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Olympiad workbook
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IOQM
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Basics of Mathematics

Olympiad workbook · IOQM · Basics of Mathematics

For a positive integer n , let n denote the perfect square integer...

For a positive integer nn, let ⟨n⟩\langle n\rangle denote the perfect square integer closest to nn. For example, ⟨74⟩=81\langle 74\rangle=81, ⟨18⟩=16\langle 18\rangle=16. If NN is the smallest positive integer such that ⟨91⟩⋅⟨120⟩⋅⟨143⟩⋅⟨180⟩⋅⟨N⟩=91.120.143.180⋅N\langle 91\rangle \cdot\langle 120\rangle \cdot\langle 143\rangle \cdot\langle 180\rangle \cdot\langle N\rangle=91.120 .143 .180 \cdot N find the sum of the squares of the digits of NN.

Correct answer

56

Step-by-step solution

∵102⋅112⋅122⋅132⋅⟨N⟩=91⋅120⋅143⋅180⋅N⇒22⟨N⟩=21N\begin{aligned} & \because \quad 10^2 \cdot 11^2 \cdot 12^2 \cdot 13^2 \cdot\langle N\rangle=91 \cdot 120 \cdot 143 \cdot 180 \cdot N \\ & \Rightarrow \quad 22\langle N\rangle=21 N \end{aligned} ∵⟨N⟩\because\langle N\rangle is a perfect square. So, ⟨N⟩\langle N\rangle must have a factor 21221^2. For least positive value of NN, let ⟨N⟩=212\langle N\rangle=21^2 So range of NN will be, 212−20≤N≤212+2121^2-20 \leq N \leq 21^2+21 ⇒22.212=21.N⇒N=21.22=462\begin{aligned} & \Rightarrow \quad 22.21^2=21 . N \\ & \Rightarrow \quad N=21.22=462 \end{aligned} Sum of squares of digits =42+62+22=56=4^2+6^2+2^2=56
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