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Mrs. Jones is celebrating her 43rd (Prime number) birthday. Over the years she has collected gold coins. The number of coins is a Prime number less than 100. She decides to give all the coins to her three children, Amy (oldest), Jack (middle one) and Lisa (youngest).

  1. First she gives each of them coins same as their individual age numbers. Turns out their ages are different, consecutive Prime Numbers.

  2. Of the remaining coins she then gives each of them a number same as the Month they were born. Turns out they were all born in the same exact month and it is also a Prime number.

  3. Of the remaining coins she gives each of them a number same as the day they were born. Turns out each one was born on a Prime Number day. (so if Amy was born on x th day of y th month and is z years old then she will get x+y+z coins).

  4. Now Mrs. Jones has no coins left.

  5. The total number of coins each kid gets is also a Prime Number. All kids end up with different number of coins

  6. Lisa the teenager ends up getting the most coins.

How many coins did Mrs. Jones have? How many did each kid get?

Two solutions possible.

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  • $\begingroup$ Do you need to specify that the three days of the month are different? (See my answer.) $\endgroup$ – Rand al'Thor Oct 26 '17 at 15:45
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    $\begingroup$ I found 3 solutions so far and am not finished. Have I made a mistake in my assumptions? $\endgroup$ – Trenin Oct 26 '17 at 16:29
  • $\begingroup$ The solutions so far have all solved by brute force exhausting all possibilities. Is there no simpler way? $\endgroup$ – Octopus Oct 27 '17 at 20:30
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    $\begingroup$ I'm voting to close this question as off-topic because it appears to be a brute force question as opposed to a puzzle, which has an 'a-ha' moment or an unexpectedly elegant solution/result. $\endgroup$ – Wen1now Oct 28 '17 at 4:02
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Complete deduction of solutions

The youngest child is a teenager (by item 6) and their ages are three consecutive primes (by item 1), so the only possibilities for their ages are:

$13,17,19$ or $17,19,23$ or $19,23,29$.

Let us denote their ages by $A_A>A_J>A_L$, their month by $M\leq12$, and their days of birth by $D_A,D_J,D_L\leq31$. All seven of these numbers are primes, as are the numbers $A_A+M+D_A,A_J+M+D_J,A_L+M+D_L$ (which are all different, with the last being the largest), and their sum, which is $\leq97$. These are our criteria.

The total number of coins each child receives is an (odd) prime, so $M$ and all the $D_{...}$ must have the same parity. So either all of these four numbers are odd primes, or they're all 2 (which would be quite a coincidence).

By item 6, Lisa gets the most coins despite her age being smallest. The difference must come from $D_L$, which must therefore be greater than $D_A$ by at least

seven (larger than one of $19-13,23-17,29-19$). But $D_A\geq3$, so $D_L\geq3+7=10$, so $D_L\geq11$. In the case that Lisa is 19, the difference must be at least eleven, so $D_L\geq3+11=14$, so $D_L\geq17$.

  • Case 1: Lisa is 19. We argue as follows.

    The sum of all three ages is $19+23+29=71$. Their month is $M\geq3$, and their days are at least $3,3,17$. Summing all of these gives $71+(3\times3)+3+3+17>100$. Contradiction.

  • Case 2: Lisa is 17. We argue as follows.

    The sum of all three ages is $17+19+23=59$. The sum of all three days is at least $3+3+11=17$. Their month can't be $M=11$ (too big), so it must be 3 or 5 or 7.

    • Case 2A: they were born in July.

      $A_L+A_J+A_A+3M=17+19+23+(3\times7)=80$, so their days must be exactly $3,3,11$ in order to have a total sum of at most 97. But then Amy gets $23+7+3=33$ coins, and 33 isn't prime. Contradiction.

    • Case 2B: they were born in May.

      $A_L+A_J+A_A+3M=17+19+23+(3\times5)=74$, so the sum of their days must be at most $23$. Lisa gets $17+5+D_L=22+(\geq11)$ coins, but $22+11=33,22+13=35,22+17=39$ are not prime, so $D_L\geq19$. Contradiction.

    • Case 2C: they were born in March.

      $A_L+A_J+A_A+3M=17+19+23+(3\times3)=68$, so the sum of their days must be at most $31$. Lisa, Jack, and Amy get $20+D_L,22+D_J,26+D_A$ coins respectively.
      If $D_L=11$, then Lisa gets 31 coins, so the others must get at most 29 and 23, contradiction.
      If $D_L=13$, then Lisa gets 33 coins - not prime.
      If $D_L=17$, then Lisa gets 37 coins, and the others can get 31 ($D_A=5$) and 29 ($D_J=7$), for a total of 97 - the maximum possible.
      Thus we have found one possibility, the only one in Case 2.

