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IMC / 2001 / Problems / Day 2, P7

IMC 2001 · Day 2 · P7

very hard

Let r,s1r, s \ge 1 be integers and a0,a1,,ar1,b0,b1,,bs1a_0, a_1, \dots, a_{r-1}, b_0, b_1, \dots, b_{s-1} be real non-negative numbers such that (a0+a1x+a2x2++ar1xr1+xr)(b0+b1x+b2x2++bs1xs1+xs)==1+x+x2++xr+s1+xr+s.\begin{align*} (a_0 + a_1 x + a_2 x^2 + \dots + a_{r-1} x^{r-1} + x^r) &(b_0 + b_1 x + b_2 x^2 + \dots + b_{s-1} x^{s-1} + x^s) = \\ &= 1 + x + x^2 + \dots + x^{r+s-1} + x^{r+s}. \end{align*} Prove that each aia_i and each bjb_j equals either 0 or 1.

Solution (official)

Multiply the left hand side polynomials. We obtain the following equalities: a0b0=1,a0b1+a1b0=1,a_0 b_0 = 1, \quad a_0 b_1 + a_1 b_0 = 1, \quad \dots Among them one can find equations a0+a1bs1+a2bs2+=1a_0 + a_1 b_{s-1} + a_2 b_{s-2} + \dots = 1 and b0+b1ar1+b2ar2+=1.b_0 + b_1 a_{r-1} + b_2 a_{r-2} + \dots = 1. From these equations it follows that a0,b01a_0, b_0 \le 1. Taking into account that a0b0=1a_0 b_0 = 1 we can see that a0=b0=1a_0 = b_0 = 1.

Now looking at the following equations we notice that all aa's must be less than or equal to 1. The same statement holds for the bb's. It follows from a0b1+a1b0=1a_0 b_1 + a_1 b_0 = 1 that one of the numbers a1,b1a_1, b_1 equals 0 while the other one must be 1. Follow by induction.

How the field did

contestants scored
182
average (of 20)
4.67
solved (≥ 80%)
12.1%
near-0 (≤ 10%)
56.6%
discrimination
0.23

Score distribution (field cohort)

Computed on contestants with a meaningful total (field cohort); discrimination is the corrected item–total correlation.

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