-1
$\begingroup$

Let $a_1, a_2, ... , a_8$ be positive integers. It was discovered that sum of any 7 numbers from this list can only yield $56, 58, 61, 63, 64, 65, $ or $66$. What is the largest number on this list?

From a hunch, I think the given isn't enough, since the sum of any 7 numbers from 8 integers would be as many as $\dbinom{8}{7} = 8$ sums, and only $7$ sums given.

I gave the benefit of the doubt and assume that one of the sums is not unique. However, even if I tried the largest sum given, $66$, as the sum that is not unique, then the sum of $a_1, a_2, ... , a_8$ would be $$\dfrac{56+58+61+63+64+65+66+66}{7} = \dfrac{499}{7} = 71.2857143$$ (See EDIT for further clarification)

and applying to the largest 7-number sum to get the smallest integer would be $$71.2857143 - 66 = 5.2857143$$

... which contradicts the word "integer".

However, it actually has an answer, as the answer sheet said that it's $15$.

Did I do wrong on some point, or the problem is utterly bad?

EDIT : To further clarify, I try to expand my work a bit

$$a_1 + a_2 + a_3 + a_4 + a_5 + a_6 + a_7 = 56$$ $$a_1 + a_2 + a_3 + a_4 + a_5 + a_6 + a_8 = 58$$ $$a_1 + a_2 + a_3 + a_4 + a_5 + a_7 + a_8 = 61$$ $$a_1 + a_2 + a_3 + a_4 + a_6 + a_7 + a_8 = 63$$ $$a_1 + a_2 + a_3 + a_5 + a_6 + a_7 + a_8 = 64$$ $$a_1 + a_2 + a_4 + a_5 + a_6 + a_7 + a_8 = 65$$ $$a_1 + a_3 + a_4 + a_5 + a_6 + a_7 + a_8 = 66$$ $$a_2 + a_3 + a_4 + a_5 + a_6 + a_7 + a_8 = 66$$

Adding all of those would be $$7 a_1 + 7 a_2 + 7 a_3 + 7 a_4 + 7 a_5 + 7 a_6 + 7 a_7 + 7 a_8= 499$$

Dividing by 7 would be $$a_1 + a_2 + a_3 + a_4 + a_5 + a_6 + a_7 + a_8= 71.2857143$$

$\endgroup$
  • $\begingroup$ You can't solve all eight. You probably can't solve any. But you can determine the relations between them all and there's no reason that won't give you the maximum possible value. $\endgroup$ – fleablood Jul 10 '16 at 8:14
  • $\begingroup$ Um, averaging the sums isn't the same thing as averaging the variables. $\endgroup$ – fleablood Jul 10 '16 at 8:16
  • $\begingroup$ @fleablood Actually, you can figure out all $8$ numbers, though obviously not their order. $\endgroup$ – Mike Jul 10 '16 at 8:20
  • $\begingroup$ But you don't know that the last sum is $66$. And what do you get on the left side if you sum all $8$ equations together? $\endgroup$ – Mike Jul 10 '16 at 8:22
  • $\begingroup$ @fleablood see my edit for clarification and a slight edit $\endgroup$ – possibility0 Jul 10 '16 at 8:24
2
$\begingroup$

I'm not sure why you're dividing by $8$, but I think you may have been at least on the right track. If one sums $7$ of the $8$ numbers, one is left out. If the duplicate sum is $x$, then $x+433$ must equal $7$ times the sum of all $8$ numbers. Using modular arithmetic, we have $433\equiv6\pmod7$, so the duplicate sum must be equivalent to $1\pmod7$. The only one that works is $64$. So the sum of all $8$ numbers is $\frac{433+64}7=71$. So our largest number must be $71-56=15$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.