SOLUTION: the tens digit of a number is 3 less than the units digit. if the number is divided by the sum of the digits. the quotient is 4, the remainder is 3, what is the original number?

Question 184479: the tens digit of a number is 3 less than the units digit. if the number is divided by the sum of the digits. the quotient is 4, the remainder is 3, what is the original number?
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
the tens digit of a number is 3 less than the units digit. if the number is divided by the sum of the digits. the quotient is 4, the remainder is 3, what is the original number?
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Let the original number be 10t+u
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Equations:
t = u - 3
(10t+u)/(t+u) = 4 + 3/(t+u)
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Simplify the 2nd equation:
10t+u = 4(t+u) + 3
10t+u = 4t+4u + 3
6t - 3u = 3
2t -u = 1
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Form the system of two equations:
t = u -3
2t -u = 1
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Substitute to solve for "u":
2(u-3)-u = 1
2u-6-u = 1
u = 7
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Substitute into t = u-3 to solve for "t":
t = 7-3 = 4
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The number is 47
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Cheers,
San H.
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