Mathematicians; I need your help!
I really need this (its extra credit; 100 points if I get all of it from 1-100) for my Math Grade. I've had finals/exams all week and just totally forgot about this, now I have about 8(?) hours before I need to hand it in.
Help away!
Thanks!
Quote:
Alright, so basically I need help.
I have to find numbers (1-100) using a number four times from 1-9.
Here's some examples.
To find '1' I would do 1 x 1 x 1 x 1
To find '2' I would do 1 x 1 + 1/1 (Fraction)
To find '3' I would do 2 x 2 - 2/2 (Fraction)
To find '4' I would do 1 + 1 + 1 + 1
To find '22' I would do 11 + 11
And so forth. I can use anything from exponents, addition, fraction, subtraction.
If you can help me that would be appreciated, better yet, what is it called?
Thanks a bunch!
http://answers.yahoo.com/question/in...7193020AAcIjEt
Re: Mathematicians; I need your help!
Do you have to use 4 numbers or up to 4 numbers?
"To find '22' I would do 11 + 11"
Not (1-9)
=x
Re: Mathematicians; I need your help!
Quote:
Originally Posted by
Organic
Do you have to use 4 numbers or up to 4 numbers?
"To find '22' I would do 11 + 11"
Not (1-9)
=x
No, but it consist of the numbers from (1-9) and is being used four times.
I have no idea why my teacher made that a rule, and told us we could do that.
Also, I can only use four numbers (the same one).
Re: Mathematicians; I need your help!
Few random easy ones. Most of these have tons of combinations.
Quote:
9= 3x3x3/3
10 =( 4x4+4)/4
11 = 11+1-1
12 = (11+1)/1
13= 11+1+1
16= 4x4x4/4
18 = 22-2-2
19= 44+4-4-4
22 = 22+2-2
25 = 5x5x5/5
26 = 22+2+2
33 = 33+3-3
36 =6x6x6/6
44 = 44+4-4
55 = 55+5-5
66 = 66+6-6
77 = 77+7-7
88 = 88+8-8
99 = 99+9-9
49 = 7x7x7/7
64 = 8x8x8/8
81 = 9x9x9/9
100 = 10x10x10/10
I'll keep adding.
I'm bored of studying, might as well take a break.
Use patterns like the ones I've bolded and just keep building.
Re: Mathematicians; I need your help!
1 = 1^4
2 = 1^2 + 1^2
3 = 1^2 + 1 + 1
4 = 1 + 1 + 1 + 1
5 = 2^2 + 2/2
6 = 2^3 - 2
7 = 3 + 3 + 3/3
8 = 2^4
9 = 4 + 4 + 4/4
10 = 2^3 + 2
...
Do you see a pattern?
The idea is to take a number and produce every number you can.
Take 1. You have
Code:
1^4 = 1
1^3 + 1 = 2
1^2 + 1 + 1 = 3
1 + 1 + 1 + 1 = 4
2:
Code:
2^4 = 16
2^3 + 2 = 10
2^3 - 2 = 6
2^2 + 2 + 2 = 6 (redundant)
2^2 - 2 - 2 = 0 (bad)
2^2 + 2/2 = 5
2^2 - 2/2 = 3 (redundant)
2 + 2 + 2 + 2 = 8
2 + 2 + 2 - 2 = 4 (redundant)
2 + 2 + 2/2 = 5 (redundant)
2(2+2+2) = 12
Be clever.
Re: Mathematicians; I need your help!
Quote:
Originally Posted by
jMerliN
1 = 1^4
2 = 1^2 + 1^2
3 = 1^2 + 1 + 1
4 = 1 + 1 + 1 + 1
5 = 2^2 + 2/2
6 = 2^3 - 2
7 = 3 + 3 + 3/3
8 = 2^4
9 = 4 + 4 + 4/4
10 = 2^3 + 2
...
Do you see a pattern?
The idea is to take a number and produce every number you can.
Take 1. You have
Code:
1^4 = 1
1^3 + 1 = 2
1^2 + 1 + 1 = 3
1 + 1 + 1 + 1 = 4
2:
Code:
2^4 = 16
2^3 + 2 = 10
2^3 - 2 = 6
2^2 + 2 + 2 = 6 (redundant)
2^2 - 2 - 2 = 0 (bad)
2^2 + 2/2 = 5
2^2 - 2/2 = 3 (redundant)
2 + 2 + 2 + 2 = 8
2 + 2 + 2 - 2 = 4 (redundant)
2 + 2 + 2/2 = 5 (redundant)
2(2+2+2) = 12
Be clever.
I'd thank you if I could, 30 more problems left!
---------- Post added at 05:29 AM ---------- Previous post was at 04:13 AM ----------
Alright, after checking and making sure they're correct, these are the problems I have left.
Would be very appreciative if someone could help me figure out those. Anyways, I have Spanish Exams/finals tomorrow, g'night people.
---------- Post added at 05:37 AM ---------- Previous post was at 05:29 AM ----------
Ah great, two of them were wrong. Also need help on 68 and 74.
Re: Mathematicians; I need your help!
Quote:
Originally Posted by
Moogra
42=22*2-2 or 33+3*3
46=22*2+2
62=4^4/4-4
68=4^4/4+4
96=33*3-3
are square roots allowed?
Yes they are, thanks so much Moogra!
Re: Mathematicians; I need your help!
9*9-root(9)-root(9)
81-3-3 = 75
9*9-9-root(9)
81-9-3 = 69
9*9+root(9)+9
81+3+9 = 93
4*4*(root(4)+root(4)) = 64.
Re: Mathematicians; I need your help!
Ah shoot my fault they aren't allowed (just called up a friend) anyways the help you provided is much appreciated. I'll just try tonfigure out the others somehow. Thanks again. G'nite
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Re: Mathematicians; I need your help!
Because sqrt(x) = x^(1/2)
If you're only allowed to use a single digit throughout the expression, 1/2 violates that immediately.
Re: Mathematicians; I need your help!
Re: Mathematicians; I need your help!
If two digits would've been allowed, any 2-digit numbers could've been put together with NUMBER+1-1.
I wonder if he ever got it finished...