Q: Can you get 50 from the digits 3 5 7 9?

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Without repeated digits, the answer is 7*6*5*4*3 = 2520

2, 3, 3, 3 and 7.

4*5*5 = 100 if digits can be repeated. 4*4*3 = 48 if not.

3+1+4+1+5+9+2+6+5+3+5+8+9+7+9+3+2+3+8+4+6+2+6+4+3+3+8+3+2+7+9+5+0+2+8+8+4+1+9+7+1+6+9+3+9+9+3+7+5+1 = 247

87531

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There are only 4 prime digits: 2, 3, 5 and 7.There are only 4 prime digits: 2, 3, 5 and 7.There are only 4 prime digits: 2, 3, 5 and 7.There are only 4 prime digits: 2, 3, 5 and 7.

Without repeated digits, the answer is 7*6*5*4*3 = 2520

2, 3, 3, 3 and 7.

(75 + 50 + 3 - 5) × 7 + 1 = 862

are the last TWO digits of 5347. 5, 3, 4, and 7 are all digits of the number 5347.

4*5*5 = 100 if digits can be repeated. 4*4*3 = 48 if not.

120 There are 6 digits in total. The numbers with 3 digits, with all digits distinct from each other, are the permutations of the 6 digits taken 3 at a time, and therefore there are 6*5*4 = 120 of them.

3+1+4+1+5+9+2+6+5+3+5+8+9+7+9+3+2+3+8+4+6+2+6+4+3+3+8+3+2+7+9+5+0+2+8+8+4+1+9+7+1+6+9+3+9+9+3+7+5+1 = 247

If you mean: -7*(5+3) --6 then it is -56+6 = -50

50 = 5*5*2 105 = 5*7*3 so LCM = 5*5*2*7*3 = 1050 1050 = 21*50 and 10 * 105 The LCM is 1050.

With repeating digits, there are 33 = 27 possible combinations.Without repeating any digits, there are 6 combinations:357375537573735753

87531