3 * 3 / ( 3 * 3 ) = 1 but that uses only four 3s, so
33 / ( 3 * 3 * 3 ) = 1 uses five 3s
51/3 is 5.3333333333333333333333 (the 3s go on forever.)
5 3's go into 16 with a remainder of 1.
1+2+34+5 = 3 + 81 + 5 = 84 + 5 = 89 ■
You do5!/(5+5+5)=8!=factorialfactorial e.g 5!=5*4*3*2*1 or 3!=3*2*1
1 + 23 + 4 + 5 +67 = 100
5 of them with a remainder of 1
(5/5)5 = 1
First, observe that 5 = 3 + 2. That uses 1 of the 3s. How do we get 2 from what's left? Add 3 and 3 to get 6, then divide that by the remaining 3 to get a 2. Done!
no, you can't. you need 5 5 5 5 and 1
(5/5-1)+3+1+4.
There would be nine 3s with a remainder of two. 3 goes into 27 nine times, so the remaining two (29-27) are left over.
51/3 is 5.3333333333333333333333 (the 3s go on forever.)
5 3's go into 16 with a remainder of 1.
subtract 1 or add -1, silly
4s-5-3s =s-5 The s and 5 can not subtract because they aren't like terms. Like terms have the same variable.
5x5=25 25-1=24
1+2+34+5 = 3 + 81 + 5 = 84 + 5 = 89 ■