# Finding Consecutive Integers

Find four consecutive odd integers such that 5 times the sum of the first and the third is 15 more than 5 times the fourth.

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What have you tried? –  Aryabhata Mar 14 '11 at 16:05
@Moron: So far, probably just posing a question! –  The Chaz 2.0 Mar 14 '11 at 16:19
Let's see who gets more up votes, @Moron or @TheChaz! (Or @muntoo :) ) –  muntoo Mar 15 '11 at 0:58
@muntoo: Now you are in the running too :-) –  Aryabhata Mar 15 '11 at 1:03

We have to find 4 odd numbers so let us take

$2n-1$ , $2n+1$ , $2n+3$ , $2n+5$

Now we have $5 ((2n-1) + (2n+3)) - 15 = 5 (2n + 5)$

You can now solve this .....

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So the answer is 3 for n? –  Barny William Mar 14 '11 at 16:14
Yes and the numbers then become 5,7,9,11 –  user8250 Mar 14 '11 at 16:27

HINT $\rm\ \ 5\ (n_1 + n_3)\ =\ 5\ n_4 + 15\: \ \iff\: \ n_1 + n_3 - n_4\ =\ 3\:.\:$ So $\rm\ n_3 - n_4 = -2\ \Rightarrow\ n_1 = \ \cdots$

The hypothesis that the $\rm\:n_i\:$ are odd is stronger than needed. To solve for the $\rm\:n_i\:$ from the given equation it suffices to know only the values $\rm\ n_{i+1} - n_i\:.$

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Let the numbers be $2m+1,2m+3,2m+5,2m+7$ then solve for $$5 \cdot \Bigl[ 2m+1 + 2m+5\Bigr] = 5(2m+7) + 15$$

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### Given:

$2n+1, 2n+3, 2n+5, 2n+7$

$5((2n+1) + (2n+5)) = 5(2n+7) + 15$

$5(4n + 6) = 10n + 35 + 15$

$20n + 30 = 10n + 35 + 15$

$10n = 20$

$n = 2$

$2n+1, 2n+3, 2n+5, 2n+7$

$2(2)+1, 2(2)+3, 2(2)+5, 2(2)+7$

$5, 7, 9, 11$

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The request was for consecutive odd integers –  Ross Millikan Mar 15 '11 at 1:15
@Ross Thanks, I should've read the question more carefully. :) How many whacks on the head do you think will fix that? –  muntoo Mar 15 '11 at 1:36
If I am the test, it will take $\omega$. And I suspect I am not alone... –  Ross Millikan Mar 15 '11 at 2:52