Arithmetic Problem

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Gamer30

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Feb 23, 2022
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Hi all,
I wish to raise this problem so hoping all Maths expertise can enlighten me.

The following is the question:
Assuming I have some sweets.
1) If I divide them into group of 6's, there r excess 2 sweets remaining.
2) If I divide them into group of 7's, I have short of 5 sweets.

So, let X be the total number of sweets and Y be the number of sweets in each group, I get:

Eq 1) X = 6Y + 2
Eq 2) X = 7Y - 5

Therefore, I get Y = 7 and X = 44 after solving the equations.

In addition, 3)If I divide the total number of sweets into group of 8's, there won't be any sweets remaining.

Hence, may I know how to solve the above problem? (There are 3 conditions to be fulfilled)

Qn is: Is the above problem solvable? If so, how do I go about solving it?
(Calling all Maths enthusiast to kindly show me how).
Your help will be greatly appreciated and thank you for your attention.
 
Gamer30 said:
Eq 1) X = 6Y + 2
Eq 2) X = 7Y - 5
Click to expand...
Those should be:

X = 6*N1 + 2​
​
X = 7*N2 - 5​
​
X = 8*N3​

With the added constraint that

N1 & N2 & N3 are positive integers​

By using a spreadsheet (brute force) we can find the solution (Qn).
 
Gamer30 said:
1) If I divide them into group of 6's, there r excess 2 sweets remaining.
2) If I divide them into group of 7's, I have short of 5 sweets.
Click to expand...
This is unclear! Do you mean that if you divide them into groups of 7, then there will be G2 groups and 2 left over?
Gamer30 said:
So, let X be the total number of sweets and Y be the number of sweets in each group, I get:

Eq 1) X = 6Y + 2
Eq 2) X = 7Y - 5
Click to expand...
"Y" can't be the number of sweets in each group (you are already using those numbers of sweets in a group (6 & 7) in your equations!

"Y" would need to be the number of groups in each condition, which is why I used G2 (above) for the number of groups in condition 2) and this should not be the same under each condition!

Your equations (by using "Y" in both) assume that each condition produces the same number of groups but that is not a fair assumption!

X = 6G1 + 2
X = 7G2 + 2 (if you did mean there was 2 left over when dividing into groups of 7)

might be a better formulation of equations but you then have 3 unknowns but only 2 equations! (So no algebraic solution is possible.)

As it happens, G1 = 7, G2 = 6 and X = 44, does fit those equations (6 × 7 = 42, plus 2 = 44 and 7 × 6 = 42, plus 2 = 44) but as soon as you introduce your third condition, that solution is no longer valid: 8 × 5 = 40, leaving a remainder of 4 so groups of 8 (with no remainders) doesn't fit your 44 sweets solution!
Gamer30 said:
Therefore, I get Y = 7 and X = 44 after solving the equations.

In addition, 3)If I divide the total number of sweets into group of 8's, there won't be any sweets remaining.

Hence, may I know how to solve the above problem? (There are 3 conditions to be fulfilled)
Click to expand...
You would need to introduce a third equation, eg: X = 8G3 but you would then have 4 unknowns and only 3 equations so it still couldn't be solved algebraically!
Gamer30 said:
Qn is: Is the above problem solvable? If so, how do I go about solving it?
(Calling all Maths enthusiast to kindly show me how).
Your help will be greatly appreciated and thank you for your attention.
Click to expand...
It might be possible to solve it by trial & error, approximation or doubling (as opposed to halving) but you need to completely re-think your logic.

Please come back and tell us what you have tried in light of this advice.

Hope that helps.
 
khansaheb said:
Those should be:

X = 6*N1 + 2​
​
X = 7*N2 - 5​
​
X = 8*N3​

With the added constraint that

N1 & N2 & N3 are positive integers​

By using a spreadsheet (brute force) we can find the solution (Qn).
Click to expand...
I don't think "X = 7*N2 - 5" is right, it seems to me it should, using your formulation, be: "X = 7*N2 + 2". 🤔
(See my post above.)
 
khansaheb said:
It is correct because 2+5 = 7
Click to expand...
The very first thing I pointed out in my first post is that his condition: "2) If I divide them into group of 7's, I have short of 5 sweets." was unclear! The only sensible way to interpret it was that he could make a whole number of groups of 7 sweets but would have 2 left over (so his final group would be 5 short of 7), therefore the correct equation to use would be: X = 7*N2 + 2 not X = 7*N2 - 5! 🤔

Hope that helps. 😊
 
The Highlander said:
he could make a whole number of groups of 7 sweets but would have 2 left over (so his final group would be 5 short of 7), therefore the correct equation to use would be: X = 7*N2 + 2 not X = 7*N2 - 5
Click to expand...

