Question 395146
1. the sum of the digits f a two-digit number is 11. when the digits are reversed, the new number is increased by 20 which is twice the original number. find the original number.
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Let the number be 10t+u.
Equations:
t + u = 11
10u+t+20 = 2(10t+u)
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Substitute for "t" and solve for "u":
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10u + 11-u + 20 = 20(11-u)+2u
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9u + 31 = 220-20u+2u
-27u = -189
u = 7
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Solve for "t":
t+u = 11
t+7 = 11
t = 4
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Original Number: 47
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2. the units digit of a two-digit number is 3 less than its tens digit. if the number is divided by the sum of its digits, the quotient is 6 and the remainder is 8. find the number.
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Equations:
u = t-3
(10t+u)/(t+u) = 6(t+u)+8
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Substitute for "u" and solve for "t":
(10t+t-3)/(t+t-3) = 6(t+t-3)+8
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(11t-3)/(2t-3) = 12t-18+8
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(11t-3)/(2t-3) = 12t-10
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11t-3 = 2(2t-3)(6t-5)
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11t-3 = 2[12t^2-10t-18t+15)
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11t-3 = 24t^2-56t+30
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24t^2-67t+33 = 0
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Comment:
This equation does not have a whole
number solution.
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Cheers,
Stan H.