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

Revision as of 01:04, 19 July 2008

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) $3 \leq b < c \leq 6$ .

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

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, $2*2*6=24\Rightarrow \mathrm{(C)}$ .

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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 *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>.
 *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>.
-*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.
+*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.
 Multiplying the possibilities for each restriction, <math>2*2*6=24\Rightarrow \mathrm{(C)}</math>.