1993 AIME Problems/Problem 4

Problem

How many ordered four-tuples of integers $(a,b,c,d)\,$ with $0 < a < b < c < d < 500\,$ satisfy $a + d = b + c\,$ and $bc - ad = 93\,$ ?

Solution

Solution 1

Let $k = a + d = b + c$ so $d = k-a, b=k-c$ . It follows that $(k-c)c - a(k-a) = (a-c)(a+c-k) = (c-a)(d-c) = 93$ . Hence $(c - a,d - c) = (1,93),(3,31),(31,3),(93,1)$ .

Solve them in tems of $c$ to get $(a,b,c,d) = (c - 93,c - 92,c,c + 1),$ $(c - 31,c - 28,c,c + 3),$ $(c - 1,c + 92,c,c + 93),$ $(c - 3,c + 28,c,c + 31)$ . The last two solutions don't follow $a < b < c < d$ , so we only need to consider the first two solutions.

The first solution gives us $c - 93\geq 1$ and $c + 1\leq 499$ $\implies 94\leq c\leq 498$ , and the second one gives us $32\leq c\leq 496$ .

So the total number of such four-tuples is $405 + 465 = \boxed{870}$ .

Solution 2

Let $b = a + m$ and $c = a + m + n$ . From $a + d = b + c$ , $d = b + c - a = a + 2m + n$ .

Substituting $b = a + m$ , $c = a + m + n$ , and $d = b + c - a = a + 2m + n$ into $bc - ad = 93$ , $bc - ad = (1 + m)(1 + m + n) - a(a + 2m + n) = m(m + n). = 93 = 3(31)$ Hence, $(m,n) = (1,92)$ or $(3,28)$ .

For $(m,n) = (1,92)$ , we know that $0 < a < a + 1 < a + 93 < a + 94 < 500$ , so there are $405$ four-tuples. For $(m,n) = (3,28)$ , $0 < a < a + 3 < a + 31 < a + 34 < 500$ , and there are $465$ four-tuples. In total, we have $405 + 465 = 870$ four-tuples.

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1993 AIME Problems/Problem 4

Problem

Contents

Solution

Solution 1

Solution 2

See also