71 and 182
At least B and 1. There may be others depending on the value of B.
41
If 'A' is 'divisible by 'B', it means that: -- If 'B' is repeatedly subtracted from 'A', then the last one fits exactly, without anything left over -- 'A' was made by piling up some whole number of 'B's, with no extras -- 'B' goes into 'A' evenly --'A'/'B'= some whole number
d, v, s, b, l
Straightforward maximization problema + b = 36, so b = 36 - a c = ab = a(36 - a) = 36a - a2Find the derivative of c with respect to a: dc/da = 36 - 2a.Set it to zero and solve for a: 36 - 2a = 0.36 = 2a18 = aa = 18b = 18
To be divisible by 36 the number must be divisible by 9 (and 4). To be divisible by 9, the sum of the digits must be divisible by 9. 5 + 5 = 10 → need an extra 8 → A + B = 8.
That a = ±b
(1) x/12=a+9, where a is an integer (2) x/18=b+9, where b is an integer 3x(1) + 2x(2) gives 3x/36 + 2x/36 = 5x/36 = 3a+27+2b+18 = 3a+2b+45 5x/36=3a+2b+45 x=36(3a+2b+45)/5 x=36(3a+2b)/5+9 as 36 is not divisible by 5 then 3a+2b should be divisible by 5 then the smallest solution is a=0 and b=0 and x=9
312 is divisible by 2 b/c the 2 at the end of 312 is an even number312 is divisible by 3 b/c 3 + 1 + 2 = 6 which is divisible by 3312 is divisible by 4 b/c 12 is divisible by 4312 is not divisible by 5 b/c the one's place isn't 0 or 5312 is divisible by 6 b/c 312 is divisible by 2 and 3312 is not divisible by 7 b/c 2*7=14, 31-14=17 which is not divisible by 7312 is divisible by 8 b/c 312/8=39 (Explanation: A way to find out how 8 is divisible its that its first three digits [the ones, tens, and hundreds place] have to be divisible by 8)312 is not divisible by 9 b/c 3 + 1 + 2 = 6 which is not divisible by 9312 is not divisible by 10 b/c the ones place is not 0
112
1/x+a+b =1/x+1/a+1/b ---- ---- ---- ----
Remark: For any two numbers a and b, b never divides a exactly if b is greater than a.So, 2 is not divisible by 7623.
At least B and 1. There may be others depending on the value of B.
It means that 'b' can go evenly into 'a'
GCF of a and b is a if b is divisible by a.87 is divisible by 29 so, GCF(29,87) = 29.
Let A, B, C are digits. Number ABC is divisible by eleven if and only if A+C-B=0 or A+C-B=11
If a is divisible by b, then LCM of a and b is a.Here 12 is divisible by 2. So, LCM of 2 and 12 is 12.