  • Case 3: Lisa is 13. We argue as follows.

    The sum of all three ages is $13+17+19=49$. The sum of all three days is at least $3+3+11=17$. Their month can't be $M=11$ (too big), so it must be 3 or 5 or 7.

    • Case 3A: they were born in July.

      $A_L+A_J+A_A+3M=13+17+19+(3\times7)=70$. Amy and Jack each get at least $17+7+3=27$ coins, so the three must get $29,31,37$ with Lisa getting the most. Lisa, Amy, and Jack get $20+D_L,24+D_J,26+D_A$ coins respectively. The only way this can work is with $D_L=17,D_J=5,D_A=5$ or $D_L=17,D_J=7,D_A=3$.
      Thus we seem to have two possibilities here, since you never specified that the three days are different.

    • Case 3B: they were born in May.

      $A_L+A_J+A_A+3M=13+17+19+(3\times5)=64$. Amy and Jack each get at least $17+5+3=25$ coins, so the three must get $29,31,37$ with Lisa getting the most. Lisa, Amy, and Jack get $18+D_L,22+D_J,24+D_A$ coins respectively, so $D_L=19$ and $D_J=D_A=7$.
      Thus we seem to have another possibility here, since you never specified that the three days are different.

    • Case 3C: they were born in March.

      $A_L+A_J+A_A+3M=13+17+19+(3\times3)=58$. Amy and Jack each get at least $17+3+3=23$ coins, so the three must get either $23,29,31$ or $29,31,37$ with Lisa getting the most. Lisa, Amy, and Jack get $16+D_L,20+D_J,22+D_A$ coins respectively, so $D_L$ must be 15 or 21, contradiction.

Thus there are four possibilities for all the different numbers mentioned in the puzzle (or only two if we require that the three days are all different). However, the actual question asked - how many coins did each child get, and what was the total - has only two solutions:

the total is $97$, and Amy, Jack, and Lisa got either $31,29,37$ or $29,31,37$ coins respectively.

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  • $\begingroup$ You have a great way of analyzing Rand al'Thor $\endgroup$ – DEEM Oct 26 '17 at 16:34
  • $\begingroup$ Your 4 possibilities appear to resolve to just 2 unique solutions to the actual questions posed.... $\endgroup$ – Hellion Oct 26 '17 at 21:35
  • $\begingroup$ There are 4 possibilities for their birthdays, but that was not the question. The question was "how many coins did each child get?" There are only 2 possibilities for that. (I used a computer to check) $\endgroup$ – Dennis_E Oct 26 '17 at 21:35
  • $\begingroup$ @Hellion and Dennis_E: good point; edited the final paragraph to fix that. $\endgroup$ – Rand al'Thor Oct 26 '17 at 23:53
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    $\begingroup$ Teenage denotes a person 13 to 19. The all have -teen at the end $\endgroup$ – DEEM Oct 27 '17 at 14:18
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Mrs. Jones has

97 coins

of which,

37 are given to Lisa, 29 given to Jack or Amy, and 31 are given to Amy or Jack (depending on which answer)

The children are aged

17, 19, and 23; or 13, 17, and 19

and were all born in

March, on the 17th, 7th, and 5th; or in July, on the 17th, 7th, and 3rd.

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If we stick to the two questions actually being asked, specifically:

How many coins did Mrs. Jones have? How many did each kid get?

Then there is only 1 solution:

Mrs. Jones had 97 coins and the children received 37, 29 and 31 coins

Unless you interpret the second question a bit more literally, then there are two possibilities:

Amy has 31 coins, Jack has 29 and Lisa has 37

or

Amy has 29 coins, Jack has 31 and Lisa has 37

No wording needs to be changed.

If you go into the actual distribution of coins per the dates within the problem then there are four solutions (which my answer previously listed as possible solutions). My previous answer was:

There are exactly 4 solutions that satisfy the wording as currently stated (I determined this programmatically), there will be exactly 2 solutions if the wording in statement #3 is changed to "each one was born on a different Prime Number day"

The two solutions that contain unique days for each are:

Lisa: 17 years old, born 03/17
Jack: 19 years old, born 03/07
Amy: 23 years old, born 03/05

and

Lisa: 13 years old, born 07/17
Jack: 17 years old, born 07/07
Amy: 19 years old, born 07/03

Additional solutions that contain non-unique days are:

Lisa: 13 years old, born 05/19
Jack: 17 years old, born 05/07
Amy: 19 years old, born 05/07

and

Lisa: 13 years old, born 07/17
Jack: 17 years old, born 07/05
Amy: 19 years old, born 07/05

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I found 3 solutions so far. Still have one more section to do later.

