3 has a value of 'ones'
0 has a value of 'tens'
9 has a value of 'hundreds'
5 has a value of 'thousands'
6 has a value of 'ten thousands'
Rounding up 6, its greater than 5 so add 1 to 7 and make all the other figures zero to give:
Answer = 800,000
Dunno what the work backward way is
seeems hard
maybe it's trial and error?
anyway
algebreaic
2n+5<13
minus 5 both sides
2n<8
divide by 2
n<4
According to the problem, it can be translated to
J + R = 97
J = 4 + 2R
solving these equations
J = 66
R = 31