Math help needed
Bookworm on Sep 18 2008 at 8:04 am | Filed under: Uncategorized
My daughter got a very challenging assignment. She did wonderfully with about 80% of it, and then hit a wall. I got intrigued and started helping her out, but I too have now hit a wall. Perhaps you can help, because this is making me very frustrated.
Here’s the deal: She was given the numbers 1, 2, 4 and 6. Using addition, subtraction, multiplication and division, and using each of those numbers only once, she has to create mathematical statements that result in the numbers 1 through 50.
Here are some examples, all of which she figured out:
1 = 6+1-2+4
3 = 2+6-4-1
8 = [(6x1)+4]-2
39 = 62+4-1
44 = [(6x2)-1]x4
As you can see, not only is she using each of the assigned numbers only once, she can also use exponents. Incidentally, you can create exponents, such as 5, by making the exponent 4+1.
Does that make sense? For those of you familiar with the old math game of Krypto, it’s like Krypto, only using four working digits, not five.
So, my daughter made fantastic progress, and I was able to help out with a few more. We got completely stumped, though, trying to create formulas that would equal 34, 38 and 43. Are those do-able or impossible? I suspected some of you might know.
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26 Responses to “Math help needed”
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38 = 2 ^ ( 4+1) + 6
*crickets*
If you can use them as numerals and not just as distinct numbers then:
43 = 42 + 1
34 = 46 – 12
I’m going to stop reading your blog if there’s going to math tests.
34 = sqrt(6^4)-2
Aaaaaaargh!(4^X2@>78)
That’s how I felt, Z, after spending an hour during the damn things and figuring out only two. In a way, it would have been easier if my daughter hadn’t done any, since I would have been more methodical in approaching the whole project than she was. As it was, I simply got confused.
That’s how I felt, Z, after spending an hour during the damn things and figuring out only two.
But those were the two she couldn’t get, right. That means they were the hard ones Book!!
Id like to know the answers when your daughter finds out!
I decided to try them all and got the answers for all but 34, 43, and 50.
I’d like to know, for example of “sqrt” is allowed.
If it is, then 34 is 6^2 – sqrt(4)
And 50 is (6-1)^2 * sqrt(4)
I still can’t see anything for 43.
What was your answer for 50?
My daughter took her page with her, so I don’t know the answer for 50. 43 seemed impossible, but I’m impressed with 34.
34 = (5^2) + 3 + 6
Never mind……
ALWAYS read the instructions twice!
Of course you should be able to use square roots as fractional exponents, but not as operations…come on, one or the other but not both. I still don’t see a solution to 34 or 43.
Book,
Did you ever get the solutions for 43 and 34?
I’m still working on it every so often and I just realized I had 34 wrong.
Spiff
I just realized I had most if not all wrong. Oh well.
Spiff
Hi Scott,
I brought it up solely because we weren’t saying:
“6^squared”… We were saying 6^2, and using up the “2″ in the list of 1,2,4,6 that was allowed.
So technically, sqrt(4) wouldn’t be allowed, but 4^(1/2) would be allowed.
For 34, you do have: (5+1)^2 – sqrt(4) by our current rules.
ps – I’d still like to see a solution for 50 that doesn’t involve sqrt()…
I believe Book and her daughter found one, and for the life of me I just can’t see it.
By the way, I apologize for what is likely far too many posts recently. I’m medically away from work for awhile and it has been far too absorbing, stimulating, enjoyable, and just plain fun to be involved in all these discussions. So very many great posts by Book, and informative and interesting discussions and debates. For the first time I’ve delved into the dialogues between Y and Sgt Dave – when normally I’ve been too busy to, since they are rather deep and involved. It’s all been worth it! Thx.
I did not get answers. My daughter hasn’t brought the assignment back home yet. I should ask, so thank you for reminding me.
