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During a particular year, exactly ten people ($A - J$) visited a certain city on five different days between January 1st and April 30th, in a non-leap year; such that on each of the five days, exactly two of the ten people visited the city.

It is also known that:

  1. $A$, who did not visit the city after $J$, visited the city $28$ days after $F$, who, in turn, did not visit the city with $B$.
  2. $J$, who visited the city in March, visited the city at least $50$ days before $C$ visited but visited the city on the same day of the week as $C$.
  3. $D$, who visited the city exactly $10$ days before $H$, visited the city with $G$.
  4. Both $E$ and $I$ visited the city on February 10th, while $E$ and $H$ visited the city on the same day of the week.

On which day did $B$ visit the city?


Source : time.com

Please tell your approach.

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2 Answers 2

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The answer is

April 28th.

We can start by figuring out who visited the city with whom.

$E$ and $I$ visited the city together; so did $D$ and $G$. $F$ didn't visit with $A$, nor did they visit with $B$. $J$ did not visit with $C$. Since $A$ didn't visit after $J$ but did after $F$, $F$ and $J$ were not together. Further, $A$ visited $28$ days after $F$. Since $C$ is in April (see below), $F$ and $C$ aren't together. Therefore $F$ and $H$ are together. We either have $A-C$ and $B-J$ or $A-J$ and $B-C$. $A-C$ must come at least before $B-J$ because $A$ was at least before $J$; but it must come strictly after $B-J$ because $J$ precedes $B$. Contradiction! So we have our pairings: $E-I$, $D-G$, $F-H$, $A-J$, and $B-C$.

Now we'll start to make some assumptions or play around a bit.

Since $J$ visited in March and at least $50$ days before $C$ and on the same day of the week as $C$, we know that $C$ must have gone in April. Further, $C$ must have gone exactly $56$ days after $J$, since $63 \hspace{0.25ex} days > 2 \hspace{0.25ex} months$. Assume that $C$ went on April 30. Then $J$ must have visited on March 5. We therefore have a five day window for $J$ and $C$: $J$ between March 1 and March 5; and $C$ between April 26 and April 30.

Let's start to make hypotheses.

$F$ and $H$ being together means that they are on the same day of the week as $E$ and $I$, and also as $A-J$ and also therefore as $C-B$. $D$ and $G$ must be 10 days before that. So from above, we know that $A-J$ must be on the same day as February 10, between March 1 and March 5. Hence $A-J$ visited on March 3. Further, $B-C$ visited exactly 56 days after $A-J$, from above. We conclude that $B$ visited on April 28th. (To round out the answer, $A-J$ went exactly 28 days after $F-H$, so this would mean $F-H$ went on February 3rd. $D-G$ went 10 days before that, on January 24th.)

To summarize:

$D-G$ went on January 24th.
$F-H$ went on February 3rd.
$E-I$ went on February 10th.
$A-J$ went on March 3rd.
$B-C$ went on April 28th (which is the answer to this puzzle).

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    $\begingroup$ Glorious answer as always, but I am just curious as to how you arrive the exact day count for $J$ to $C$? $\endgroup$
    – Taco
    Commented Oct 10, 2018 at 15:49
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    $\begingroup$ @PerpetualJ: You note that J and C visited on the same day of the week. So the day count is a multiple of 7. Then the difference has to be at least 50, so it's going to be 56, 63, 70, etc. But if J visited in March, then the maximum difference is between March 1 and April 30: a 60 day difference. So the day count is a multiple of 7 between 50 and 60 - hence it must be 56. $\endgroup$
    – El-Guest
    Commented Oct 10, 2018 at 15:51
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    $\begingroup$ Simple oversights; got to love them. :/ Good find! +1 $\endgroup$
    – Taco
    Commented Oct 10, 2018 at 15:55
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    $\begingroup$ You had the right idea and right way of thinking for sure! Good on you as well, @PerpetualJ! $\endgroup$
    – El-Guest
    Commented Oct 10, 2018 at 15:58
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    $\begingroup$ Glad to help @sam! Hope this is correct! $\endgroup$
    – El-Guest
    Commented Oct 11, 2018 at 2:57
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The Answer

$B$ visited on:

Saturday, April 21st, 2018.


Pairs and Dates

$E$ and $I$

Visited on Saturday, February 10th, 2018.

$D$ and $G$

Visited on Wednesday, January 24th, 2018.

$J$ and $A$

Visited on Saturday, March 3rd, 2018.

$C$ and $B$

Visited on Saturday, April 21st, 2018.

$F$ and $H$

Visited on Saturday, February 3rd, 2018.


Reasoning

$A$ not after $J$ but $28$ days after $F$ and $F$ didn't visit with $B$.

This tells us that $A$ cannot visit after $J$ nor before $28$ days after $F$ has visited. Thus the smart thing to do is to pair $A$ with $J$ since this is not technically after $J$ (which threw me off as I was avoiding the pairing until I read the phrase more carefully).

$J$ in March; at least $50$ days prior to $C$ on the same day of the week.

This tells us that $C$ comes after $J$ and that $J$ is in March. This limits the available dates for $J$ and $C$ to land on. $J$ is limited to the 1st and 11th, while $C$ is limited to the 21st and 27th of April.

$D$ and $G$ visited together, exactly $10$ days prior to $H$.

This statement is pretty obvious. With the follow on statement regarding $E$ and $I$ we know that $H$ is on a Saturday, and thus $D$ and $G$ visit on a Wednesday 10 days prior.

$E$ and $I$ visited together on February 10th and $H$ visited on the same day of the week.

This gives us a fully taken date, which is a Saturday, and thus allows us to understand that $H$ also comes on a Saturday.

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  • $\begingroup$ You're so close, but I think your math for the 50 days might be slightly off...? $\endgroup$
    – El-Guest
    Commented Oct 10, 2018 at 15:40

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