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The number of integral solutions of the equation $$\left[\frac{k}{1}\right] + \left[\frac{k}{2}\right] + \left[\frac{k}{3}\right] = 100$$, where $$[x]$$ is the greatest integer less than or equal to $$x$$, is
For integers $$k$$, $$f(k) = k + \left[\frac{k}{2}\right] + \left[\frac{k}{3}\right]$$ is strictly increasing, and $$f(k) \approx \frac{11k}{6}$$, so $$k$$ is near 55. Now $$f(54) = 54 + 27 + 18 = 99$$, $$f(55) = 55 + 27 + 18 = 100$$ and $$f(56) = 56 + 28 + 18 = 102$$. So $$k = 55$$ is the only integral solution, and the number of solutions is 1.
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