So we will use the following tool: the primes under 100.

2,3,5,7,11,13,17,19,23,29,31,37,41,43,47,53,59,61,67,71,73,79,83,89,97

We have 3 possibilities for their ages based on the fact that lisa is a teenager. These are:

  1. 13, 17, 19
  2. 17, 19, 23
  3. 19, 23, 31

Note that they cannot be born in February since Lisa would need to have a larger day than the others and she would then have prime+2+prime as her number of coins, which can never be prime.

Lets start with the last possibility first.

19, 23, 31

The sum of their ages is 73. At most, this leaves 24 coins to doll out. We know Lisa is supposed to get the most, so she needs to get 13 more coins than Amy. If they were born in march, then 9 coins are now gone and we have 15 left. If we give 3 to Amy and 7 to Jack, that doesn't leave enough for Lisa.

Hence, this is not an option.

17, 19, 23

This one is a little more doable since the total is 59 which leaves 38 coins at most. But Lisa still needs 5 more coins than Amy.

Assume month is May (5)

The totals so far are:

  • Amy - 23+5=28
  • Jack - 19+5=24
  • Lisa - 17+5=22
  • Total: 74 leaving at most 23

If Amy's day is 13 her total would be 41. But Lisa would need at least a day of 19 which uses up more coins than we have. Thus Amy's day needs to be less than 13. But of those values, all are composite except the 3rd. (28+11=39, 28+7=35, 28+5=33). Thus, Amy is 23 born on May 3rd.

We now have 20 more coins. Jack cannot be 11 or more since Amy would need at least 3 more to be the highest and their aren't enough coins. If Jack was born on the 7th, then Amy could be born on the 13th or 11th, but both totals make Amy's coins composite (35 and 33 respectively). If Jack was born on the 5th, then Amy again needs to be born on the 11th or 13th to be high enough; neither of which work.

Thus, there is no solution.

Assume month is march(3)

The totals so far are:

  • Amy - 23+3=26
  • Jack - 19+3=22
  • Lisa - 17+3=20
  • Total: 68 leaving at most 29

If Amy's day is 11th or later, then Lisa's needs to be at least 17, which leaves only 1 coin for Jack, so this is impossible. Thus, Amy must be born on the earlier than the 11th and the 5th is the only option that makes her total prime. Thus Amy is born on the 5th of March and we now have at most 24 coins left.

We know Lisa's day needs to be at least 13 to be more than Amy, but she also needs to be at least 3 more than Jack's day. Thus, Jack's day cannot be 11 or more because Lisa would need at least 17 which we don't have the coins for.

The only possibility for Jack is the 7th since the others are composite for him. Thus Jack was born on the 7th of March, and this leaves 17 coins for Amy.

If she is born on the 17th, we get our first solution!!

Solution 1:
Amy: 23, born on the 5th of March gets 31 coins
Jack: 19, born on the 7th of March gets 29 coins
Lisa: 17, born on the 17th of March gets 37 coins
Total: 97 coins

We know Amy needs at least 13 coins, but that gives a total of 33, which is composite.

There are no other options, so there are no other solutions.

13, 17, 19

The total now is 49 which gives lots of options.

Lets say they were born in November. This takes up lots of coins and leaves us with only 16. Lisa needs to be born at least 7 days later than Lisa, and the lowest possibilities are Amy and Jack born on the 3rd, and Lisa on the 11th, which is still too high.

Thus, they must be born before November.

Assume Month is July (7)

The totals are:

  • Amy - 19+7=26
  • Jack - 17+7=24
  • Lisa - 13+7=20
  • Total: 70 leaving at most 27

Lisa needs to be 7 days later than Amy and 5 days later than Jack.

Options for Amy are 3, 5, 11. If the 11th, then Lisa would need to be 19 which leaves no coins for Jack.

Jack's options are 5, 7, 13. If the 13th, then Lisa would need 19, which leaves none for Amy.

Lisa's options are 11, 17, 23. If the 23rd, then this leaves too few for Jack and Amy.

Coincidentally, there are 2 solutions which use all 97 coins again.

