I found a bunch of coins: Quarters (25 cents), Dimes(10 cents) and Nickels(5 cents). The combined amount was a whole two digit number of dollars: that is, no fractional amount. I divided the coins into 3 groups: Nickels, Dimes and Quarters. One randomly selected group went to John, the second one went to Raj and the third one went to Mina. Each one got all the coins from individually separated groups. So John either got all the Nickels or all the Dimes or all the Quarters and so on.

Then I asked them: “Can you tell me a specific math related thing about your coins?”

John: Let us say the number of coins I have is X. If I multiply the individual digits in that number and then I multiply the product by 6, I get X.

Raj: Well, I have Y number of coins. If I multiply the individual digits in that number and then I multiply the product by 3, I get Y.

Mina: I have Z number of coins:If I multiply the individual digits in that number and then I multiply the product by 2, I get Z.

So what type of coin group did Raj get? How many of those coins? What was the total $ number I had?

Can be linked to How many coins did Mrs. Jones have?


2 Answers 2


Here is a solution

Suppose that each of $X,Y,Z$ is a two-digit number (they are obviously not one-digit numbers).

Let $X = 10a + b$.
Then $6ab = 10a+b \Rightarrow b = \frac{10a}{6a-1}$
and since $a$ is coprime with $6a-1$, it must be that $6a-1$ divides $10$.
Hence $6a-1 = 1,2,5$ or $10$ and the only one that works is $6a-1=5$ which means that $a=1$ and $b=2$ so that $X=12$.
A similar line of reasoning leads us to $Z=36$ while $Y$ may be $15$ or $24$.

Now, since we have a whole number of dollars overall, it must be that the number of quarters and the number of nickels are both odd or both even. This means that $15$ would not be included in our current framework and we will only consider $Y=24$.
This means that each of $X,Y,Z$ is divisible by $4$ and the quarters will make up a whole number of dollars on their own. So, for a solution to exist, the dimes and nickels must also make up a whole number of dollars and there are two possibilities here: the number of dimes is $12$ and the number of nickels is $36$ or the number of dimes is $24$ and the number of nickels is $12$.

In the first case, the total number of dollars is only $9$ while in the second case, it is $12$ so this is a possible solution.

In this case Raj gets $24$ dimes.
The total number of coins is $36+24+12 = 72$ and the dollar amount is $12$.

What I have not considered

Whether $X,Y,Z$ can have three or four digits, yielding another possible solution. The maximum of the three is less than $2000$ so it would suffice to check up to this value but I think such occurrences are rare - I still have to check if there are any more possibilities.

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    – DrD
    Commented Jan 11, 2022 at 16:10

Assume first that $x,y,z$ are all positive; this means that none of them can have a $0$, since they are divisible by each of their digits and the corresponding divisors, and $x,z$ can have no $5$ for that would imply being divisible by $10$ and hence containing a $0$. $x,y,z<2000$ since $2000$ nickels is $100$ dollars, and none of them can be single-digit for obvious reasons.

We first work out what $z$ can be. If it has two digits $ab$ then $10a+b=2ab$, but since $z$ is even we have $b=2c$ and $5a+c=2ac$ or $a=\frac c{2c-5}$. Trying the possibilities for $c$, which range from $1$ to $4$ inclusive, shows that $z$ may only be $36$.

If $z$ has three digits $abc$ then $100a+10b+c=2abc$; we may apply the $c=2d$ reduction and rearrange to get $a=\frac{5b+d}{2bd-50}$. Clearly $d\ne1,2$ since that forces the denominator to be negative; $d=3$ forces $b=9$ and $d=4$ forces $b=8$ (note that $b,d$ must have the same parity). In no case does $a$ evaluate to a digit. If $z$ has four digits the first one is a $1$, and pretty much the same procedure shows there are no solutions since the numerator is at least $500$ and the denominator at most $22$. Hence $z=36$.

Exactly the same trial-and-error approach may be used on $x$, which shows that it must be $12$, and on $y$ which shows it must be $15$ or $24$. Now check all six permutations of coins and both possibilities for $y$ to see whether the sum of all coins is an integer number of dollars at least ten – only $x$ nickels, $y$ dimes and $z$ quarters works, with $12$ dollars in all, so Raj has $24$ dimes.


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