+9479 \(0 \ >\ \dfrac{1}{1-\frac{10}{x}}\ >\ 1 - \dfrac{5}{y}\)
So...
\(\dfrac{1}{1-\frac{10}{x}}\ >\ 1 - \dfrac{5}{y}\\~\\~\\ \dfrac{x}{x-10}\ >\ 1 - \dfrac{5}{y}\\~\\~\\ \dfrac{x}{x-10}-1\ >\ - \dfrac{5}{y}\\~\\~\\ \dfrac{x}{x-10}-\dfrac{x-10}{x-10}\ >\ - \dfrac{5}{y}\\~\\~\\ \dfrac{10}{x-10}\ >\ -\dfrac5y\)
And we know 0 < x < 10 because that is the only way \(\dfrac{1}{1-\frac{10}{x}}\) can be less than 0.
So x - 10 is negative, and so when we multiply both sides by (x - 10) , flip the sign.
\(10\ <\ -\dfrac5y(x-10)\)
And we know 0 < y < 5 because that is the only way \(1-\dfrac5y\) can be less than 0.
So y is positive, and so when we multiply both sides by y , don't flip the sign.
\(10y\ <\ -5(x-10)\\~\\ y\ <\ -\frac12(x-10)\\~\\ y\ <\ -\frac12x+5\)
And so we have these three inequalities:
\(1.\qquad0\ {<}\ x\ {<}\ 10\\~\\ 2.\qquad0\ {<}\ y\ {<}\ 5\\~\\ 3.\qquad y\ {<}\ -\frac12x+5\)
We can see on a graph that the intersection of these is a triangle:
https://www.desmos.com/calculator/2xloyva03v
All the points that lie within the shaded triangle are solutions to the inequality.
And we can see there are 16 pairs of integers (x, y) that satisfy the inequality.
+9479 1. Each of the digits 1,2,3,4,5,6 is used exactly once in forming the 3-digit integers X and Y. How many possible values of X+Y are there if |X-Y|=111?
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| I'm not as sure of this one but here's my guess: | |
| We can let Y < X without changing the number of possible values of X + Y |
|
| For every value of Y such that |X - Y| = 111 there is one unique corresponding value of X | |
| So if we just count the possible values of Y, that will be the number of possible values of X + Y |
|
| And since Y < X , the possible digits of Y are only: 1, 2, 3, 4, 5 | |
| (If 6 was a digit of Y then there is not a possible corresponding X which makes |X - Y| = 111) |
|
| the number of possibilities of Y = 5 * 4 * 3 = 60 | |
| So there are 60 different pairs each with a different sum. |
|
| And so there are 60 different possibilities of X + Y |
To help see this...here is the beginning of the list of pairs:
123, 234
124, 235
125, 236
132, 243
134, 245
135, 246
142, 253
143, 254
.
.
.
+9479 2. The product of all digits of positive integer M is 105 How many such M's are there with distinct digits?
I tried 6 for the problem above but it wasn't right... Any thoughts on why?
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I think the answer is not 6 because 1 can also be a digit of M.
105 = 3 * 5 * 7
So 5, 3, and 7 must be digits of M, while 1 might be a digit of M.
So the possiblites are:
357
375
537
573
735
753
In addition to:
| 1357 1375 1537 1573 1735 1753 | ____ | 3157 3175 3517 3571 3715 3751 | ____ | 5137 5173 5317 5371 5713 5731 | ____ | 7135 7153 7315 7351 7513 7531 |
the total number of possibilities = 3! + 4! = 6 + 6*4 = 6 + 24 = 30