Well you might find this a little gratifying Book. I wrote a cheesy little program to “brute force” the answers (just try every combination of +, -, *, /, ^ on a random sequence of “1, 2, 4, 6). And here are the results:
1 = (((1 ^ 2 [1])^ 4 [1])^ 6 [1])
2 = (((2 + 6 [8])/ 4 [2])/ 1 [2])
3 = (((4 – 1 [3])^ 2 [9])- 6 [3])
4 = (((1 * 4 [4])- 6 [-2])^ 2 [4])
5 = (((4 + 6 [10])/ 1 [10])/ 2 [5])
6 = (((6 / 2 [3])+ 4 [7])- 1 [6])
7 = (((4 + 6 [10])- 1 [9])- 2 [7])
8 = (((6 – 2 [4])/ 1 [4])+ 4 [8])
9 = (((1 + 6 [7])+ 4 [11])- 2 [9])
10 = (((4 ^ 2 [16])- 6 [10])/ 1 [10])
11 = (((2 – 1 [1])+ 6 [7])+ 4 [11])
12 = (((6 + 4 [10])+ 2 [12])* 1 [12])
13 = (((1 + 2 [3])+ 6 [9])+ 4 [13])
14 = (((1 + 6 [7])* 4 [28])/ 2 [14])
15 = (((6 * 2 [12])- 1 [11])+ 4 [15])
16 = (((6 * 1 [6])- 2 [4])* 4 [16])
17 = (((2 * 6 [12])+ 4 [16])+ 1 [17])
18 = (((6 + 4 [10])- 1 [9])* 2 [18])
19 = (((6 + 4 [10])* 2 [20])- 1 [19])
20 = (((6 + 4 [10])* 2 [20])* 1 [20])
21 = (((4 + 6 [10])* 2 [20])+ 1 [21])
22 = (((4 ^ 2 [16])+ 6 [22])* 1 [22])
23 = (((4 ^ 2 [16])+ 1 [17])+ 6 [23])
24 = (((1 ^ 2 [1])* 6 [6])* 4 [24])
25 = (((1 ^ 6 [1])+ 4 [5])^ 2 [25])
26 = (((6 + 1 [7])* 4 [28])- 2 [26])
27 = (((6 * 4 [24])+ 1 [25])+ 2 [27])
28 = (((6 – 1 [5])+ 2 [7])* 4 [28])
29 = (((6 – 1 [5])^ 2 [25])+ 4 [29])
30 = (((4 + 2 [6])- 1 [5])* 6 [30])
31 = (((6 ^ 2 [36])- 4 [32])- 1 [31])
32 = (((6 ^ 2 [36])- 4 [32])^ 1 [32])
33 = (((6 + 2 [8])* 4 [32])+ 1 [33])
35 = (((2 + 4 [6])* 6 [36])- 1 [35])
36 = (((2 + 6 [8])+ 1 [9])* 4 [36])
37 = (((4 + 2 [6])* 6 [36])+ 1 [37])
39 = (((6 ^ 2 [36])- 1 [35])+ 4 [39])
40 = (((6 ^ 2 [36])/ 1 [36])+ 4 [40])
41 = (((6 ^ 2 [36])+ 1 [37])+ 4 [41])
42 = (((2 + 1 [3])+ 4 [7])* 6 [42])
44 = (((6 * 2 [12])- 1 [11])* 4 [44])
45 = (((6 + 1 [7])^ 2 [49])- 4 [45])
46 = (((4 * 6 [24])- 1 [23])* 2 [46])
47 = (((6 * 4 [24])* 2 [48])- 1 [47])
48 = (((4 * 1 [4])* 6 [24])* 2 [48])
49 = (((2 * 4 [8])* 6 [48])+ 1 [49])
50 = (((4 * 6 [24])+ 1 [25])* 2 [50])
HA HA HA HA! 34, 38 and 43 don’t show up!! Doesn’t mean that they don’t exist…it just means that they are more “non-trivial”. Mine program was not clever enough to alter grouping as per Jon’s most excellent answer for 38 above.
The numbers in “[]” are just “the result so far” so that I could confirm that I wasn’t making an error.
NOTE: Obviously there is more than one way to get a lot of the numbers.
That was a fairly fun hour long diversion.
Arg! Can’t get those last 2.
I’ve tried these patterns…can anyone see a pattern that I’ve missed?
01, 02, 03 represent operators. The “__” are the slots for the 1, 2, 4, 6
I. (( __ O1 __ ) O2 __ ) O3 __
II. ( __ O1 __ ) O2 ( __ O3 __ )
III. ( __ O1 (__ O2 __ )) O3 __
IV. __ O1 (__ O2 ( __ O3 __ ))
V. __ O1 (( __ O2 __ ) O3 __ )
PS: Jon your clever one showed up in pattern II.
38 = (2^(1+4))+6
One neat thing…all 5 of the patterns will get you at least 46 of the required results. Here is an example that uses more of the patterns. Tried tossing a random sign change in front of each of the arguments. No joy. Ok…I really should probably go do something else.
