Q: How do you get to the number 5 using these numbers.3 4 5 10?

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4/.4 + 4/4 = 10 + 1 = 11.

10 raised to power 4

(3x4+5+6+7)/10=3

10,000 * * * * * WRONG! That is the number of permutations, NOT the number of combinations. The number of combinations denoted by nCr = n!/[r!*(n-r)!] = 10!/[4!*6!] = 10*9*8*7/(4*3*2*1) = 210

can do it without using the 3, is that allowed? 4 x 6 x 10 = 240 + 6 + 8 = 254

Related questions

1 - 1/4get common denominators4/4 - 1/4subtract the top numbers3/4

Many ways, but one example is: 4+4+(4/4)+(4/4)= 10.

The closest you can get by: Using Each Number Once With Powers: 10 th the second power - 4 th the third power=46 Using each Number Once Without Powers:(4*10)-3+2 Using an indefinite amount of each number:4-*10+(3*2)-(2+2+2)

4/.4 + 4/4 = 10 + 1 = 11.

3 - 4 + 5 * 6 + 7 > 10

10 raised to power 4

(3x4+5+6+7)/10=3

9+1-4(2) 10-8=2

(9 + 1)^(4/2) = 10^2 = 100.

6*(9 + 5 - 10) = 6*(14 - 10) = 6*4 = 24

10,000 * * * * * WRONG! That is the number of permutations, NOT the number of combinations. The number of combinations denoted by nCr = n!/[r!*(n-r)!] = 10!/[4!*6!] = 10*9*8*7/(4*3*2*1) = 210

can do it without using the 3, is that allowed? 4 x 6 x 10 = 240 + 6 + 8 = 254