The problem is as follows:

A bank manager calculates the greatest common divisor of the numbers that represent the amounts of money in usd that Jenny and Gwen have, using Euclid's algorithm, the successive quotients are $3$; $1$ and $2$ respectively. If the first division was made by excess and also the least common multiple of the quantities they have is 315, how many dollars represents the greatest savings among the two girls?

The alternatives given in my book are as follows:

$\begin{array}{ll} 1.&\textrm{105 usd}\\ 2.&\textrm{98 usd}\\ 3.&\textrm{112 usd}\\ 4.&\textrm{77 usd}\\ \end{array}$

The official solution according to my precalculus workbook is as follows:

(Comment: The author arranged the information in this sort of weird table which I don't understand)

$\begin{array}{|c|c|c|c|} \hline & \textrm{3(excess)} & 1 & 2 \\ \hline a=7d & b=3d & 2d & d\\ \hline & 2d & 2d& 0\\ \hline \end{array}$

Assume the numbers representing the savings for Jenny and Gwen are $a$ and $b$.

From the given information in the problem:

Use the least common multiple:



Then $d=15$

Therefore $a=7\cdot 15 =105$

Therefore the greatest saving it is $105$ usd.

The answer is option 1.

This is the part where it ends the official solution.

Now my problem is I'm stuck here at the very beginning. Does it exist a way to reverse engineering what it was intended to be done here by this method?

From what I do recall the Euclid's algorithm of the division it goes as this:


In an inexact division the $d$ represents the divisor and the $q$ the quotient and the $r$ represents the remainder.

I'm assuming that the second row in the table seems to point the first term of the algorithm. But from then on I don't get it.

A division by excess implies that the remainder is like as follows:


But I don't know if this was used here?. The part where it mentions the least common multiple seems to imply primes because it places as $(7d)(3d)=315$

But the prime factorization of $315$ is equal to $3^2 \cdot 5 \cdot 7$. But from these there can be different combinations such as:

$3\cdot 7$ or $3 \cdot 5$

Since it implies that both must be the same I mean $d$ there isn't sufficient $5$'s to get $15$ on both sides of the multiplication from the prime factorization of $315$.

All and all, can someone help me here?. I'm just lost. Perhaps does it exist a better argument or method of solution for this problem?. I don't know how is it established that the greatest of both savings is $105$. As it is mentioned the sort of answer which it would help me the most is a step-by-step explanation of the solution.

  • $\begingroup$ Just like your prior word problem from this book, the exercise and solution are horrendous. You shouldn't have to "reverse engineer" what is intended in the exercise from an even more cryptic "solution". Again, I strongly recommend that you find a higher quality textbook. $\endgroup$ Feb 13, 2021 at 8:41
  • $\begingroup$ Are you translating this from another language? $\endgroup$ Feb 13, 2021 at 8:48

1 Answer 1


Note that the calculations which agree with the answer to this question are $$105=3\times45-30$$ $$45=30+15$$ $$30=2\times 15$$

We can work this out as follows, where $a$ is the larger of the two amounts $a,b$. $$a=3\times b-R$$ $$b=R+S$$ $$R=2\times S$$

Solving in terms of $S$ gives $$R=2S,b=3S,a=7S.$$ The LCM of $7S$ and $3S$ is $315$ and so $$21S=315.$$

Therefore $S=15$ and $a=105$.

  • $\begingroup$ Where do you see "the following calculations in the book"? It seems you are making a guess at what is intended. $\endgroup$ Feb 13, 2021 at 8:38
  • $\begingroup$ Those are your guesses at what calculations are intended. They don't occur in the cryptic solution. $\endgroup$ Feb 13, 2021 at 8:44
  • $\begingroup$ Your first sentence is ambiguous and seems to imply that you are presenting the calculations given in the book. I suggest you reword it to avoid that ambiguity (there is already too much ambiguity in this book's problems). Something like "my interpretation of the calculations hinted in the solution are as follows..." or somesuch. $\endgroup$ Feb 13, 2021 at 8:55

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