Question

determinati numerele naturale a,b,c stiind ca a < b < c si ca 9×a +9×b+9×c= 72

Answer 1

$a < b < c$

$9a + 9b + 9c = 72$

$9(a + b + c) = 72 \: | \div 9$

$a + b + c = 8$

$0 + 1 + 7 = 8$

$0 < 1 < 7$

$0 + 2 +6 = 8$

$0 < 2 < 6$

$0 + 3 + 5 = 8$

$0 < 3 < 5$

$1 + 2 +5 = 8$

$1 < 2 < 5$

$1 + 3 + 4 = 8$

$1 < 3 < 4$