Solve the poblem in base 5
2 ∗ 5 1 5 + 4 ∗ 5 1 4 + 5 1 3 + 3 ∗ 5 1 2 + 2 ∗ 5 1 1 + 2 ∗ 5 1 0 + 4 ∗ 5 9 + 3 ∗ 5 8 + 5 7 + 2 ∗ 5 6 + 3 ∗ 5 5 + 4 ∗ 5 4 + 5 3 + 5 2 + 1 9
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It Is very clear to get the following expressions: 5 = 1 0 5 2 = 1 0 0 2 × 5 1 5 = 2 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 4 × 5 1 4 = 4 0 0 0 0 0 0 0 0 0 0 0 0 0 0 5 1 3 = 1 0 0 0 0 0 0 0 0 0 0 0 0 0 3 × 5 1 2 = 3 0 0 0 0 0 0 0 0 0 0 0 0 2 × 5 1 1 = 2 0 0 0 0 0 0 0 0 0 0 0 2 × 5 1 0 = 2 0 0 0 0 0 0 0 0 0 0 4 × 5 9 = 4 0 0 0 0 0 0 0 0 0 . . . . . . . . . . . . . . . . . . . . . . . . . . . 1 9 = 4 5 So: R e s = 2 4 1 3 2 2 4 3 1 2 3 4 1 1 3 4
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Let the number to be found be in based 5 be N 5 , then we have:
N 5 = 2 ˙ 5 1 5 + 4 ˙ 5 1 4 + 3 ˙ 5 1 3 + 5 1 2 + 2 ˙ 5 1 1 + 2 ˙ 5 1 0 + 4 ˙ 5 9 + 3 ˙ 5 8 + 5 7 + 2 ˙ 5 6 + 3 ˙ 5 5 + 4 ˙ 5 4 + 5 3 + 5 2 + 1 9 = 2 ˙ 5 1 5 + 4 ˙ 5 1 4 + 3 ˙ 5 1 3 + 1 ˙ 5 1 2 + 2 ˙ 5 1 1 + 2 ˙ 5 1 0 + 4 ˙ 5 9 + 3 ˙ 5 8 + 1 ˙ 5 7 + 2 ˙ 5 6 + 3 ˙ 5 5 + 4 ˙ 5 4 + 1 ˙ 5 3 + 1 ˙ 5 2 + 3 ˙ 5 1 + 4 ˙ 5 0 [ 1 9 = 1 5 + 4 ] = 2 4 1 3 2 2 4 3 1 2 3 4 1 1 3 4 5