Maths test question on number of ways to pay

Question

My daughter got given another maths question which is confusing me...

Q. John and Alice have bought the same calculator, but have paid for it in different ways. The price is between $80$ and $90$ Euros and to pay for it they have used the same number of $1$€ coins as notes. (Possible notes being €$50$, €$20$, €$10$ and €$5$). How much was the calculator and how did each of them pay for it.

I tried to work this out by making an equation.

$m$ = number of coins

$A$ = number of $5$€ notes

$B$ = number of $10$€ notes

$C$ = number of $20$€ notes

$D$ = number of $50$€ notes

So $$ 50D + 20C + 10B + 5A + m = X, \quad 80 \le X\le 90 $$ and since $$ m = A + B + C + D $$ so $$ 51D + 21C + 11B + 6A = X, \quad 80\le X\le 90 $$ Each variable could be zero or non zero.

If $A$ were non zero, then the answer would need to be divisible by $6$, so that just leaves the answers $84$ and $90$.

If $B$ were non zero, then the answer would need to be divisible by $11$, so that just leaves the answer $88$.

If $C$ were non zero, then the answer would need to be divisible by $21$, so that just leaves the answer $84$.

If $D$ were non zero, then the answer would need to be divisible by $51$, which is impossible...so $D$ must be zero.

At this point I thought I was getting somewhere, but I did a quick test with $D =1$ and found that actually there are two ways to pay using a $50$€ note:

1. $50 + 20 + (3 \cdot 5) + (5 \cdot 1) = 90$

2. $50 + (3\cdot 10) + 5 + (5 \cdot 1) = 90$

In both cases, the number of coins = number of notes.

So these tests seemed to contradict my mathematical result and now I am just completely confused. Can somebody please help me?

Thanks :-)

Adrian

Answer 1

When you say, "If A were non-zero then the answer would need to be divisible by $6$," this is not true. You can conclude only that $X-51D-21C-10D$ is divisible by $6$. The same goes for your next 3 assertions.

Otherwise, you're on the right track. You also know that $D=0$ or $1$, because the calculator costs less than $90$ euros. Similarly $C<5$ and $B<9$ and $A<18.$ So there are not many possibilities to check.

EDIT: The most number of notes happens when $D=B=C=0$ so $80\leq 6A \leq 90$. (By the way, it's not clear whether these inequalities are strict. I read the problem so that this would be $80 < 6A < 90.$) So this forces $m\leq 15$ (or $14,$ if you use my interpretation of "between".) This means $A+B+C+D\leq 15.$

Then there are lots of cases that get eliminated as you go. $D=0$ or $1$. If $D=1$, then $C$ can be only $0, 1 or 2$ and $B$ must be less than $4$. Etc. It might be do-able in 15 minutes.

Answer 2

You are wrong in the assumption that if ( will simplify the actual equation a bit for simplicity) $$ X+Y=Z $$ and $Z$ is divisible by $5$ for example, you cannot conclude that $X$ is divisible by $5$ neighter that $Y$ is divisible by $5$, the only thing you can conclude that the sum $X+Y$ is divisible by $5$.