/sci/ - Science & Math

56 57 58

56 + 57 + 58

Jesus, dude. Divide by three.

n + (n+1) +(n + 2) = 171

o = n+1

p = n+2

Do the algebra.

Okay, One-fifth of a number plus one-sixth of a the same number is 33. Bet you can do that. that shit was hard.

Don't do OP's work for him guys.

I'll help if you're willing to learn how to solve these mathematically.

>>1942356
i have the answers dude, im just testing you guys to see if you know calculus.

i didnt ask a question, i just said find the answer.

>>1942363
>calculus
>this is 8th grade algebra

>>1942366
>implying this is even 8th grade
>implying this isn't 6th grade

>>1942366
it's okay if you don't know the answer i can help you get them. the first one you just use trial and error til you get the answer....

first you would do 1,2,3 = 6 up until you get 3 consecutive that add up to 171.

>>1942375
>implying I wasn't in the stupid class until I got my shit together in college

>>1942376

USE ALGEBRA YOU HUGE FAGGOT

You are literally wasting your time if you are guessing the solutions. The purpose of these exercises is to teach you basic algebra. To get something out of these exercises you should solve them algebraically.

>>1942375
This is dividing a number by 3. More like 2nd grade, at the most.

>>1942402
Third or fourth for integer division, not second.

Okay, my bad for posting. Just thought i could help out a little bit. Go math! ;_;

>>1942402

By dividing by three you only find the approximate solution, which isn't useful for more complicated problems.

When you do math it's always easier to learn a generalized method than to learn a thousand special cases.

HOLY SHIT, I REMEMBER LEARNING THIS STUFF IN GRADE 5.

Not even joking.

>>1942418
Find X consecutive integers whose sum is N.

First integer is N/X - (X-1)/2.

In this case, divide by 3 and subtract 1. Is that too complicated for you?

NUMBERS MAKE MY BRAIN HURT

171/3 = 57

57 + 57 + 57 = 171
(57 +(-1))+ 57 +(57 + 1) = 171
56 + 57 + 58 = 171

OP is in elementary school; this is a homework thread.

>>1942426
>mfw that's a special case that you will never use unless you are given this specific problem

>>1942436
> mfw you're retarded

Find 5 consecutive integers whose sum is 120.
120/5 - (5-1)/2 = 22.
22+23+24+25+26 = 120.

Find 4 consecutive integers whose sum is 62.
62/4 - (4-1)/2 = 14.
14 + 15 + 16 + 17 = 62.

>>1942418
3 consecutive numbers, 1 will be divisible by 3.

>>1942456

How often do you actually find x number of consecutive integers that add to a known number?

Never.

The general case applies to every possible application of algebra ever.

Me: 1
You: 0

>>1942469
See: (>>1942426)

>>1942473
The most general case for this problem is "Find X consecutive integers whose sum is N." What are you smoking?

>>1942324
>Asks for people to prove an existential statement
>Surprised by the results

>>1942482

Try solving this with that formula >>1942338

Protip: you won't get the correct answer

>>1942495
I think we all know that "formula" is not correct. I was referring to "First number = N/X - (X-1)/2"

>>1942426

What if (X-1)/2 is a non-integer, but N/X is an integer? Is there just NO SUCH THING?

56 57 58
ofit
pr

>>1942511
If N is truly the sum of X consecutive integers, they will both be non integers. For example, the example of 14+15+16+17: 62/4 = 15.5, 3/2 = 1.5.

(n/3 - 1), (n/3), (n/3 + 1)

>>1942640
WOW YOU ARE GENIUS.

A bit late to the party, there, buckaroo.

>>1942324
n+(n+1)+(n+2)=171
3n +3 = 171
3n = 168
n = 56

=> {56,57,58}

\thread

>>1942338
1/5x+1/6x=33
11/30x=33
x=90

That's nearly as easy as OP, you are either stupid or a troll.

>>1942662
i know how to do these problems... why would i ask them if i didn't already know the answer. I was just trying to see if you guys could do them.

but obviously everyone thinks they're too good for algebra.

>mfw OP who can't do basic algebra outsmarted /sci/ and had you guys do his homework

> mfw 32 replies

>>1942456
k consectutuve numbers that when added together equal z

n+(n+1)+(n+2)....(n+k-1) = z
=>
k*n + 1+2+3+....+(k-1) = z
=>
k*n + (k-1)*(k-1+1)/2 = z
=>
k*n + (k-1)*k/2 = z
=>
n = z/k - (k-1)/2

>>1942426
I said that over an hour ago:
>>1942689

And they didn't believe me. Disappointing to say the least.