help!~math!?

Question

Math question?
You are equipped with two 2's, three 3's, and the ability to combine them using addition, subtraction, Multiplication, division, and exponentiation

Your job is to create all of the integers from 37 to 67 (inclusive)

You may use any number of parentheses to control the order of operations and you dont have to use all five numbers each time.

Help, is there any trick to this or just plugging in cuz im having a har time

Answer 1

This is what I've got so far

37 = (3 * 2)^2 + (3 / 3)
38 = ((3 * 3) - 3)^2 + 2
39 = ((3 * 3) + 2 + 2) * 3
40 = ((3 + 3) * 3 + 2) * 2
41 = (3 + 3)^2 + 2 + 3
42 = (3 * 2)^2 + 3 + 3
43 = (2^3 * 2) + 3^3
44 = (3 + 3)^2 + 2^3
45 = (3 * 2)^2 + (3 * 3)
46 = ((3 * 3) - 2)^2 - 3
47 = ((3^3 - 2) * 2) - 3
48 = (2^3 + 2^3) * 3
49 = ((3 + 3 + 3) - 2)^2
50 = ((3^3 - 3) * 2) + 2
51 = (2^3 * 3 * 2) + 3
52 = ((3 * 3) - 2)^2 + 3
53 = ((3^3 - 2) * 2) + 3
54 = (3 - 2) * (3^3 * 2)
55 = (2^(3 + 3)) - 3^2
56 = (3 * 3 * 3 * 2) + 2
57 =
58 = ((3^3 + 3) * 2) - 2
59 = (2^(3 + 3)) - 3 - 2
60 = ((3 * 3 * 2) + 2) * 3
61 = (3 + 3 + 2)^2 - 3
62 = ((3^3 + 3) * 2) + 2
63 = (2^(3 + 3)) - 3 + 2
64 = 2^((3 + 3) * (3 - 2))
65 = (2^(3 + 3)) + 3 - 2
66 = ((3 * 3) + 2) * 2 * 3
67 = (3 + 3 + 2)^2 + 3

57 is giving me a hard time

I guess the trick was that I would come up with a formula that gave me a number, then I would just change the sign and make other numbers. At the end, I had to work out the ones that were left out. For example:
(2^(3 + 3)) - 3 - 2 = 59

So I changed the last part to + 2 and it gave me 63
(2^(3 + 3)) - 3 + 2 = 63

Then I changed the last part to + 3 and it gave me 65
(2^(3 + 3)) + 3 - 2 = 65

And so on

I hope this helped

Kia

PS. If I figure out 57, I will update this

Answer 2

As you have 2's and 3's let's use 2+3 as a divisor and find how many of them go into each number and you can use again 2's and 3's to fill the reminder.

ex:

for 37 : 37 ÷ 5 = 7 R 2 , so you need 7 2's and 7 3's and another 2(which is a reminder).

for 38: 38 ÷ 5 = 7 R 3, so you need 7 2's and 7 3's and another 3(which is a reminder).


Like this you can solve all the way upto 67.
If the reminder is 1 (we use 3-2)
if The reminder is 2 (we use 2)
if the reminder is 3 (we use 3)
if the reminder is 4 ( two 2's)
if the reminder is 0 - you are done.