{}$ can be $0$, $a = 1$. Then, $a_1 = 1 = 2\lfloor{}\sqrt{1+c}\rfloor{}+1$, and $\lfloor{}\sqrt{1+c}\rfloor{}$ = 0, and $c = -1$ because $c$ is an integer. $b+c+d = 2+(-1)+1 = \boxed{\text{(C)}\ 2}$

-e_power_pi_times_i

See also

1980 AHSME (Problems • Answer Key • Resources)
Preceded by
Problem 24 		Followed by
Problem 26
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30
All AHSME Problems and Solutions

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