Question 203493
 the tens digit of a 3 digit number is 3. the sum of its digits is 18. if the 10s and 100s digits are reversed, the sum of the new number and the original number is 1008. find the original number.
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Let the original number be 100h + 10t + u
Then the reversed form is 100t + 10h + u
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Equations:
t = 3
h + t + u = 18
110h + 110t + 2u = 1008
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Make use of t = 3 and you get:
h + u = 15
110h + 2u = 678
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Modify the 1st equation:
2h + 2u = 30
110h + 2u = 678
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Subtract the 1st from the 2nd to solve for "h":
108h = 648
h = 6
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Since h+u=15, u = 9
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The original number is 639
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Cheers,
Stan H.