1995 AHSME Problems/Problem 11

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≥6 if a≥6. Therefore, a=4,5.

For condition (ii), the condition is put on d; it must be a multiple of 5. Therefore, d=0,5.

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.

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
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1995 AHSME Problems/Problem 11

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