Q: How do you get 24 using the numbers 2 4 8and 9 once?

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(5+7)*(5-3)=12*2=24

24

24

(4+5-6)x8=24

24 of them.

24

It is: 24

4*3*2 = 24 of them.

(7 + 5 - 9) x 8 = 24

24

5! / 4 - 2 * 3

24 different numbers.

3= 21/7 21/7 + 3/7 =24/7(7)= 24

The LCM is: 24

If you want 4-digit numbers, there are 24 of them.

Assuming each of the given digits can be used only once, the answer is 24. If not, the answer is infinity.

(6 x 13) - (6 x 9) = 78 - 54 = 24

For the values 24, 9, 8 the LCM is: 72

To get 24 using 1 3 4 and 6 once you do 6 times 4

Assuming it's required that all the numbers must be used once and only once: (2 + 8 - 9) x 4! = 4! = 24 Factorials are always handy! I imagine other solutions will follow.

(7+6-5) x 3 = 8 x 3 = 24

7455 equal 24

The least common multiple of 8 , 3 = 24

add 2+5+6+6+2+4 which equals 24.but you get the 2 and 4 from the 24 that's in the problem.

It could be that if: 3+6+7+x = 24 The: x = 8

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