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Algebra Difficulty 6.4 National olympiad Prove it Japan

Let a,b,c,d,e,f,g,h,ia, b, c, d, e, f, g, h, i be the distinct integers lying in between 11 and 99 (both 11 and 99 inclusive). Let NN be the maximum of the three numbers a×b×ca \times b \times c, d×e×fd \times e \times f and g×h×ig \times h \times i. Determine the minimum value the number NN can take.

Solution

Let us write p=abcp = abc, q=defq = def and r=ghir = ghi.
We first show that it is possible to make the maximum of the three numbers p,q,rp, q, r to be no more than 7272. Indeed, if we consider the following, we see that this is possible:
1×8×9=72,2×5×7=70,3×4×6=72. 1 \times 8 \times 9 = 72, \quad 2 \times 5 \times 7 = 70, \quad 3 \times 4 \times 6 = 72.
Next, we will show that the maximum of the three numbers p,q,rp, q, r must be at least 7272. Since all the numbers a,b,c,d,e,f,g,h,ia, b, c, d, e, f, g, h, i lie in between 11 and 99 and are distinct, we see that the product pqrpqr of the three numbers p,q,rp, q, r must be a constant, which from the example above must equal 72×70×7272 \times 70 \times 72. In particular, we have 703<pqr70^3 < pqr. Thus the maximum of p,q,rp, q, r must be greater than 7070. Since 7171 cannot be written as a product of integers lying in between 11 through 99, we see that it is impossible to have any of p,q,rp, q, r to be 7171. Therefore, the maximum of p,q,rp, q, r has to be at least 7272.

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