how many two digit counting numbers are there where the tens place digit is greater than the unit digit
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tens digit greater than their unit digit
t>uthe sum of their digits equal to twice their differencet+u = 2(t-u) t+u = 2t-2u 3u = t
Since 1 ≦ t ≦ 9 1 ≦ 3u ≦ 9 1/3 ≦ u ≦ 3 1 ≦ u ≦ 3 So u = 1, 2 or 3 If u = 1, then t = 3u = 3(1) = 3, so the number is 31 If u = 2, then t = 3u = 3(2) = 6, so the number is 62 If u = 3, then t = 3u = 3(3) = 9, so the number is 93 So there are three solutions: 31, 62, and 93 Edwin
t>uthe sum of their digits equal to twice their differencet+u = 2(t-u) t+u = 2t-2u 3u = t
Since 1 ≦ t ≦ 9 1 ≦ 3u ≦ 9 1/3 ≦ u ≦ 3 1 ≦ u ≦ 3 So u = 1, 2 or 3 If u = 1, then t = 3u = 3(1) = 3, so the number is 31 If u = 2, then t = 3u = 3(2) = 6, so the number is 62 If u = 3, then t = 3u = 3(3) = 9, so the number is 93 So there are three solutions: 31, 62, and 93 Edwin
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