Question-98077

Question Number 98077 by Lekhraj last updated on 11/Jun/20

Commented by MJS last updated on 11/Jun/20

it only makes sense if a, b are digits and ab doesn′t mean a×b but 10a+b: a, b ∈N∧1≤a≤9∧0≤b≤9 now try all possible pairs 1!+0!=10^2 1!+1!=11^2 ... 9!+9!=99^2

$$\mathrm{it}\:\mathrm{only}\:\mathrm{makes}\:\mathrm{sense}\:\mathrm{if}\:{a},\:{b}\:\mathrm{are}\:\mathrm{digits}\:\mathrm{and}\:{ab} \\ $$$$\mathrm{doesn}'\mathrm{t}\:\mathrm{mean}\:{a}×{b}\:\mathrm{but}\:\mathrm{10}{a}+{b}: \\ $$$${a},\:{b}\:\in\mathbb{N}\wedge\mathrm{1}\leqslant{a}\leqslant\mathrm{9}\wedge\mathrm{0}\leqslant{b}\leqslant\mathrm{9} \\ $$$$\mathrm{now}\:\mathrm{try}\:\mathrm{all}\:\mathrm{possible}\:\mathrm{pairs} \\ $$$$\mathrm{1}!+\mathrm{0}!=\mathrm{10}^{\mathrm{2}} \\ $$$$\mathrm{1}!+\mathrm{1}!=\mathrm{11}^{\mathrm{2}} \\ $$$$… \\ $$$$\mathrm{9}!+\mathrm{9}!=\mathrm{99}^{\mathrm{2}} \\ $$

Commented by Lekhraj last updated on 11/Jun/20