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

Mathematics
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I think I can work on the first part, but have no idea on the second "converse part'
Can you show your work on the first part?
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

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Other answers:

I am sorry, I should have mentioned k is an integer.
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.
Ok?
for the converse...just give a specific counter example
2=3*0+2 2=6k+5 -3=6k k=-1/2 which is not an integer

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