1. **State the problem:** Find three-digit numbers such that the number itself is 21 times the sum of its digits. 2. **Define variables:** Let the three-digit number be $100a + 10b + c$ where $a$, $b$, and $c$ are digits and $a \neq 0$. 3. **Write the equation:** The sum of the digits is $a + b + c$. The problem states: $$100a + 10b + c = 21(a + b + c)$$ 4. **Expand and simplify:** $$100a + 10b + c = 21a + 21b + 21c$$ $$100a - 21a + 10b - 21b + c - 21c = 0$$ $$79a - 11b - 20c = 0$$ 5. **Rewrite:** $$79a = 11b + 20c$$ 6. **Analyze constraints:** - $a$ is from 1 to 9 (since it's a three-digit number) - $b$ and $c$ are from 0 to 9 7. **Check integer solutions:** For each $a$ from 1 to 9, find $b$ and $c$ such that $11b + 20c = 79a$. 8. **Trial for $a=1$:** $$11b + 20c = 79$$ Try $c=0$ to 9: - $c=3$: $11b = 79 - 60 = 19$ (no integer $b$) - $c=4$: $11b = 79 - 80 = -1$ (invalid) No solution. 9. **Trial for $a=2$:** $$11b + 20c = 158$$ Try $c=0$ to 9: - $c=9$: $11b = 158 - 180 = -22$ (invalid) - $c=7$: $11b = 158 - 140 = 18$ (no integer $b$) - $c=8$: $11b = 158 - 160 = -2$ (invalid) No solution. 10. **Trial for $a=3$:** $$11b + 20c = 237$$ Try $c=0$ to 9: - $c=9$: $11b = 237 - 180 = 57$ => $b=57/11=5.18$ (no) - $c=6$: $11b = 237 - 120 = 117$ => $b=117/11=10.63$ (no) - $c=3$: $11b = 237 - 60 = 177$ => $b=177/11=16.09$ (no) No solution. 11. **Trial for $a=4$:** $$11b + 20c = 316$$ Try $c=0$ to 9: - $c=8$: $11b = 316 - 160 = 156$ => $b=156/11=14.18$ (no) - $c=7$: $11b = 316 - 140 = 176$ => $b=176/11=16$ (no) No solution. 12. **Trial for $a=5$:** $$11b + 20c = 395$$ Try $c=0$ to 9: - $c=5$: $11b = 395 - 100 = 295$ => $b=295/11=26.81$ (no) - $c=10$ invalid No solution. 13. **Trial for $a=6$:** $$11b + 20c = 474$$ Try $c=0$ to 9: - $c=7$: $11b = 474 - 140 = 334$ => $b=334/11=30.36$ (no) No solution. 14. **Trial for $a=7$:** $$11b + 20c = 553$$ Try $c=0$ to 9: - $c=3$: $11b = 553 - 60 = 493$ => $b=493/11=44.82$ (no) No solution. 15. **Trial for $a=8$:** $$11b + 20c = 632$$ Try $c=0$ to 9: - $c=6$: $11b = 632 - 120 = 512$ => $b=512/11=46.55$ (no) No solution. 16. **Trial for $a=9$:** $$11b + 20c = 711$$ Try $c=0$ to 9: - $c=1$: $11b = 711 - 20 = 691$ => $b=691/11=62.82$ (no) No solution. 17. **Conclusion:** No three-digit number satisfies the condition exactly with digits $a,b,c$ in their ranges. 18. **Check for possible error:** Re-examining the problem, the only possible solution is when $a=1$, $b=8$, $c=7$: $$100(1) + 10(8) + 7 = 187$$ Sum of digits: $1 + 8 + 7 = 16$ $$21 \times 16 = 336 \neq 187$$ No. 19. **Try $a=3$, $b=6$, $c=3$:** $$363$$ Sum: $3+6+3=12$ $$21 \times 12 = 252 \neq 363$$ No. 20. **Try $a=4$, $b=5$, $c=9$:** $$459$$ Sum: $4+5+9=18$ $$21 \times 18 = 378 \neq 459$$ No. 21. **Try $a=7$, $b=1$, $c=8$:** $$718$$ Sum: $7+1+8=16$ $$21 \times 16 = 336 \neq 718$$ No. 22. **Try $a=9$, $b=9$, $c=9$:** $$999$$ Sum: $9+9+9=27$ $$21 \times 27 = 567 \neq 999$$ No. 23. **Final step:** Since no digit combination satisfies the equation, the only possible three-digit number is 0, which is invalid. **Answer:** There is no three-digit number that is exactly 21 times the sum of its digits.