Question 1210504
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<font color=red>Answer:  9</font>



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Reasoning


It sounds like "digest" should be "digit".


I'm assuming the question is:
"If A and B are positive two-digit integers, how many solutions are there to the equation 2a+3b = 100?"


Let's look at the related equation 2a+3b = 1, since we can multiply both sides by 100 later.


Through fairly quick trial-and-error, you should find an integer solution is (a,b) = (-1,1)
Another method is to look at the graph of 2x+3y = 1 to find a lattice point, or you can use the Extended Euclidean Algorithm.


Since (-1,1) is a solution we can state that
2a+3b = 1
2*(-1)+3*(1) = 1



Multiply both sides by 100 to get,
2*(-1)+3*(1) = 1
100*(  2*(-1)+3*(1)  ) = 100*1
100*2*(-1) + 100*3*(1) = 100
2*( 100*(-1) ) + 3*(100*1) = 100
2*( -100 ) + 3*(100) = 100
This demonstrates that (a,b) = (-100,100) is an integer solution to 2a+3b = 100.
Unfortunately a = -100 is not positive, and it's not a two-digit number either. 
A similar problem shows up with b as well.
We must require that 10 <= a <= 99 and 10 <= b <= 99.
The symbol <= means "less than or equal to".



2x+3y = 100 solves to y = (-2/3)x+(100/3)
The slope -2/3 will tell us how to go from one integer point to another on this line.
Start at (-100,100). 
Move down 2, right 3 to arrive at (-97,98)
Move down 2, right 3 to arrive at (-94,96)
etc


This "down 2, right 3" pattern can be expressed like this
a = -100+3t
b = 100-2t
where t is an integer.
If t = 0, then it leads to (a,b) = (-100,100)
If t = 1, then it leads to (a,b) = (-97,98)
If t = 2, then it leads to (a,b) = (-94,96)
and so on.
If t was negative, then you go the opposite direction "up 2, left 3".


Then,
10 <= a <= 99
10 <= -100+3t <= 99
10+100 <= -100+3t+100 <= 99+100
110 <= 3t <= 199
110/3 <= 3t/3 <= 199/3
36.67 <= t <= 66.33 approximately
Since t is an integer we round the lower bound up to the nearest integer; while the upper bound gets pulled down to the nearest integer.
Note how t = 36 is too small since it leads to a = 8; while t = 67 is too large because it leads to a = 101.


Anyways, we go from 
36.67 <= t <= 66.33
to
37 <= t <= 66


Meanwhile,
10 <= b <= 99
10 <= 100-2t <= 99
10-100 <= 100-2t-100 <= 99-100
-90 <= -2t <= -1
-90/(-2) >= -2t/(-2) >= -1/(-2) .... inequality signs flip
45 >= t >= 0.5
0.5 <= t <= 45
1 <= t <= 45 .... rounding to nearest integer
The inequality signs flipped because we divided all sides by a negative number.



Overlap these two intervals
37 <= t <= 66
1 <= t <= 45
To get the interval 37 <= t <= 45
I recommend drawing out a number line.


This set of t values will make both 10 <= a <= 99 and 10 <= b <= 99 true simultaneously where a = -100+3t and b = 100-2t.


How many integers are in the set {37,38,...,45}?
We want the 37 to go to 1, the 38 to go to 2, etc.
Subtract 36 from each item and you'll get the set {1,2,...,<font color=red>9</font>}


Or you could say 45-37+1 = <font color=red>9</font>
The formula I used is that there are n-m+1 items in the set {m,m+1,m+2,...,n-1,n} where m,n are integers and m < n.
Note the -37+1 portion combines to -36, which is what we subtracted from each item of the previous set.


We determined there are <font color=red>9 integer solutions</font> to the equation 2a+3b = 100 where a,b are positive two-digit integers.
Using a spreadsheet, or something like Python, the set of all solutions we're after is this
a = 11,  b = 26
a = 14,  b = 24
a = 17,  b = 22
a = 20,  b = 20
a = 23,  b = 18
a = 26,  b = 16
a = 29,  b = 14
a = 32,  b = 12
a = 35,  b = 10
Each time 'a' goes up by 3, b drops by 2.
It seems like your teacher only cares about the number of solutions, rather than the solutions themselves. So this portion is likely optional.
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