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Question Number 206806 by Frix last updated on 26/Apr/24

$$\mathrm{Given}\mathrm{is}\mathrm{a}\mathrm{square}\mathrm{with}\mathrm{side}\mathrm{length}15.$$$$\mathrm{We}\mathrm{need}\mathrm{to}\mathrm{find}\mathrm{exactly}17\mathrm{smaller}\mathrm{squares}$$$$\mathrm{to}\mathrm{fill}\mathrm{the}\mathrm{big}\mathrm{one}.\mathrm{How}\mathrm{many}\mathrm{solutions}\mathrm{are}$$$$\mathrm{possible}?$$$$(\mathrm{Note}:{\mathrm{it}}^{\prime }\mathrm{s}\mathrm{not}\mathrm{enough}\mathrm{to}\mathrm{find}\mathrm{squares}\mathrm{with}$$$$\mathrm{the}\mathrm{sum}\mathrm{of}\mathrm{their}\mathrm{areas}\mathrm{being}225,\mathrm{they}\mathrm{must}$$$$\mathrm{fit}\mathrm{into}\mathrm{the}15\times 15\mathrm{square}.\mathrm{Example}\mathrm{with}$$$$3\mathrm{squares}:2\times 2+5\times 5+14\times 14=225\mathrm{but}$$$$\mathrm{you}\mathrm{cannot}\mathrm{fit}\mathrm{these}\mathrm{in}\mathrm{a}15\times 15\mathrm{square})$$

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