1 = (1+6) – (4+2)
2 = (2+6) / (-4*-1)
3 = (2-1) – (-6+4)
4 = (1^4) + (-6/-2)
5 = (-6-4) / (2*-1)
6 = (-2+(4-1))*6
7 = (-1^6) + (4+2)
8 = (2*4) * (1^-6)
9 = (2+6) + (1^-4)
10 = (-6+-4) / (1+-2)
11 = (6+-1) – (-2-4)
12 = (4–2) – (-6*1)
13 = (-1–6) + (-2*-4)
14 = (-4*(-1*2))–6
15 = (1–4) * (-6/-2)
16 = (-2–6) * (-4*-1)
17 = (1+4) + (6*2)
18 = (4-(2+-1))*6
19 = (2*(6+4))+-1
20 = (-2–6) * (1–4)
21 = (-1+-2) – (-4*6)
22 = (-2^4) – (6*-1)
23 = (-2+(4*6))+1
24 = (-4*-6) / (2-1)
25 = (1^-2) – (-6*4)
26 = (-6*-4) + (1*2)
27 = (1-(4*-6))–2
28 = (6+(-1–2))*4
29 = 4+((-6+1)^2)
30 = (-1-2) * (-4-6)
31 = -4+(-1+(6^2))
32 = (-1*4) * (-2-6)
33 = 1-(-4*(2–6))
35 = (-6*(-2-4))+-1
36 = (-6*-1) ^ (-4/-2)
37 = (6*(2–4))+1
38 = 6-(-2^(4–1))
39 = -1-(-4-(-6^2))
40 = (6^2) – (-4*1)
41 = 1+(4+(-6^2))
42 = (-1-(2*-4))*6
44 = -4*(1+(2*-6))
45 = -4+((-1-6)^2)
46 = (1-(-4*-6))*-2
47 = (-6*(4*-2))-1
48 = (-4*6) * (1*-2)
49 = (6*(-2*-4))+1
50 = -2*(-1+(6*-4))
I am very, very impressed. And, given the negative numbers involved, I’m not at all surprised that we weren’t able to get all of them. I’m only impressed that we got as many as we did!
Alright…Book you have to close out the comments on this thread before I get in trouble.
The negatives were just a flailing attempt to find the last elusive pair.
Here is yet another randomly generated cleaned up list without all of the useless negatives.
1 = (4+1) – (6-2)
2 = (1^6) / (2/4)
3 = (6*1) / (4/2)
4 = (6+2) – (4*1)
5 = (6+4) / (1*2)
6 = (4+2) * (1^6)
7 = (6-1) – (2-4)
8 = (4+6) – (2^1)
9 = (4^2) – (1+6)
10 = (2-4) * (1-6)
11 = (6-1) + (2+4)
12 = (4-2) * (6^1)
13 = (6+4) + (2+1)
14 = (1*6) + (2*4)
15 = (4*2) + (6+1)
16 = (2^6) / (4/1)
17 = (4+1) + (6*2)
18 = (4-(2-1))*6
19 = (2*(6+4))-1
20 = (4+6) * (2^1)
21 = (4^2) + (6-1)
22 = (4*6) – (2^1)
23 = (4*6) – (1^2)
24 = (4*6) ^ (2-1)
25 = (6*4) – (1-2)
26 = (2^1) + (6*4)
27 = (2+1) + (6*4)
28 = (6*(1+4))-2
29 = 4+((1-6)^2)
30 = (6+4) * (1+2)
31 = (4*(6+2))-1
32 = (2+6) / (1/4)
33 = (4*(6+2))+1
35 = (6^2) – (1^4)
36 = (4-1) * (6*2)
37 = (6^(4/2))+1
38 = (2^(4+1))+6
39 = (4+(6^2))-1
40 = (1+4) * (6+2)
41 = (1+(6^2))+4
42 = (4+(1+2))*6
44 = 4*((2*6)-1)
45 = (((6 + 1)^ 2)- 4)
46 = 2*((4*6)-1)
47 = (2*(4*6))-1
48 = (6*1) * (2*4)
49 = (6+(1^4))^2
50 = 2*(1+(4*6))
Good god, an *easy* answer for 50!
There I was, thinking, how can I get 1,4, and 6 to get me 25, and I couldn’t see it?
Some days it doesn’t pay to even crawl out of bed. (Some week in this case!)
Cycling through the operators and groupings via a program was inspired, jlibson!
You guys are getting scary! You really need to try out Krypto. My sister used to be something of a Krypto champion back in the days, although it’s a skill you lose if you don’t practice it for 30+ years.
@jlibson,
WTF? You are some kind of computer genius or somthing. I’ve been doing this with pen, paper and a calculator for nearly a week.
Good work dude!
Spiff
Book!
The answers for #43 and #34 please!!
For pities’ sake!!!!