Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 1993 based on Integers.

Integer – AIME I, 1993


Find the number of four topics of integers (a,b,c,d) with 0<a<b<c<d<500 satisfy a+d=b+c and bc-ad=93.

  • is 107
  • is 870
  • is 840
  • cannot be determined from the given information

Key Concepts


Integers

Digits

Algebra

Check the Answer


But try the problem first…

Answer: is 870.

Source
Suggested Reading

AIME I, 1993, Question 4

Elementary Algebra by Hall and Knight

Try with Hints


First hint

Let k=a+d=b+c

or, d=k-a, b=k-c,

or, (k-c)c-a(k-a)=k(c-a)-(c-a)(c+a)

=(a-c)(a+c-k)

=(c-a)(d-c)=93

Second Hint

(c-a)(d-c)=(1,93),(3,31),(31,3),(93,1)

solving for c

(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)

Final Step

taking first two solutions a<b<c<d<500

or,\(1 \leq c-93, c+1 \leq 499\)

or, \(94 \leq c \leq 498 \) gives 405 solutions

and \(1 \leq c-31, c+3 \leq 499\)

or, \(32 \leq c \leq 496\) gives 465 solutions

or, 405+465=870 solutions.

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