Every number besides 0 can be safely divided into 3, 5, and 7, although most of the resultant quotients will have a non-zero remainder.
If you require all three quotients to have a remainder of zero, then the divisor must be the reciprocal of any non-zero integer.
If you require the divisor itself to be a non-zero integer, then there are only two numbers that can be divided into 3, 5, and 7 without a remainder, 1 and -1.
Any element of the set of numbers of the form 210*k, where k is an integer, is evenly divisible.
35 and its multiples.
630 is one possible answer.
Any multiple of 210.
315
Any multiple of 140.
975
35 and all of its integral multiples
The LCM of (5, 6, 7) is 210.
4, 5 5, 6 6, 7 7, 8 8, 9 9, 10
7 and 5 are prime numbers, but 3*2=6
no need for 3 as it already goes in to 6 so 5 6 7 go in to 210 210 = 5 x 42 = 6 x 35 = 7 x 30
Go decimal: 5/6 = 0.83, 6/7 = 0.86 so 5/6 is smaller. Alternatively, 5/6 = 35/42, 6/7 = 36/42 5/6<6/7 83.313333<85.11428
51
5 6 -8 7 6 -5 5 5 5 -5 6 -6 6 5 6 -8 7 6 -5 6 6 6 -6 -7 7 7 -8 6 6 -7 -6 6 5 6 7 -6 7 -8 7 -7 6 5 6 -8 7 6 -5 6 6 6 -6 -7 7
5+6+7 , 5+7+6 , 6+5+7, 6+7+5 , 7+5+6, and 7+6+5
It does not go in equally. 7 times 5 is 35 7 times 6 is 42
41 ÷ 7 = 5 remainder 6
1. As 7 is a prime number.
210
go 5-6=-1go -1-7=-8 (-1+-7)=-8aswer is -8
It will certainly go ito 7! = 1*2*3*4*5*6*7 = 5040, although that is not the smallest positive integer that they will go into.