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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Question 164602: 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 the number?
Answer by Fombitz(32388) About Me  (Show Source):
You can put this solution on YOUR website!
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%2BB=1%2B4A
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"square of the sum of the digits is 72 more than the number"
3.A%5E2%2BB%5E2%2BC%5E2=A%2A100%2BB%2A10%2BC%2B72
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Use eqs.1 and 2 and substitute for B and C in eq. 3 and solve for A.
3.A%5E2%2BB%5E2%2BC%5E2=A%2A100%2BB%2A10%2BC%2B72
A%5E2%2B%284A%29%5E2%2B%281%2B4A%29%5E2=A%2A100%2B%284A%29%2A10%2B%281%2B4A%29%2B72
A%5E2%2B16A%5E2%2B1%2B8A%2B16A%5E2=100A%2B40A%2B1%2B4A%2B72
33A%5E2%2B8A%2B1=144A%2B73
33A%5E2-136A-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.