SOLUTION: I have a word problem that has been driving me crazy all night. I believe I have my setups, but I'm not sure. A number between 300 and 400 is forty times the sum of its digits.

Question 172704: I have a word problem that has been driving me crazy all night. I believe I have my setups, but I'm not sure. A number between 300 and 400 is forty times the sum of its digits. The tens digit is 6 more than the units digit. Find the number. Here is my setups: 300 + 10t + u = 40 (h + t + u), t = u + 6. Somewhere in figuring I get lost. Thanks for any help you can give.
Found 2 solutions by stanbon, solver91311:
Answer by stanbon(48558) (Show Source):
You can put this solution on YOUR website!
A number between 300 and 400 is forty times the sum of its digits. The tens digit is 6 more than the units digit. Find the number.
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The number must look like 3tu where t is the tensdigit and u is the units digit.
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Equations:
300 + 10t + u = 40(3+t+u)
t = u + 6
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Simplify the first equation:
10t + u = -300 +120 + 40t + 40u
-30t -39u = -180
10t + 13u = 60
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Substitute t = u+6 to solvefor "u":
10(u+6) +13u = 60
10u + 60 + 13u = 60
3u =0
u = 0
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Therefore t = 0+6 = 6
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Number: 360
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Cheers,
Stan H.

Answer by solver91311(12126) (Show Source):
You can put this solution on YOUR website!
Your setup is just right, except for one small detail. As you have constructed your equations, you have two equations in three variables and such a system never has a unique solution. What you may have missed is that the number must be between 300 and 400, so the hundreds digit is either 3 or 4. If the hundreds digit is 4, then the number must be 400, but . That means that the hundreds digit must be 3.

Now you can write your equations:

and

Now if you simplify and put the equations in standard form, you should have a fairly straightforward system to solve.

Write back if you need more help on this.