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September 1, 2014

September 1, 2014

Posted by **Rob** on Wednesday, September 13, 2006 at 8:56pm.

Can anyone help???

We're asked:

Determine whether there are 2 consecutive odd integers such that 5 times the first exceeds three times the second by 54.

Let the first odd number be 2k+1, then the next odd number is is 2(k+1)+1=2k+3.

Then 5*(2k+1)=3*(2k+3)+54 so

10k+5=6k+9+54 thus

4k=58

Since 4 does not divide 58 there is no k to satisfy the integers, so the answer is no, the integers don't exist.

The closest k's we could use are 14 and 15.

Consider k=14, then 2k+1=29 and 2k+3=31

We find 5*29=145, 3*31=93 and 145-93=52

Consider k=15, then 2k+1=31 and 2k+3=33

We find 5*31=155, 3*33=99 and 155-99=56

Check that the problem was entered correctly and that I read it correctly.

Since 5x = 3(x + 2) + =54, 2x = 60 making x = 30 and (x + 2) = 32.

I guess not.

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