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
(55-5)/(5+5)
101
It's done by: 5+5/5-5/5 = 5
ccsndf
it is 1 7 tens 1 fives
(55-5)/(5+5)
55----------------------------------------------------------------------------------------------From Rafaelrz.You can make, 5! = 120, five digit numbers using 1,2,3,4 and 6.
You add up all the numbers except one of the fives and then take that total and subtract it by the five you didn't use
(5/5)5 = 1
101
(5 cubed - 5 - 5 + 5) / 5= 24
55/5 = 11 - (5+5) = 1
It's done by: 5+5/5-5/5 = 5
ccsndf
120
it is 1 7 tens 1 fives
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.