Help please!! Of all the pairs of consecutive even integers whose sum is greater than 44, find the pair whose sum is the least.
x + y > 44 let x = 2n let y=2n+2 2n+2n+2 > 44 solve for n
I dont know how to solve for n
@Kittykatkat14 Did you understand the explanation given by @amistre64 ?
No, sorry
you know what are consecutive even integers?
Yes 2,4,6 ... Right?
Cool, suppose we assume we have one even integer as 2n, what would be the consecutive integer for 2n?
4n?
If you have 4, what's the next even integer?
2n+2
6
Yes, so what's the sum of them ?
sum of 2n and 2n+2
Sum of what?
2n+2n+2 this is the sum of consecutive integers, isn't it?
Yes
Now we have the condition, that the sum of consecutive integers should be greater than 44. How would you write that in an equation form? particularly an inequality
I was looking for least
Yeah, but first we'll write equation for all greater than 44 \[2n+2n+2>44\] Is that correct?
Oh okay, yes
Sum of 2 consecutive even integers is greater than 44, but the sum should be least. We are obviously looking for a+b =46 where a and b are consecutive even integers.
so we have \[4n+2>44\] Yes @shubhamsrg has given us a nice hint so we have now 4n+2=46 Did you understand ?
Yes i understand, so it would be 22 and 24?
yes :)
How bout least?
a+b=42?
This is the least sum, which is greater than 44
Greater or less than?
Of all the pairs with sum greater than 44, find the ones with the least sum? so we know it has to be greater than 44 but least 46 satisfies both conditions
It's like find the least integer greater than 10? it has to be 11
Is it less than 46 or 44? Nothing says that less than 46?
greater than 44, but least what's the next even integer after 44?
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