It is well known that if the digits of a number add to a multiple of 3 then so does the number (the same property holds for 9). We look for a number such that
2_ _ is a minimum, and divides 9
so we can see the least number possible must be 207
The graph for the real numbers is the number line.
1) decide on a scale - say for example 1 cm = 1 unit
2) draw and label a number line
3) use the magnitude of the number to determine how far from 0 to place a point
4) use the sign of the number to determine whether to plot it left (for negative) or right of 0
Expanding and simplifying...
(a+1)(b+1)(c+1)=
(a+1)(bc+b+c+1)=
abc+ab+ac+bc+a+b+c+1=
ab+bc+ac+a+b+c+2
Since their product is 1, the only possibilities are a,b,c={-1,1} (why?)
Furthermore, there must be an odd number of positives. To find the minimum we can ignore the case a=b=c=1 - that would be the maximum.
Case1
if say a=b=1, c=-1, then
ab+bc+ac+a+b+c+2=
1-1-1+1+1-1+2=
2
Case2
if say c=1, a=b=-1, then
ab+bc+ac+a+b+c+2=
1-1-1-1-1+1+2=
0
Minimum value is 0
If a number can be written as a fraction its decimal expansion will eventually begin to repeat and it is rational. If not it is irrational. Examples: pi is irrational and 0.125=0.12500000...=1/8 is rational.
3^2 < 11 < 4^2
so that means
3 < sqrt(11) < 4
and since 11 is closer to 9, sqrt(11) is closer to 3.
This process can be refined as much as we like to get as close to sqrt(11) as our resources allow, but since sqrt(11) is irrational, we can never locate it's exact position
we try the first digit as 9 because it is a perfect square and the greatest 3-digit numbers have 9 in their thousands place the last one is not prime, so the middle number must be the prime so we choose 7 (greatest prime less than 9) 6 is not a prime, it is our only last digit possible if we require our number to not be a multiple of 3 so our number is 976
which has the following solution So as you can see there is no real pair of numbers that can do both at the same time but if we allow the numbers to be complex, then it is possible
Careful thought will convince you that this task is impossible. The square of an odd number is also odd. This implies that the sum of the squares of any even number of odd numbers must be even. But 85 is odd...this is a contradiction to the the above true statements :)