128 = 7 * 18 + 2 = 7 * 19 - 5

The original problem states
Gamer30 said:
2) If I divide them into group of 7's, I have short of 5 sweets
Click to expand...
When something is short - in general it refers to lopping off (subtract)
 
Last edited by a moderator:
Gamer30 said:
The following is the question:
1) If I divide them into group of 6's, there r excess 2 sweets remaining.
2) If I divide them into group of 7's, I have short of 5 sweets
3) If I divide the total number of sweets into group of 8's, there won't be any sweets remaining.
Click to expand...
What is the Question? Where is it stated?
 
Gamer30 said:
The following is the question:
Assuming I have some sweets.
1) If I divide them into group of 6's, there r excess 2 sweets remaining.
2) If I divide them into group of 7's, I have short of 5 sweets.

Qn is: Is the above problem solvable? If so, how do I go about solving it?
(Calling all Maths enthusiast to kindly show me how).
Your help will be greatly appreciated and thank you for your attention.
Click to expand...
This is a standard kind of problem, for which the standard question at the end, which you haven't stated, is,

What is the smallest possible number of sweets I might have?​

In this form, it is solvable.

A way to find the solution fairly easily, within the realm of (elementary) arithmetic, is to consider what will happen if they give me two sweets to hold temporarily, before dividing them into groups. In particular, this sort of problem is often assigned after teaching about the LCM.

There is a slight language issue in the problem; at least in my dialect, we wouldn't say "I have short of 5 sweets", but something like "I am short 5 sweets" (But then, we also wouldn't say "there r excess 2 sweets remaining", but "there are 2 extra sweets". I imagine the 7k-5 is implied rather than 7k+2 to make the trick a little harder to see; both, of course, are equivalent, with different k.
 
What is really needed is for @Gamer30 to come back and say what the three conditions are exactly in precise unequivocal language. It doesn't have to be good English it just has to be Mathematically clear and not open to interpretation.
 
The Highlander said:
The very first thing I pointed out in my first post is that his condition: "2) If I divide them into group of 7's, I have short of 5 sweets." was unclear! The only sensible way to interpret it was that he could make a whole number of groups of 7 sweets but would have 2 left over (so his final group would be 5 short of 7), therefore the correct equation to use would be: X = 7*N2 + 2 not X = 7*N2 - 5! 🤔

Hope that helps. 😊
Click to expand...
Pardon me for the tricky language used.
To make the question easier, in simple form, maybe I would say for condition 1 and 2, the number of sweets in EACH GROUP must be the same.
For instance, eg. If I say in each group, there r 7 sweets, therefore, u would notice that the number of sweets in each group for condition 1 and 2 MUST be the SAME thru'out.
Hence, if I were to solve the Eq1 and 2, I would get the total number of sweets to be 44 right?
So, using 44 as the total number of sweets, for condition 2, if I were to divide the sweets by 7, in each group, there would be 7, resulting in LACK of 5 sweets. (So, maybe I shouldn't use the word SHORT of. Sorry for the wrong interpretation)
In that case, if the condition 3 (divide by 8) were to come into picture, would I still be able to solve it?
Kindly express your answer and views.
Thanks.
 
Assuming your work is correct:
You found a solution for X when there are just two constraints. Let's say for a moment that you found several solutions for the case with two constraints.
Now you add a third constraint. The solution to this new problem must obviously satisfy the first 2 constraints (as well as the third). So in the end, if there is solution, it must be one or more of the solutions you got just using the first 2 constraints.

Now in your solution you found just one value for X that satisfies the first 2 constraints. Since this value doesn't satisfy the 3rd constraint it follows that there is no solution when you have all three constraints.
 
khansaheb said:
That's a very tricky statement. I almost misunderstood it at the first reading (of response 12)!!

Yes .... What have you tried?
Click to expand...
Hi, I have tried out and found that 128 is an answer that fits into these 3 conditions but the number of sweets in each group is different already. (If 128 / 6 = 21 R2, 128 / 7 = 18 R2, lack of 5, 128 / 8 = 16 R0)
From the above, number of sweets in each group is 21, 18, 16 respectively. (Correct me if I'm wrong)
Thus, in this case, the number of solution might be more since the constraints would be more lenient. (No need that number of sweets in each group MUST be the SAME).

Hi, Steven G, is my deduction (number of solution is more in the above case) correct?
Do kindly input your valuable comment.
Thank you.

Note: The above R implies remainder.
 
khansaheb said:
X = 6*N1 + 2X = 7*N2 - 5X = 8*N3
With the added constraint that

N1 & N2 & N3 are positive integers
Click to expand...
That is neither a stated nor an implied constraint.

Your equations are:

X = 6 * Y1 + 2

X = 7 * Y2 - 5 ..............and

X = 8 * Y1 .................... with implied constraint

Y1, Y2 & Y3 are integers

The smallest value of 'X' satisfying above equations and constraint is:

X = 128

This is the smallest value of X, but not unique.

The next X to satisfy the equations and the implied constraint would be 296
 
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