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Difference between revisions of "1995 AHSME Problems/Problem 11"

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==Solution==
 
==Solution==
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For condition (i), the restriction is put on a; N<4000 if a<4, and N≥6 if a≥6. Therefore, a=4,5.
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*For condition (i), the restriction is put on <math>a</math>; <math>N<4000</math> if <math>a<4</math>, and <math>N \ge 6</math> if <math>a \ge 6</math>. Therefore, <math>a=4,5</math>.
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*For condition (ii), the restriction is put on <math>d</math>; it must be a multiple of <math>5</math>. Therefore, <math>d=0,5</math>.
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*For condition (iii), the restriction is put on <math>b</math> and <math>c</math>. The possible ordered pairs of <math>b</math> and <math>c</math> are <math>(3,4)</math>, <math>(3,5)</math>, <math>(3,6)</math>, <math>(4,5), (4,6),</math> and <math>(5,6),</math> and there are <math>6</math> of them. Alternatively, we are picking from the four digits 3, 4, 5, 6, and for every combination of two, there is exactly one way to arrange them in increasing order, so we have <math>\binom{4}{2} = 6</math> choices for <math>b</math> and <math>c</math> when we consider them together.
  
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For condition (ii), the condition is put on d; it must be a multiple of 5. Therefore, d=0,5.
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Multiplying the possibilities for each restriction, <math>2 \cdot 2 \cdot 6=24\Rightarrow \mathrm{(C)}</math>.
  
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For condition (iii), the condition is put on b and c. The possible ordered pairs of b and c are (3,4), (3,5), (3,6), (4,5), (4,6), and (5,6), and there are 6 of them.
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==See also==
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{{AHSME box|year=1995|num-b=10|num-a=12}}
  
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Multiplying the possibilities for each restriction, <math>2*2*6=24\Rightarrow \mathrm{(C)}</math>
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[[Category:Introductory Combinatorics Problems]]
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{{MAA Notice}}
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==See also==
 

Latest revision as of 09:37, 30 March 2023

Problem

How many base 10 four-digit numbers, $N = \underline{a} \underline{b} \underline{c} \underline{d}$, satisfy all three of the following conditions?

(i) $4,000 \leq N < 6,000;$ (ii) $N$ is a multiple of 5; (iii) $3 \leq b < c \leq 6$.


$\mathrm{(A) \ 10 } \qquad \mathrm{(B) \ 18 } \qquad \mathrm{(C) \ 24 } \qquad \mathrm{(D) \ 36 } \qquad \mathrm{(E) \ 48 }$

Solution

  • For condition (i), the restriction is put on $a$; $N<4000$ if $a<4$, and $N \ge 6$ if $a \ge 6$. Therefore, $a=4,5$.
  • For condition (ii), the restriction is put on $d$; it must be a multiple of $5$. Therefore, $d=0,5$.
  • For condition (iii), the restriction is put on $b$ and $c$. The possible ordered pairs of $b$ and $c$ are $(3,4)$, $(3,5)$, $(3,6)$, $(4,5), (4,6),$ and $(5,6),$ and there are $6$ of them. Alternatively, we are picking from the four digits 3, 4, 5, 6, and for every combination of two, there is exactly one way to arrange them in increasing order, so we have $\binom{4}{2} = 6$ choices for $b$ and $c$ when we consider them together.

Multiplying the possibilities for each restriction, $2 \cdot 2 \cdot 6=24\Rightarrow \mathrm{(C)}$.

See also

1995 AHSME (Problems • Answer Key • Resources)
Preceded by
Problem 10 		Followed by
Problem 12
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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