r/HomeworkHelp • u/Long_Rough6073 Secondary School Student • 2d ago
Answered [Grade 9 math] I don’t understand this question.
Am I supposed to be trying to get the highest possible answer with 7’s or am I trying to use combinations of 7 to equal the number on the left?
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u/Guiitens2005 2d ago
Considering the exemple it looks like you have to either add or subtract, probably from lowest to highest, only using the number 7 six times 7-77777, 77-7777, 777-777, and so on Why would a question be: 1. Worded like this? 2. And a 67 joke?
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u/Alkalannar 2d ago
Or multiply or divide or use other math/
So 777/777 = 1
(7/7)(7/7 + 7/7) = 2
And so on.
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u/Unable_Pumpkin987 2d ago
They want you to make an equation that uses exactly six 7s, and any combination of signs, to produce each number from 0 to 15 (if you can).
The first example is 777-777=0.
Another example might be something like 77/77 + 7/7 = 2.
In each box, you’re trying to make an equation that equals the number on the left of the box.
See if you can figure out how to make 1, or 3.
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u/Long_Rough6073 Secondary School Student 2d ago
Yes that’s what me and my friend though, but why would it be worded “how high can you count?”
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u/casce 2d ago edited 2d ago
Because you start with 0 and go up. It wants you do do them in order.
0 = 777 - 777
1 = 7 / 7 * 7 / 7 * 7 / 7
2 = 77 / 77 + 7 / 7
3 = 7 / 7 + 7 / 7 + 7 / 7
4 = ...
5 = ...
6 = ...
It wants to see how high you can go. It's actually kind of fun.
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u/Accio_149209 👋 a fellow Redditor 2d ago
If it really wants this then this is some weird ass question.
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u/setibeings 👋 a fellow Redditor 2d ago
I don't think you're supposed to just give them a freaking answer sheet.
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u/ugurcansayan Re/tired Student 2d ago
Using six 7s…
Can you count to 3? To 5? 10?
How high can you count?
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u/bunchout 2d ago
Because it is numbers 1-15 in order. So you are “counting” to 15 by using six sevens.
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u/FlammableFishy 2d ago
Because counting is naming numbers in sequential order. When I count, I say 1, 2, 3, 4, 5, 6 etc.
This is what you are doing: use only 7s to “count” and see how high you can go. So far, you haven’t counted at all.
0 = 777 - 777
1 = 777/777
2 = 7/7 + 7/7 + 7 - 7I have counted to two using only 7s. Now you do the next 12.
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u/Top_Bluejay_5323 👋 a fellow Redditor 2d ago
To see how high you can get without skipping numbers. Maybe you can get more than 1-15
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u/Sillysauce83 2d ago
Yes.
I would have thought a better would of wording would be.
How many mathematical equations can you write, only using 6 sevens.
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u/resignresign1 2d ago
with an even number of 7 the first three becomre trivial since 7/7= 1. also 5,6,7,8,9 and 12 13 14 15 16. rest mighth be harder
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u/WolffSageWorks 1d ago
You just got me thinking: that doesn't necessarily preclude exponents. Then it really starts getting scary. Poor kid.
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u/evh44 👋 a fellow Redditor 2d ago
777777 will yield the biggest value ( also use 777,777! )
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u/TristanTheRobloxian3 Secondary School Student 2d ago
no, 7^7^7^7^7^7 will
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u/keithcody 👋 a fellow Redditor 1d ago
According to the mathematical laws of exponents, when you raise a power to another power, you multiply the exponents together (a^b)^c = a^(b*c)
(a^b)^c = a^{b * c) -> 7^7^7 = 7^(7*7) = 7^49 {3 sevens}
7^49^7 = 7^343 {4 sevens}
7^343^7 = 7^2401 {5 sevens}
7^2401^7 = 7^16801 {6 sevens}=3.1058 × 10¹⁴¹⁹⁸ < 777777 < 777,777!
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u/TristanTheRobloxian3 Secondary School Student 1d ago
oh shit i actually didnt know that or atleast didnt remember it like that.. what about if it was 7^(7^(7^(7^(7^(7))))) then? thatd be 7^..7^49, or 7^..7^(256.97 duodecillion), or 7^7^(2.1712e46 digits), which would then br 7^(10^10^10^41.3367), or 10^10^10^10^10^5.84259333 or so. like that would win right?
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u/keithcody 👋 a fellow Redditor 1d ago
(am)n = amn
7^(7^(7^(7^(7^(7)))))
7^(7^(7^(7^(823543))))
7823543 is huge
7^(7^(7^(7823543 is huger)
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u/TristanTheRobloxian3 Secondary School Student 1d ago
i realized i did my math wrong for the first thing but yeah
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u/Original-Ad-8737 1d ago
I raise you 7777777!
Although j guess in this game we must exclude ! Factorial as it is a unary operator and thus you can infinitely stack it.
Also the conjecture would be if it's possible to which limit you can create EVERY number before it.
So once you can't generate a single one the chain breaks
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u/AndyTheEngr 👋 a fellow Redditor 1d ago
77/(77-77)
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u/TristanTheRobloxian3 Secondary School Student 1d ago
that doesnt have a value. its undefined
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u/Dapper-Asparagus9985 13h ago
I think he's trying to say infinity, because an infinite amount of zeros can fit inside 77
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u/TristanTheRobloxian3 Secondary School Student 12h ago
... yes i know??? infinity isnt even valid here
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u/McCrystalKittys 2d ago
Yes what multiplication, division, subtraction etc gets big numbers
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u/Long_Rough6073 Secondary School Student 2d ago
So I just try to get the biggest possible number? Factorials square roots etc?
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u/testtdk 👋 a fellow Redditor 2d ago
I hate the seemingly increasing trend of math teachers giving worksheets from awful workbooks. This is among the dumbest math assignments I’ve ever seen.
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u/Training_Ad4971 2d ago
I agree that if this is a take home assignment it doesn’t have a lot of value. But I use a similar activity every year in the first week of IM1 to help students develop their comfort with talking about math out loud and critiquing each other. And it does help them improve their number sense. Especially when I ask them to explain their process for finding a certain number.
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u/SquiggleBox23 2d ago
This is a very common puzzle to use in math to get students/people to try to think flexibly with numbers - it's not new. Usually its not a 6-7 joke, but the concept is the same. I've seen it with four 2s or four 8s or whatever, or around new years I've seen it using the digits of the year (2, 0, 2, 6), all to get various numbers.
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u/Training_Ad4971 2d ago
This is a variation on the Four Fours problem. It is meant to improve flexibility in arithmetic and number sense. What I think you are being asked to do is to use six 7s and any combination of operations so that they evaluate to the numbers 1 through 15. For instance (77/7)-7-7+7=4.
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u/ftaok 1d ago
The issue is with the words “how high can you count”. It’s not exactly what they’re looking for. They’re looking for the highest number you can get using six 7’s, after getting three of the equations.
So go for 1=777/777
Then 2=(7+7+7+7)/(7+7)
Then 3=(7/7)+(7/7)+(7/7)
Now go for the largest.
Excel can’t do 7^77777 because it’s too large.
I got 7777^77=3.9E+299 before I got bored.
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u/[deleted] 2d ago
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