I am new to stack exchange.I have a problem.For which I am unable to find an answer on the net.The question goes like this

"The sum of six different multiples of 3 is 66.

All the numbers are natural numbers.

The largest number among them is? "






  • $\begingroup$ Have you made any attempts so far? $\endgroup$
    – TomGrubb
    Dec 19, 2016 at 1:41
  • 1
    $\begingroup$ Hint: compare to "6 different natural numbers add up to 22, then the largest can be $\cdots$". $\endgroup$
    – dxiv
    Dec 19, 2016 at 1:43

3 Answers 3


Well, $3+6+9+12+15+18=3\cdot(1+2+3+4+5+6)=3\cdot21=63.$ What does that tell you, since we want them all to be distinct?


You want the other numbers to be as small as possible, so pick 3,6,9,12,15 and subtract the sum from 66

  • 1
    $\begingroup$ That's a bit too informal in my opinion. Why do we want the other numbers to be as small as possible? If the sum was $99$, your solution would say the answer must be $54$... $\endgroup$
    – TMM
    Dec 19, 2016 at 12:51
  • $\begingroup$ Yeah, you're right, I guess I assumed the fact that no other configuration fits the question was self-explanatory... $\endgroup$
    – dasaphro
    Dec 19, 2016 at 20:55
  • $\begingroup$ Sure, if $3, 6, 9, 12, 15, x$ is not the unique solution, then there is no unique solution to the question, which is (presumably) a contradiction with the fact that this is a constructed exercise of some kind. But it's a small step to actually argue that $3, 6, 9, 12, 15, x$ is indeed the unique solution, without having to rely on the correctness of the question. It's better to show $P$ than to show $Q \implies P$ ;) $\endgroup$
    – TMM
    Dec 19, 2016 at 22:02

Try dividing all of the numbers in this problem by $3$ (save the number of numbers in the sum). You can do this because they're all multiples of $3$, so you identify each number with how many $3$s it has. You can rephrase this problem as:

Six different natural numbers sum to $22$. The largest of them is: $7, 10, 9$, or $8$.

Now you can say that the five smallest numbers are at least $1,2,3,4,$ and $5$. What conclusion can you make now, if the six sum to $22$?


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