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Write the cubes of 5 natural numbers which are of the form 3n + 1 (e.g. 4, 7, 10, …) and verify the following:“The cube of a natural number of the form 3n+1 is a natural number of the same from i.e. when divided by 3 it leaves the remainder 1’' |
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Answer» As we know that the first 5 natural numbers in the form of (3n + 1) are 4, 7, 10, 13 and 16 The cube of 4, 7, 10, 13 and 16 43 = 4 × 4 × 4 = 64 73 = 7 × 7 × 7 = 343 103 = 10 × 10 × 10 = 1000 133 = 13 × 13 × 13 = 2197 163 = 16 × 16 × 16 = 4096 Hence , all these cubes when divided by ‘3’ leaves remainder 1. \(\therefore\) “The cube of a natural number of the form 3n+1 is a natural number of the same from i.e. when divided by 3 it leaves the remainder 1’ is true. |
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