AMC10 2016 B

Answer (A): Note that for any $x$, $$ f(x+1)=\sum_{k=2}^{10}\big(\lfloor kx+k\rfloor-k\lfloor x+1\rfloor\big) =\sum_{k=2}^{10}\big(\lfloor kx\rfloor+k-k\lfloor x\rfloor-k\big)=f(x). $$ This implies that $f(x)$ is periodic with period $1$. Thus the number of distinct values that $f(x)$ assumes is the same as the number of distinct values that $f(x)$ assumes for $0\le x<1$. For these $x$, $\lfloor x\rfloor=0$, so $$ f(x)=\sum_{k=2}^{10}\lfloor kx\rfloor, $$ which is a nondecreasing function of $x$. This function increases at exactly those values of $x$ expressible as a fraction of positive integers with denominator between $2$ and $10$. There are $31$ such values between $0$ and $1$. They are $$ \frac12,\frac13,\frac23,\frac14,\frac34,\frac15,\frac25,\frac35,\frac45,\frac16,\frac56,\frac17,\frac27,\frac37,\frac47,\frac57,\frac67,\frac18,\frac38,\frac58,\frac78,\frac19,\frac29,\frac49,\frac59,\frac79,\frac89,\frac1{10},\frac3{10},\frac7{10},\frac9{10}. $$ Thus $f(0)=0$ and $f(x)$ increases $31$ times for $x$ between $0$ and $1$, showing that $f(x)$ assumes $32$ distinct values.