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?
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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.