Solution 2:
Amy: 19, born on the 3rd of July gets 29 coins
Jack: 17, born on the 7th of July gets 31 coins
Lisa: 13, born on the 17th of July gets 37 coins
Total: 97 coins

And

Solution 3:
Amy: 19, born on the 5th of July gets 31 coins
Jack: 17, born on the 5th of July gets 29 coins
Lisa: 13, born on the 17th of July gets 37 coins
Total: 97 coins

There is another solution with 89 coins where everyone uses the minimum, but Amy and Jack both get 29, which is not allowed.

Assume the Month was May(5)

TBD... this is taking too long...

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  • $\begingroup$ Trenin. Why did you omit 29 as a prime number? $\endgroup$ – DEEM Oct 26 '17 at 18:13
  • $\begingroup$ @DEEM Where did I omit it? It is in the list. $\endgroup$ – Trenin Oct 26 '17 at 18:47
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    $\begingroup$ @Trenin Your 3rd possibility - 19, 23, 31 - ignores 29. $\endgroup$ – Apep Oct 26 '17 at 19:56
  • $\begingroup$ What is the actual question that you need to solve for? $\endgroup$ – Octopus Oct 26 '17 at 21:10
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Brute force solver found

four possibilities:

Testing ages L:13,J:17,A:19
Got possible solution: month=5, days are L:19,J:7,A:7, coins are L:37,J:29,A:31, total coins=97
Got possible solution: month=7, days are L:17,J:5,A:5, coins are L:37,J:29,A:31, total coins=97
Got possible solution: month=7, days are L:17,J:7,A:3, coins are L:37,J:31,A:29, total coins=97
Testing ages L:17,J:19,A:23
Got possible solution: month=3, days are L:17,J:7,A:5, coins are L:37,J:29,A:31, total coins=97

Since the question is "How many coins did Mrs. Jones have? How many did each kid get?", the solutions are
coins are L:37,J:29,A:31, total coins=97
coins are L:37,J:31,A:29, total coins=97

Lisa should be described as "a teenager", not "the teenager" because there are no solutions with ages L:19 J:23 A:29.

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One of the answers is:

She has a total of 97 coins.

How does this add up?

Lisa gets 37 coins, Jack gets 31 coins and Amy gets 29 coins.

Explanation:

Ages- 13,17 and 19.
They are born on the 17th, 7th and 3rd of July respectively.

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If Lisa is 'the' teen then she's 19 and the others are 23 and 29 (and her mom had Amy at 14).

Their age total is 71.

Prime Numbers above 71 with suiting 3 * min prime month/day (min is 12 or 3 * 2 * 2) are 89 and 97 leaving 18 and 26 remaining coins

The only 3 * prime month subtractions are months = 2, 3 and 5

This leaves permutations (age total, max starting coins, month, Lisa day, Jack day, Amy day): (71,89,2,7,3,2) (71,89,2,7,2,3) (71,89,3,5,2,2) (71,97,2,13,5,2) (71,97,2,11,7,2) (71,97,2,13,2,5) (71,97,2,11,2,7) (71,97,3,13,2,2) (71,97,3,11,3,3) (71,97,3,7,5,5) (71,97,5,7,2,2) (71,97,5,5,3,3)

If Lisa is 19 and Amy is 29, the Lisa needs > 10 more 'day' coins than Amy If Lisa is 19 and Jack is 23, the Lisa needs > 4 more 'day' coins than Jack

It looks like -- only these work: (71,97,2,13,5,2) (71,97,3,13,2,2)

First: Ages(19, 23, 29) mapped {+ month = 2} = (21, 25, 31) & mapped {+ day born} = (34, 30, 33) Second: Ages(19, 23, 29) mapped {+ month = 3} = (22, 26, 32) & mapped {+ day born} = (35, 28, 34)

Either solution ... definitively, 97 is the starting point. Solution 1: Lisa = 34, Jack = 30, Amy = 33 Solution 2: Lisa = 35, Jack = 28, Amy = 34

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Used trial-an-error to get my answer. My answer is different from the rest in terms of the children's age. So it seems like there are more than 2 possible solutions.

How many coins did Mrs. Jones have?

97

How many did each kid get?

Amy: 31
Jack: 23
Lisa: 43

Age of the children:

Amy: 13
Jack: 7
Lisa: 3

Birth dates of the children:

Amy: 7 Nov
Jack: 5 Nov
Lisa: 29 Nov

Calculation for number of coins received by the children:

Amy: 13 + 11 + 7 = 31
Jack: 7 + 11 + 5 = 23
Lisa: 3 + 11 + 29 = 43

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  • $\begingroup$ Lisa is specified to be a teen. $\endgroup$ – Rubio Oct 31 '17 at 21:30

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