I like that the problem calls for odd integers. If it's a series of even integers, it's intuitive that we add two to get the next value in the series. When it's odd integers, that same rule applies: just add 2. But for a moment, it might not _feel_ right, since 2 is one of the more famous even numbers and we're trying really hard to think about _odd_ numbers. That's the beauty of it, because the same prompt could be given for consecutive even numbers, and the approach would be exactly the same. In fact the approach here is not specific to odd or even numbers. But if the prompt with these specific numbers called for even numbers, we'd run into a problem because the solution would of course give us this same odd number as the answer.
I like to think of adding/multiplying evens and odds together as the best way to introduce the idea of a group, even if it's not part of one. It's a very relatable form of "pseudo" abstract algebra that almost anyone can understand. It's missing key elements to be a proper group, but you can reason about the shape of an answer at a general case, which is an important stepping stone towards more advanced topics.
I think this is a great video, however, I decided the first term to be "x", 2nd term to be "x+1", 3rd "x+3", x+5, x+7. I can see why it went wrong for me. I learn more from the mistakes that I made. Thanks for the great video. The mistake I made, I should stayed with x+1 as my 1st, x+3 as my second.... etc, or x, x+2, x+4...etc, instead of mixing odd and even numbers.
@@mrhtutoringprofessor, an explanation of why x+2 or x-2 should be the second term might be more illuminating. x+2, x-2 x+4, x-4 x+6, x-6 x+8, x-8 x+10, x-10 In both cases you get 2 true statements of 2 odd numbers but the answer is different. I approached the question from a programming perspective. This could be a viral video in makings.
@@Kedatgahbelu12 You'll get the same consecutive integers as 5,7,9,11,13 if you set the first term as x+2 and x+4, x+6,x+8,x+10. (x=3 in this set up and 5th term is x+10=13) But you should set the second group as x-10 x-8 x-6 x-4 x-2 (x=15 in this set up and 5th term is 15-2=13) We're not looking for x but the 5th term from the 5 consecutive integers(5,7,9,11,13)
@@user-iu8uk5tc9s my point was, setting the right parameters are super important. As long as the differences are two and consective odd numbers, it should work. I was just pointing out the mistakes so other people wouldn't be embarrassed to get the question wrong, as long as they give it a good try. The majority the people that hate math, because they have this great fear of looking stupid in front of other people. Quiting is the absolute worst thing you can do. Thanks for your inputs. I did catch my mistakes and have learned from my mistake. 🙏🏽
I believe this is incorrect. It’s saying the sum of the first and third integers are less, so 19 is on the side of the first and third integers x+x+4-19=3x+18 2x-15=3x+18 x=-33
Think about it with another example. 5 is 19 less than 24. According to you, 5-19 = 24. This is not true, as we got -24 = 24, which is incorrect. According to the tutor, we get 5=24-19. This is correct, as we get 5=5. Therefore it is clear that the -19 should correctly be on the side of the equals sign which would otherwise be more than the other side.
It says five ODD consecutive integers, and you've done even. Instead it should be x+1, x+3 , x+5 , x+7, x+9. X is equal to 18 and answer is x+9 therefore answer is 27.
If the 5 numbers are consecutive odd numbers, they will each be 2 apart. As long as x is odd, the rest of the numbers are odd as well. X represents the first odd number, so the second number will be 2 more than that, the third number will be 4 more, etc.
Excellent set up but the answer you arrived is a bit off. It should be 13. With your set up x =4 is even but x+9 =13. X=4 obviously is incorrect as 4 is an even number but 13 is certainly an even one. You were a lot closer than I was, mixing even and odd numbers, completely throwing off, incorrect series.
@@alexanderflores6438 yes but x is not on the list of chosen integers. So it doesn't matter. List of integers of mine starts with x + 1. Even + Odd = Odd. From my working x is 18 (even indeed) but first integer of choice is 18 + 1 which is 19, therefore all integers of choice being odd and consecutive!
@@dylanjayabahu2878 yes but exclude the possibility of x being an even number we can add and odd to it to make it odd. As X is a variable we cannot specify that it is to be chosen as an odd, so it could still be even.
The prompt "5 consecutive odd integers" does not imply that we start at 1. For example, we could have these five consecutive odd integers: 63, 65, 67, 69, 71 In this case, the first and third integers would be 63 and 67, and so on.
What will be the first term? x+1 1st, x+3 2nd ....etc or did you had x-1 as the first term? x-1, x+1, x+3, x+5, x+7? Sure you will get the 5th series to be equal to 13, but x equal to 6. A 6 and 13 would have a false and true test on the boolean (odds) tests. Both of them have to be true to pass that test. Good thinking processes, but as soon as you find the first false statement, you shouldn't waste your time on the second test. This is NOT a mutually exclusive situations. Both must occur in the correct series. Odd # on the first test, and another odd # on the second test.
