SOLUTION: Find a three digit number whose tens digit is 4 times the hundred’s digit and the unit’s digit is one more than the ten’s digit. The square of the sum of the digits is 72 more than

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Let H be the hundred digit.
Then the tens digit is 4H
The ones digit is 4H+1
The "number is (H*100)+(4H*10)+(4H+1) = (100H)+(40H)+(4H+1)= 144H + 1
The sum of the digits is H+4H+4H+1 = 9H+1
You are told the square of the sum of the digits is 72 more than the number itself
So


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H = -4/9 or 18/9=2
Since the number is an integer, the answer must be H=4
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