Question 164602
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.{{{B=4A}}}
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"unit’s digit is one more than the ten’s digit"
2.{{{C=1+B=1+4A}}}
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"square of the sum of the digits is 72 more than the number"
3.{{{A^2+B^2+C^2=A*100+B*10+C+72}}}
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Use eqs.1 and 2 and substitute for B and C in eq. 3 and solve for A.
3.{{{A^2+B^2+C^2=A*100+B*10+C+72}}}
{{{A^2+(4A)^2+(1+4A)^2=A*100+(4A)*10+(1+4A)+72}}}
{{{A^2+16A^2+1+8A+16A^2=100A+40A+1+4A+72}}}
{{{33A^2+8A+1=144A+73}}}
{{{33A^2-136A-72=0}}}
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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.