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1995 AHSME Problems/Problem 11

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

See also

1995 AHSME (Problems)
Preceded by
Problem 10
Followed by
Problem 12
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