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Eunice sold several cakes. If each cake sold for either exactly 17 or 14 Jan 2024, 10:02
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KarishmaB
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Hi, VeritasKarishma, is there any method wherein you know when to stop testing answers? I got (2,6) as a correct pair giving 17*6+19*2 as 140; but how do I know with certainty that there can be no other pair to satisfy the conditions?

Yes, only one solution needs to be found using hit and trial. After that, you can figure out all the rest.
Check here: https://anaprep.com/algebra-integer-sol ... variables/
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So, if I got it right what's in the article, I should proceed like this:

- guess for a solution: in this case i would solve the equation 17x + 19y = 140 as (x,y)=(6,2) (not so much a lucky guess if you start with x, time consuming).
- then only possible other solutions would be to increase or decrease x by 19 and same with y by 17. Both would take me out of range, so no need to check for other solutions.

Is this right?
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Eunice sold several cakes. If each cake sold for either exactly 17 or 14 Jan 2024, 23:03
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Freddie92
KarishmaB
Kritisood
Hi, VeritasKarishma, is there any method wherein you know when to stop testing answers? I got (2,6) as a correct pair giving 17*6+19*2 as 140; but how do I know with certainty that there can be no other pair to satisfy the conditions?

Yes, only one solution needs to be found using hit and trial. After that, you can figure out all the rest.
Check here: https://anaprep.com/algebra-integer-sol ... variables/


So, if I got it right what's in the article, I should proceed like this:

- guess for a solution: in this case i would solve the equation 17x + 19y = 140 as (x,y)=(6,2) (not so much a lucky guess if you start with x, time consuming).
- then only possible other solutions would be to increase or decrease x by 19 and same with y by 17. Both would take me out of range, so no need to check for other solutions.

Is this right?
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Yes, absolutely. Keep in mind that when x is being increased, y should be decreased and vice versa (since the total sum should remain a constant).
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Eunice sold several cakes. If each cake sold for either exactly 17 or 24 Mar 2025, 02:50
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How do you know from this reasoning that b =2 and a = 6 is the only possible answer?
manishtank1988
Hello Friends,
My 2 cents...

In such scenarios i cheat: i use S-1 info in S-2 and try to solve - if i am able to solve viola we have to answer otherwise NOT

For example, here - we know that a + b = 8 and we need to find b therefore a = 8 -b in other equation:
a = 17 cents
b = 19 cents
17a+19b = 140
17*(8-b) + 19b = 140
136 - 17b + 19b = 140
2b = 4
b = 2 | we have our answer!!!

PLEASE NOTE: i ONLY do it in such scenarios as usually we SHOULD NOT carry data from S-1 to S-2.
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Eunice sold several cakes. If each cake sold for either exactly 17 or 08 Jul 2025, 00:07
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I'm trying to find a more efficient general manner of approaching solving linear equations and I wanted to check if this makes sense: When we have a linear equation e.g., 17x+19y=140 and we need to determine the solution or determine the number of valid combinations present, you can follow the strategy

1. Use statement 1 to find the solution quickly,
2. Then plug the values in and determine what the max values of x and y can be (17 * 8<140 and 19*7<140)
3. Now determine if we can adjust the solution say the solution we got from statement 1 is x = 6 and y = 2
a. Notice that if we reduce x: 6->5 and y: 2->3 then the amount reduces by 17 and increases by 19 (a difference of 2 which is excess)
b. So we need an amount by which the decrease in 17 is negated by an increase in 19
c. That would be the LCM of 17 and 19 which is greater than 140, therefore there is only one solution

If you cannot determine the solution from statement 1's assistance, then you can right out the values of 19 till 140 and 17 till 140 and see what combination adds to 140.
- This can be made more efficient by determining the units' digits that add to 10
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Eunice sold several cakes. If each cake sold for either exactly 17 or 08 Jul 2025, 02:47
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Hi Bunuel KarishmaB

I was thinking of the following theory when solving this question, please tell me if it makes sense or was it a lucky fluke that helped me:

Because the LCM of 19 and 17 is greater than 140, we can't sell fewer of either cakes and try to make up for the exact revenue lost by selling more of the other (given that max revenue is 140). And hence, if x and y can only be positive integers, 17x + 19y = 140 will have only one set of values for x and y. And hence, without further calculations we know statement 2 will suffice.
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Eunice sold several cakes. If each cake sold for either exactly 17 or 16 Oct 2025, 07:26
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I thought that since there are two variables, I would need 2 equations. Hence chose C. Can someone help me understand how I can get a hint that one equation would be enough? I understand trial and error but how do I know I have to do that and not use the 2 equations like we generally do? Please help! Suffering big time in DI now.
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Eunice sold several cakes. If each cake sold for either exactly 17 or 16 Oct 2025, 08:22
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Robo_123
I thought that since there are two variables, I would need 2 equations. Hence chose C. Can someone help me understand how I can get a hint that one equation would be enough? I understand trial and error but how do I know I have to do that and not use the 2 equations like we generally do? Please help! Suffering big time in DI now.
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Generally, such linear equations (ax + by = c) have infinitely many solutions. However, since x and y represent the number of cakes, they must be non-negative integers, making the equation a Diophantine equation (one whose solutions must be integers only). For such equations, it is necessary to check manually, through trial and error, whether there is only one combination that works.
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Eunice sold several cakes. If each cake sold for either exactly 17 or 26 Dec 2025, 23:34
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Bunuel, for Stmnt 2 a Diophantine Equation - can we determine that the no of solutions is either 0 or 1 by reducing equation to Ax + By = C, C/A x B?
The equation being 17x + 19y = 140
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