<PRE><B>Question 1189249</B>
&lt;br&gt;
Let x be the number of pencils in a large bag; let y be the number in a small bag.  Since the total number of pencils is 224, we have&lt;br&gt;
19x+3y=224&lt;br&gt;
where x and y have to be positive integer values.&lt;br&gt;
We can solve the problem informally, using trial and error.  The total of 224 pencils, minus the number of pencils in the 19 large bags, equals the number of pencils in the 3 small bags; that means the number of pencils in the small bags is a multiple of 3.&lt;br&gt;
So try some numbers....&lt;br&gt;
10 pencils in each large bag makes 19*10=190 pencils, leaving 224-190=34 pencils in the small bags.  But 34 is not a multiple of 3.  So try something else.&lt;br&gt;
11 pencils in each large bag makes 19*11=209 pencils, leaving 224-209=15 pencils in the small bags.  15 IS a multiple of 3: 3*5=15.&lt;br&gt;
So we have one POSSIBLE solution: 11 pencils in each large bag and 5 pencils in each small bag.&lt;br&gt;
Technically, we need to verify that this is the only solution.  However, I will leave that part for the formal algebraic solution below.&lt;br&gt;
But for now, we have....&lt;br&gt;
(One possible) ANSWER: The sum of the numbers of pencils in a large bag and in a small bag is 11+5 = 16.&lt;br&gt;
While a problem like this can be solved informally, it is useful to know the formal mathematical method.&lt;br&gt;
This is a Diophantine equation -- a problem in which we have two variables but only one equation; but we can find the solution (or possible a family of discrete solutions) knowing that the values of the variables are non-negative integers.&lt;br&gt;
Here is the standard method for solving a linear Diophantine equation like in this problem.&lt;br&gt;
Our equation is&lt;br&gt;
{{{19x+3y=224}}}&lt;br&gt;
Step 1: Solve the equation for one variable in terms of the other.  You should be able to finish solving the problem by solving for either variable.  I will show the solution solving for y in terms of x.&lt;br&gt;
{{{19x+3y=224}}}&lt;br&gt;
Isolate the y term:&lt;br&gt;
{{{3y = -19x+224}}}&lt;br&gt;
When we divide by 3, we want to express as much of the result as possible in whole numbers.  In this problem, it might look like this:&lt;br&gt;
{{{3y = (-21x+222)+(+2x+2)}}}
{{{y = (-7x+74)+(2x+2)/3}}}&lt;br&gt;
In that equation, we know y is an integer, and x is an integer so (-7x+74) is an integer; and that means {{{(2x+2)/3 = (2(x+1))/3}}} must be an integer.&lt;br&gt;
Since {{{(2(x+1))/3}}} must be an integer, (x+1) must be a multiple of 3, so the possible values of x are 2, 5, 8, 11, ....&lt;br&gt;
We look at those possible values of x and find the one(s) that make y a positive integer:&lt;br&gt;&lt;pre&gt;
   x  y=(-7x+74)+(2(x+1)/3)
  --------------------------
   2   (-14+74)+2 = 60+2 = 62
   5   (-35+74)+4 = 39+4 = 43
   8   (-56+74)+6 = 18+6 = 24
  11   (-77+74)+8 = -3+8 = 5
  14   (-98+74)+10 =  -24+10 = -14&lt;/pre&gt;
We see that, mathematically, there are several ways to get 224 pencils with 19 large bags and 3 small bags:&lt;br&gt;
(1) 2 in each large bag and 62 in each small bag: 19(2)+3(62)=38+186=224
(2) 5 in each large bag and 43 in each small bag: 19(5)+3(43)=95+129=224
(3) 8 in each large bag and 24 in each small bag: 19(8)+3(24)=152+72=224
(4) 11 in each large bag and 5 in each small bag: 19(11)+3(5)=209+15=224&lt;br&gt;
However, in the context of the problem, there should be more pencils in a large bag than in a small bag; that leaves only one solution: 11 pencils in each large bag and 5 in each small bag.&lt;br&gt;
ANSWER: (formally) There are 11 pencils in each large bag and 5 in each small bag&lt;br&gt;

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