55----------------------------------------------------------------------------------------------From Rafaelrz.You can make, 5! = 120, five digit numbers using 1,2,3,4 and 6.
There are 59C5 = 59*58*57*56*55/(5*4*3*2*1) = 5,006,386 combinations.
11 Proof; 22 / 11 = 2 55 / 11 = 5 If you try the same process with larger numbers (12-22) one of your answers will not be a whole number.
No, 2/12 is only one-fourth as large as 2/3.using 6, the LCD:2/12 = 1/6 (divide both numbers by 2)2/3 = 4/6 (multiply both numbers by 2)or using 12, the existing denominator2/12 = 2/122/3 = 8/12 (multiply both by 4)*if both numerators are the same (2), the larger fraction is the one with the smaller denominator (3, not 12).
No 5291 is not a prime using 2 numbers. It is a prime using three numbers.
2
2 x 2 x 3
Using the equation 55 x 2 = 110, the numbers are called as follows: multiplicand x multiplier = product.
55----------------------------------------------------------------------------------------------From Rafaelrz.You can make, 5! = 120, five digit numbers using 1,2,3,4 and 6.
Well - 12-4 is 8 BUT assuming you have to use all the numbers... Add 12 to 4, then divide by 2 !
(5+7)*(5-3)=12*2=24
91/6 - 33/4 To calculate this we need to... convert both numbers to improper (or 'top-heavy') fractions to find the Lowest Common Multiple (LCM) of the two denominators re-calculate the two numbes using the LCM 91/6 ..... 6 x 9 = 54 ..... 54 + 1 = 55 ..... 91/6 = 55/6 33/4 ..... 3 x 4 = 12 ..... 12 + 3 = 15 .... 33/4 = 15/4 The factors of 6 are 2 and 3, and the factors of 4 are 2 and 2 - so the LCM must be 2 x 2 x 3 = 12 55/6 = 110/12 15/4 = 45/12 110 - 45 = 65 65/12 = 55/12
There are 59C5 = 59*58*57*56*55/(5*4*3*2*1) = 5,006,386 combinations.
You can make 4*3*2/2 = 12 numbers.
There are not 55. Only these: 1, 2, 3, 4, 6, 7, 8, 12, 14, 21, 24, 28, 42, 56, 84, 168.
-11
11 Proof; 22 / 11 = 2 55 / 11 = 5 If you try the same process with larger numbers (12-22) one of your answers will not be a whole number.