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's call the hundreds' digit A, tens' digit B, and ones' digit C.
The number is ABC, or numerically A*100+B*10+C.
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"tens digit is 4 times the hundred’s digit"
1.
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"unit’s digit is one more than the ten’s digit"
2.
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"square of the sum of the digits is 72 more than the number"
3.
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Use eqs.1 and 2 and substitute for B and C in eq. 3 and solve for A.
3.




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This quadratic equation does not have an integer solution.
Please check the problem set-up because something is wrong.
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The only possible number choices are 001, 145, and 289.
The "sum of the squares" for those are : 1, 42, 149.
While "72 more than the number" is: 73, 217, 361.
Clearly the last two lines aren't equal.
There is no solution the way the problem is set up.
Please check and re-post.