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2bornot2b

  • 2 years ago

Show that an integer of the form 6k+5 is also of the form 3k+2, but not conversely

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  1. 2bornot2b
    • 2 years ago
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    I think I can work on the first part, but have no idea on the second "converse part'

  2. Mani_Jha
    • 2 years ago
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    Can you show your work on the first part?

  3. 2bornot2b
    • 2 years ago
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    So if an integer is of the form 6k+5, then it is equivalent to 6k+3+2, which is equal to 3(2k+1)+2. Now if k is an integer 2k+1 is also an integer, so it is of the form 3k+2

  4. 2bornot2b
    • 2 years ago
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    I am sorry, I should have mentioned k is an integer.

  5. Mani_Jha
    • 2 years ago
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    Yes, then just try the reverse. 3k+2=6k+5-3k-3 =6k-3(k+1)+5 6{k-(k+1)/2}+5 =6(k-1)/2+5 (k-1)/2 may not be an integer if n is even. See that the expression does not come out to be in the form 6n+5, where n is an integer. So the converse is true sometimes, but not always.

  6. Mani_Jha
    • 2 years ago
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    Ok?

  7. Zarkon
    • 2 years ago
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    for the converse...just give a specific counter example

  8. Zarkon
    • 2 years ago
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    2=3*0+2 2=6k+5 -3=6k k=-1/2 which is not an integer

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