N = 100A + 10B + C = C × (10A + B) or C × (10B + A) = 10AC + BC , or , 10BC + AC Because C < 10, that means 100A + 10B + C = 10BC + AC 100A + 10B = 10BC + AC – C 10 × (10A + B) = C × (10B + A – 1) Therefore, either C = 5 and 20A + 2B = 10B + A – 1 OR C = 2 and 5 × (10A + B) = 10B + A – 1 OR A = 1 and 10 × (10 + B) = C × (10B) First case : 19A = 8B – 1 ===> No solutions ( A = 1 and A = 3 not applicable ) Second case : 49A = 5B – 1 ===> No solutions (since B < 10 too) Third case : 10 + B = BC 10 = B × (C – 1) ===> (B, C) = { (2, 6) , (5, 3) } Answer : There are two firm numbers, 126 and 153

In the case where $(1) \;\; 100A + 10B + C = (10B + A)C$ we obtain $(2) \;\; 10 \mid (A - 1)C$. Condition (2) is satisfied when $C$ is even and $A=6$, which inserted in equation (1) result in $600 + 10B + C = 10BC + 6C$ $2BC - 2B + C = 120$ $(2B + 1)(C - 1) = 7 \cdot 17$ $(2B + 1,C - 1) = (17,7)$ $B=C=8$ Conclusion: $688$ is a firm number (since $8 \cdot 86 = 886$).

Thank you, Solar ! Solar found all three firm numbers : 688 = 8 × 86 = 2^4 × 43 153 = 3 × 51 = 3^2 × 17 126 = 6 × 21 = 2 × 3^2 × 7