9
$\begingroup$

Question: For 3 non negative integers $a, b, c$; if $28a + 30b + 31c = 365$ what is the value of $a +b +c$ ?

How I approached it : I started immediately breaking it onto this form on seeing it :

$28(a +b +c) +2b +3c = 365 .......(1)$ $30(a +b +c) -2a +c = 365 .......(2)$ $31(a +b +c) -b -3a = 365 .......(3)$

And then I find out that

$365 = 28*13 + 1......(1')$; $365 = 30*12 + 5......(2')$ $365 = 31*11 + 24......(3')$

Now as we see (1) and (1') as well as (3) and (3') or even equations $2$ and $2'$ do not combine quiet congruently, so I meet with a dead end here.

my issue : how should I approach such problems where we are given no other equations or data? Basically I am asking what are a few ways to get a solution for this problem.

$\endgroup$
  • 3
    $\begingroup$ $28\times 2 + 30\times 7 + 31\times 3 = 359\ne 365$ $\endgroup$ – Joey Zou Aug 11 '16 at 21:25
  • 12
    $\begingroup$ This is a trick question, you are supposed to observe that these are the lengths of months and 365 is the length of a normal year. $\endgroup$ – f'' Aug 11 '16 at 21:31
9
$\begingroup$

More direct path to $a+b+c = 12$:

Write $x=a+b+c$. In particular the last equation implies $31 x \ge 365$ so $x>11$. On the other hand, the first equation is $28x + 2b + 3c = 365$. If either $b$ or $c$ is nonzero, this means $28x < 364$ so $x < 13$. And $b=c=0$ is not possible because $365$ is not divisible by $28$.

added: to be fair I should complete the proof... $x=12$ means $2b + 3c = 29$. $c$ must be odd and less than $10$, so it can be $1, 3, 5, 7, 9$. Substitute in $2b+3c = 29$, then in $a+b+c = 12$ and keep only the solutions with non-negative $a$.

$\endgroup$
2
$\begingroup$

Your equation (2) should say $$ 30(a+b+c) - 2a + c = 365. $$ Note that this implies $$ c- 2a \equiv 5 \mod 30. $$ I now claim that, in fact, $c-2a = 5$. This would imply that $30(a+b+c)+5=365$, or that $a+b+c = 12$.


Since we know that $c-2a\equiv 5\mod 30$, it suffices to show that $-25< c-2a < 35$ to conclude that $c-2a = 5$. The upper bound is easy, since $$31c\le 365\implies c\le 11\implies c-2a\le 11 < 35$$ as $a$ is nonnegative. To get the lower bound, we note that $$ 28a\le 365\implies a\le 13.$$ Furthermore, if $a=13$, then $28\times 13 + 30b+31c = 365\implies 30b+31c = 1$, which is clearly impossible for nonnegative $b$ and $c$. Thus, $a\le 12$, and hence $$ c-2a\ge 0 - 2(12) = -24 >-25 $$ since $c$ is also nonnegative. Hence, $-25 < c-2a < 35$, and combined with the fact that $c-2a\equiv 5\mod 30$, we conclude that $c-2a = 5$, as desired.


Using the fact that $c-2a = 5$, we can also solve for the possible solutions. Substituting $c = 2a+5$ into the original equation yields \begin{align} 28a+30b+31(2a+5) &= 365 \\ \implies 28a + 30b + 62a + 155 &= 365 \\ \implies 90a + 30b &= 210 \\ \implies 3a + b &= 7. \end{align} We thus see that \begin{align} a = 0 &\implies b = 7, c = 5 \\ a = 1 &\implies b = 4, c = 7 \\ a = 2 &\implies b = 1, c = 9. \end{align} If $a\ge 3$, then $b<0$. Hence, we conclude that $(0,7,5)$, $(1,4,7)$, and $(2,1,9)$ are the only solutions to the problem.

$\endgroup$
2
$\begingroup$

The sum of $a,b,c$ is greater than $11$ because $31(11) = 341 < 365.$

The sum has to be less than $14$ because $28(14) = 406 > 365$.

It also cannot be $13$ because, although $28(13) = 364 < 365$, we can only swap out a $30$ or $31$ for one of the $28$'s, and this puts us over $365$.

So, the sum is $12$.

The value of $c$ must be odd because the other two terms must be even, and the sum is odd. Let's check each odd value of $c$ such that $0 \leq c \leq 12$.

$c=11$ doesn't work because an additional $28$ or $30$ puts us over $365$.

$c=9$ can work. $31(9) = 279$, and $365-279 = 86$. $28(2) + 30 = 86,$ so $(a,b,c) = (2,1,9)$ is a solution.

$c=7$ works a bit by inspection: Seven months have $31$ days, four have $30$, and one has $28$, which total $365$ days. So $(1,4,7)$ is also a solution.

$c=5$ works also. $31(5)=155$, and $365-155= 210$, which is $30(7)$. So $(0,7,5)$ is a third solution.

$c=3$ doesn't work because $365-3(31) = 272$, which is greater than $30(9)$.

$c=1$ doesn't work either, because $365-31 = 334 > 30(11)$.

So, three solutions: $(2,1,9), (1,4,7), (0,7,5).$

$\endgroup$
  • $\begingroup$ 1 + 6 + 7 = 14. I think you mean 1 + 4 + 7. $\endgroup$ – gnasher729 Aug 11 '16 at 22:08
  • $\begingroup$ @gnasher729 LOL thanks. Apparently I don't inspect very well. $\endgroup$ – John Aug 15 '16 at 15:45
1
$\begingroup$

I'm not sure what you meant by combining (2) and (2'). However, here is a possible continuation.

Equations (1) and (1') imply that $a+b+c<14$. Equations (3) and (3') imply that $a+b+c>11$. If $a+b+c=13$, then (2) yields $2a-3c=25$, or $a\geq 13$, whence $a=13$, $b=0$, and $c=0$, which do not form a solution. That is, $a+b+c=12$ must hold. It is easy to see that $$(a,b,c)=(1,4,7)+t(-1,3,-2)$$ with $t\in\mathbb{Z}$ are the only integer solutions to $28a+30b+31c=365$ and $a+b+c=12$. For nonnegative-integer solutions, $t\in\{-1,0,+1\}$ are the only possibilities, giving three triples $(a,b,c)=(2,1,9)$, $(a,b,c)=(1,4,7)$, and $(a,b,c)=(0,7,5)$.

$\endgroup$
  • $\begingroup$ But then how come my triad too agree with the given conditions? $\endgroup$ – Arnav Das Aug 11 '16 at 21:33
  • $\begingroup$ Your triad is not a solution. $\endgroup$ – Batominovski Aug 11 '16 at 21:35
  • $\begingroup$ Ummm but why is that ? Am not able to grasp it $\endgroup$ – Arnav Das Aug 11 '16 at 21:38
  • $\begingroup$ Joey Zou already posted a comment on that. $\endgroup$ – Batominovski Aug 11 '16 at 21:39
  • 2
    $\begingroup$ Actually $(0,7,5)$ and $(2,1,9)$ are also solutions, although their sums are also $12$, of course. $\endgroup$ – Joey Zou Aug 11 '16 at 21:49

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.