Question 1210504
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If 'a' and 'b' are positive {{{highlight(cross(2-digest))}}} <U>2-digit</U> integers, 
how many solutions are there of the equation 2a+3b=100.
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&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;The post by @MathLover1 isn't a solution to the problem because it doesn't answer the problem's question.

&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;The question "how many" remains unanswered, or she shifts the calculation to the reader.

&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;Therefore, her post is a talk about a solution, but not the solution itself.

&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;I came to deliver the solution as it should be.



<pre>
In  equation

    2a + 3b = 100,


the term '2a' is an even number and right side '100' is an even number, too.

Hence, the term '3b' must be even number.  It implies that number 'b' must be even number.

Then '3b' is a multiple of 6.


Thus, '3b' should be multiple of 6 and 'b' itself should be a two-digit positive integer number.


The smallest such number '3b' is 30.


So, the numbers '3b'  form the set < 30, 36, 42, . . . >.


We want to determine the maximum possible value of '3b'.


It should allow 'a' in equation  '2a = 100 - 3b'  to be a 2-digit positive integer number.

So, the greatest possible value of '3b' is not more than 80.


Thus, our set for '3b' is  the sequense  < 30, 36, 42, . . . , 78 >.


It contains  {{{(78-30)/6+1}}} = 9 numbers.


So, equation  2a + 3b = 100  has 9 solutions in positive integer numbers (a,b),
such that 'a' and 'b' are 2-digit positive integer numbers.


<U>ANSWER</U>.  There are 9 solutions under imposed conditions.
</pre>

Solved.