Question: "Bart, Lisa, and Maggie have a bag with three balls in it. Each ball has a positive integer on it. Each of the three children randomly picks two balls from the bag, announces the product of the two numbers, and then replaces the balls in the bag. The numbers they announce are 36, 42, and 90. Unfortunately, Bart is not very good at arithmetic, and the number he announced was 2 different than the correct product, but the two girls correctly computed their products. What are the possible values of the numbers on the balls? (Note that we are not told which of the three numbers was announced by Bart.)"


"We assign variables to the three different balls; $a$, $b$, and $c$, in which $a>b>c$. Since $90$ is the greatest number, it is $ab$: Since $42$ is the second greatest number, it is $ac$: Since $36$ is the smallest number, it is $bc$. Since $90$ and $42$ both have a common factor, a, we can find the prime factorization and take the GCF:

$90 = 2*3^2*5$

$36 = 2^2 * 3^2$

$a$ must be $6$. We can find $b$ by considering $90$ and $36$: $b$ must be $18$; but since it must be less than $a$, this does not work. Is the problem messed up? We must remember that Bart is not very good at arithmetic; $42$ must be the number he messed up. We can consider $2$ choices: $40$, or $44$. $44$ does not work because $44$ is , and it is ac. $36$ is bc, and $11$ does not divide into $36$. Also, if $11$ were $a$, it would not divide into $90$ anyway. So we now know that the $3$ numbers are $90$, $40$, and $36$.

$40 = 2^3*5$

$a$ must be $10$. Also, we can find out that $b$ is $9$, because $ab$ is $90$, and $a$ is $10$. We can also apply this to $36$. $b$ is $9$, so $9c=36$, and $c=4$. So we have $a=10$, $b=9$, and $c=4$."

I don't really understand this solution. I'm okay with saying that the $a$ is 6 because of the GCF and that $b$ is 18 because of the GCF, but how does this solution imply that the problem number is $42$?

  • $\begingroup$ I don't even understand the previous steps. For example, why is $a = \gcd(ab,ac)$? In general we only know that $a$ divides $\gcd(ab,ac)$. $\endgroup$ – Trevor Wilson Sep 29 '13 at 18:26
  • $\begingroup$ Also it seems to me that the part of the solution before "is the problem messed up?" is useless: we are given that one of the products is in fact "messed up," so why do any calculations on the assumption that the products are correct? $\endgroup$ – Trevor Wilson Sep 29 '13 at 18:29
  • $\begingroup$ @TrevorWilson: we are given that two of the products are correct. We need to make use of that. I agree that the proposed solution fails because it assumes the products are correct. $\endgroup$ – Ross Millikan Sep 30 '13 at 3:31

The argument makes a few unjustified leaps and is in general very poorly presented, but a correct argument along similar lines is possible.

Let the numbers on the balls be $a,b$, and $c$, where $a\ge b\ge c$, so that the three correct products are $ab,ac$, and $bc$, with $ab\ge ac\ge bc$. If $a=b$, then $ac=bc$, and if $b=c$, then $ab=ac$; in either case two of the correct products are equal. If they belonged to the girls, we’d see two equal products, and if one belonged to Bart, we’d see two products that differed by $2$; in fact we see neither of these, so it must be that $a>b>c$. Thus, $90$ is $ab$, $ab-2$, or $ab+2$; $42$ is $ac$, $ac-2$, or $ac+2$; and $36$ is $bc$, $bc-2$, or $bc+2$.

Notice that among the reported products of $90,42$, and $36$ only $42$ has $7$ as a factor. This is impossible if $42$ really is $ac$: $7$ is prime, so it would have to be a factor of either $a$ or $c$, and then it would be a factor of one of the other two products as well. Thus, $42$ must be a mistake: Bart’s product is really either $40$ or $44$, and the products $90=ab$ and $36=bc$ are correct.

Since $b>c$, the possibilities for $bc=36$ are $36\cdot 1$, $18\cdot 2$, $12\cdot 3$, and $9\cdot 4$. However, $b$ must be a factor of $90=ab$, so $b$ must be $18$ or $9$. If $b=18$, then $a=\frac{90}{18}=5<b$, which is impossible, so $b=9$, $c=4$, and $a=\frac{90}9=10$.


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