RMO 2014 Questions, Solutions, Discussions

The highest exponents of \(2\) dividing \(32!\) and \(16!\) are given by \[ v_2 (32!) = 16 + 8 + 4 + 2+ 1, \quad v_2(16!) = 8 + 4+ 2+ 1. \] So the highest power of \(2\) dividing the product of the integers \(17, 18, 19, \dots, 32\) is \(2^{16}\). Now if it were possible to write these numbers in a \(4\times 4\) grid in the above-mentioned manner, then the product of the numbers in each of the \(2 \times 2\) subgrids with blue boundary (see Fig. 1) would be divisible by \(2^4\). Note that one such subgrid would contain \(32\), which implies that the product of \(17, 18, \dots, 32\) is divisible by \(2^4\cdot 2^4\cdot 2^4 \cdot 32 = 2^{17}\), which is impossible. Hence it is not possible to write the integers \(17, 18, \dots, 32\) in a \(4\times 4\) grid satisfying the given conditions.