0: (5x5)-(5x5)
1: (5:5)+5-5
2: (5:5)+(5:5)
3: (5+5+5):5
4: [(5x5)-5]:5
5: [(5-5)x5]+5
6: [(5x5)+5]:5
7: [(5+5):5]+5
10: 5+5+5-5
11: (5:5)+5+5
15: (5x5)-5-5
20: 5+5+5+5
24: (5x5)-(5:5)
25: (5x5)+5-5
26: (5x5)+(5:5)
30: [(5:5)+5]x5
35: (5x5)+5+5
45: [(5+5)x5]-5
50: (5x5)+(5x5)
55: [(5+5)x5]+5
75: (5+5+5)x5
120: (5x5x5)-5
130: (5x5x5)+5
625: 5x5x5x5
how do i make 17 using only 2 4 6 8
No. Not if the numbers are to be used only once. There is only one 5 (or a multiple of 5) in the numbers available. Using times and divide cannot produce any more of them. On the other hand 100 is divisible by 5*5 so at least two fives are required. If the numbers can be used more than once then 2*2*5*5 is one possible solution.
Google "Four Fours" and you will find a number of solutions, presented more comprehensively than we can.
total 6 they are 3412,4312,1324,3124,1432,4132
Four five nine
101
how do i make 17 using only 2 4 6 8
No. Not if the numbers are to be used only once. There is only one 5 (or a multiple of 5) in the numbers available. Using times and divide cannot produce any more of them. On the other hand 100 is divisible by 5*5 so at least two fives are required. If the numbers can be used more than once then 2*2*5*5 is one possible solution.
1
Have a pair of dice where one has only sixes and one only fives, for example.Have a pair of dice where one has only sixes and one only fives, for example.Have a pair of dice where one has only sixes and one only fives, for example.Have a pair of dice where one has only sixes and one only fives, for example.
You can't unless something is missing in your question.
4*6+1
5 divided by 5, plus 5
All of them, but it does depend on what operators are allowed.
Only four numbers:20233032
Google "Four Fours" and you will find a number of solutions, presented more comprehensively than we can.
145.992 using only four significant figures is 146.0