1, 3, 5, 15, 37, 111, 185, 555.
60
15 x 1/3 = 5
5x.6.Five 5's to get 35 - 5/5 - 5/5 = 5 - 1 - 1 = 3
9+9+9+9=36-5-5-5-5=16+3+3+3-3=22+1-1-1/1=21
The factors of 15 are 1, 3, 5, 15 The factors of 37 are 1, 37
1 + 2 + 3 + 4 + 5 + 6 + 7 + 2 + 3 + 4 = 37 note: there are many other variations/answers
37 ÷ 5 = 7 remainder 2 → You can make 7 teams, but have 2 children left out. However, if you meant you have a pool of 37 children and need to make a team consisting of 5 children from that 37, then how many different teams (or groups) of 5 children you can make then: This is a combination as the order of selection doesn't matter, only which 5 children are chosen. Here you can make: 37 C 5 = 37!/((37-5)!5!) = (37 × 36 × 35 × 34 × 33)/(5 × 4 × 3 × 2 × 1) = 435,897 different teams
The Greatest Common Factor (GCF) of 5, 7, and 3 is 1.
37/5 : 6*37/6 = 37/5 : 37 = 1/5 : 1 = 1 : 5
5 multiplied by 7 plus 3 minus 1
3 + 2/3 + 5 + 7/12 = 37/4 it is 37 fourths 9 and 1/4
The factors of 555 are 1, 3, 5, 15, 37, 111, 185, and 555. The prime factors of 555 are 3, 5, and 37 The prime factorization of 555 is 3 x 5 x 37.
4 multiplied by 1 3/7 is 5 5/7.
There are infinitely many. Start with 1 + 1 + 1 + 1 + ... 37 times then, 2 + 1 + 1 + 1 + ... with 36 1s and 3 + 1 + 1 + 1 + ... etc then you can use fractions, and irrational numbers. Next, try, 37 * 1 or 37 * 1 * 1 or 37 * 1 * 1 * 1 and go on for ever. or 37*2*.5 and 37 * 2 * 2 * .5 *.5 etc. And then there are equations involving fractions, subtractions, powers and trigonometric functions - eg 37*sin(pi/2) or 74*sin(pi/6) I hope that, by now, you have got the point that the question needs tightening up if you want a more meaningful answer.
Remember BODMAS/PODMAS (or variants). 3 + 7 * 5 - 1 = 3 + 35 - 1 = 38 - 1 = 37
The factors of 555 are: 1 3 5 15 37 111 185 555 The prime factors are: 3 x 5 x 37