The number of elements in the
set $$ S=\left\{ x:x\in [0,100] \text{ and } \int_{0}^{x} t^{2} \sin(x-t)dt=x^{2}\right\}$$ is _________

We need to find the number of elements in $$S = \{x : x \in [0, 100] \text{ and } \int_0^x t^2 \sin(x-t)\,dt = x^2\}$$.

Using the identity $$\sin(x-t) = \sin x \cos t - \cos x \sin t$$, we write the integral as $$\int_0^x t^2 \sin(x-t)\,dt = \sin x \int_0^x t^2 \cos t\,dt - \cos x \int_0^x t^2 \sin t\,dt$$.

To evaluate $$\int_0^x t^2 \cos t\,dt$$ we apply integration by parts twice, taking $$u = t^2$$ and $$dv = \cos t\,dt$$ to obtain $$\int t^2 \cos t\,dt = t^2 \sin t - \int 2t \sin t\,dt$$.

In the integral $$\int 2t \sin t\,dt$$ we set $$u = 2t$$ and $$dv = \sin t\,dt$$, giving $$\int 2t \sin t\,dt = -2t\cos t + \int 2\cos t\,dt = -2t\cos t + 2\sin t$$ and hence $$\int t^2 \cos t\,dt = t^2 \sin t + 2t\cos t - 2\sin t + C$$ so that $$\int_0^x t^2 \cos t\,dt = x^2 \sin x + 2x\cos x - 2\sin x$$.

For $$\int_0^x t^2 \sin t\,dt$$ we use integration by parts with $$u = t^2$$ and $$dv = \sin t\,dt$$, yielding $$\int t^2 \sin t\,dt = -t^2 \cos t + \int 2t\cos t\,dt$$.

Since $$\int 2t\cos t\,dt = 2t\sin t - \int 2\sin t\,dt = 2t\sin t + 2\cos t$$ we have $$\int t^2 \sin t\,dt = -t^2\cos t + 2t\sin t + 2\cos t + C$$ and thus $$\int_0^x t^2 \sin t\,dt = -x^2\cos x + 2x\sin x + 2\cos x - 2$$.

Substituting these results into the expression for the integral gives $$\text{LHS} = \sin x(x^2\sin x + 2x\cos x - 2\sin x) - \cos x(-x^2\cos x + 2x\sin x + 2\cos x - 2)$$.

Expanding and simplifying leads to $$= x^2\sin^2 x + 2x\sin x\cos x - 2\sin^2 x + x^2\cos^2 x - 2x\sin x\cos x - 2\cos^2 x + 2\cos x = x^2(\sin^2 x + \cos^2 x) - 2(\sin^2 x + \cos^2 x) + 2\cos x = x^2 - 2 + 2\cos x$$.

Setting this equal to $$x^2$$ gives $$2\cos x = 2$$, so $$\cos x = 1$$ and hence $$x = 2k\pi, \quad k = 0, 1, 2, \ldots$$.

Requiring $$2k\pi \leq 100 \implies k \leq \frac{50}{\pi} \approx 15.915$$, we find $$k = 0, 1, 2, \ldots, 15$$, giving 16 elements.

The answer is $$\mathbf{16}$$.