A faster way would be: a+b+c+d in [1,36], which implies abcd in [1955,1990]. So we must have that a=1 and b=9. To find c and d, we can reduce the equation to cd+c+d=81. Or, more mathematically, 11c+2d = 81. We know 2d in [0,18], so 11c in [63,81]. But this means c can only be 6 or 7. But 2d is even, so 6 falls off. So we have c=7 and d=2.
A faster way would be:
a+b+c+d in [1,36], which implies abcd in [1955,1990]. So we must have that a=1 and b=9.
To find c and d, we can reduce the equation to cd+c+d=81. Or, more mathematically, 11c+2d = 81. We know 2d in [0,18], so 11c in [63,81]. But this means c can only be 6 or 7. But 2d is even, so 6 falls off. So we have c=7 and d=2.
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a=1,b=9,cd+c+d=81 c=7 d=2 n=1972
1972
But different Method, gross method but quick
abcd+a+b+c+d=1991
a+b+c+d=
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a+b+c+d 1900, therefore a=1 and b=9
1+9+0+0
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