3. Solve the rebus UHA = LCM(UX, UA, HA). Here U, X, A are three different digits. Two-digit and three-digit numbers cannot start with zero. Recall that the LCM of several natural numbers is the smallest natural number that is divisible by each of them.
Answer. $150=\operatorname{HOK}(15,10,50)$. Solution. Since UHA is divisible by UX, A $=0$, that is, the number has the form UX0. Since UX0 is divisible by U0 and X0, UX is divisible by U and X, from which it follows that X is divisible by U, and 10 U is divisible by X. Let $\mathrm{X}=\mathrm{M} \cdot \mathrm{U}$. Then $10 \mathrm{~V}$ is divisible by $M \cdot U$, from which 10 is divisible by M, that is, $M=2$ or $\mathrm{M}=5$. The condition $\mathrm{X}=2$ U is satisfied only by the numbers 120, 240, 360, 480, and the condition $\mathrm{X}=5 \mathrm{~V}$ - only by the number 150. Verification shows that only the number 150 fits.