Passing through 1,-4 and -2,5
1 answer:
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Let's say our first integer is "a".
how to get the next consecutive EVEN integer? well, just add or subtract 2 from it, therefore, the second consecutive integer will be "a + 2".
and the next after that, will then be (a + 2) + 2, or "a + 4".
so those are are 3 integers, a a + 2 a+4
notice that, from any even or odd integer, if you hop twice either forwards or backwards, you'll land on another even or odd integer respectively.
2 + 2 is 4, or 8 + 2 is 10 some even ones
3 + 2 is 5, or 13 + 2 is 15, some odd ones
![\bf \stackrel{\textit{3 times the first}}{3a}~~=~~\stackrel{\textit{26 less than twice the sum of the others}}{2[~(a+2)+(a+4)~]~~~-26} \\\\\\ 3a=2[~2a+6~]-26\implies 3a=4a+12-26\implies 3a=4a-14 \\\\\\ 0=a-14\implies 14=a](https://tex.z-dn.net/?f=%5Cbf%20%5Cstackrel%7B%5Ctextit%7B3%20times%20the%20first%7D%7D%7B3a%7D~~%3D~~%5Cstackrel%7B%5Ctextit%7B26%20less%20than%20twice%20the%20sum%20of%20the%20others%7D%7D%7B2%5B~%28a%2B2%29%2B%28a%2B4%29~%5D~~~-26%7D%0A%5C%5C%5C%5C%5C%5C%0A3a%3D2%5B~2a%2B6~%5D-26%5Cimplies%203a%3D4a%2B12-26%5Cimplies%203a%3D4a-14%0A%5C%5C%5C%5C%5C%5C%0A0%3Da-14%5Cimplies%2014%3Da)
what are the other two consecutive integers? well, a + 2 and a + 4.
how to get the next consecutive EVEN integer? well, just add or subtract 2 from it, therefore, the second consecutive integer will be "a + 2".
and the next after that, will then be (a + 2) + 2, or "a + 4".
so those are are 3 integers, a a + 2 a+4
notice that, from any even or odd integer, if you hop twice either forwards or backwards, you'll land on another even or odd integer respectively.
2 + 2 is 4, or 8 + 2 is 10 some even ones
3 + 2 is 5, or 13 + 2 is 15, some odd ones
what are the other two consecutive integers? well, a + 2 and a + 4.
Answer:
Step-by-step explanation:
We are given
Let's assume it can be factored as
now, we can multiply right side
and then we can compare it
now, we can compare coefficients
now, we can find all possible factors of 48
and then we can assume possible prime factors of 48
Since, we have to find the largest value of n
So, we will get consider larger value of r because of 5r
and because n is negative of 5r+s
so, we will both n and r as negative
So, we can assume
r=-48 and s=-1
so, we get