Question 1146503: There is quite a pile of playing cards sitting at my house. A discovery I have
made is as follows. If all the cards are dealt to all the people, then:
if 2 people are playing cards one card will be left over.
if 3 people are playing cards 2 cards will be left over.
if 4 people are playing cards 3 cards will be left over.
if 5 people are playing cards. 4 cards will be left over.
if 6 people are playing cards 5 cards will be left over.
If seven people are playing cards no cards will be left over.
if I have the minimum amount of cards to satisfy the above conditions how many
cards do I have?
Found 2 solutions by Edwin McCravy, ikleyn:
Answer by Edwin McCravy(20060)   (Show Source):
You can put this solution on YOUR website!
There must be positive integers A,B,C,D,E,F, such that
2A+1 = 3B+2 = 4C+3 = 5D+4 = 6E+5 = 7F = the answer
We will solve linear Diophantine equations by the standard rules, which are:
1. Find the least coefficient of a variable in absolute value, say it's P.
2. Write all other numerical integers in terms of their nearest multiple of P.
3. Divide all terms by P.
4. Move all fractions to one side, and all non-fractions to the other side.
5. Set both sides equal to a new integer, say, Q, making 2 equations.
6. This will give two new Diophantine equations.
6. For the side that has the fractions, clear it of its fractions and solve.
7. Substitute to find F in terms of the new integer.
6E + 5 = 7F
6E + 6 - 1 = 6F + F
E + 1 -1/6 = F + F/6
E + 1 - F = F/6 + 1/6 = G
E + 1 - F = G; F + 1 = 6G
F = 6G - 1
E + 1 - (6G - 1) = G
E + 1 - 6G + 1 = G
E + 2 - 6G = G
E = 7G - 2
6E + 5 = 7F
6(7G - 2) + 5 = 7F
42G - 12 + 5 = 7F
42G - 7 = 7F
6G - 1 = F
5D + 4 = 7F
5D + 4 = 7(6G - 1)
5D + 4 = 42G - 7
5D + 11 = 42G
5D + 10 + 1 = 40G + 2G
D + 2 + 1/5 = 8G + 2G/5
D + 2 - 8G = 2G/5 - 1/5 = H
D + 2 - 8G = H; 2G/5 - 1/5 = H
2G - 1 = 5H
2G - 1 = 4H + H
G - 1/2 = 2H + H/2
G - 2H = H/2 + 1/2 = J
G - 2H = J; H + 1 = 2J
H = 2J - 1
G - 2(2J - 1) = J
G - 4J + 2 = J
G + 2 = 5J
G = 5J - 2
F = 6G - 1
F = 6(5J - 2) - 1
F = 30J - 12 - 1
F = 30J - 13
4C + 3 = 7F
4C + 3 = 7(30J - 13)
4C + 3 = 210J - 91
4C + 94 = 210J
4C + 92 + 2 = 208J + 2J
C + 23 + 2/4 = 52J + 2J/4
C + 23 - 52J = 2J/4 - 2/4 = J/2 - 1/2 = K
C + 23 - 52J = K; J - 1 = 2K
J = 2K + 1
C + 23 - 52(2K + 1) = K
C + 23 - 105K - 52 = K
C - 29 - 104K = K
C - 29 = 105K
C = 105K + 29
4C + 3 = 7F
4(105K + 29) + 3 = 7F
420K + 116 + 3 = 7F
420K + 119 = 7F
We can stop here, since 7F is the answer, and it's odd, so if we divide by 2,
it will leave remainder 1. The minimum positive value for 7F is when K = 0.
Answer = 420(0) + 119 = 119
Checking:
7)119 gives 17 with remainder 0
6)119 gives 19 with remainder 5
5)119 gives 23 with remainder 4
4)119 gives 29 with remainder 3
3)119 gives 39 with remainder 2
2)119 gives 59 with remainder 1
Edwin
Answer by ikleyn(52852)  (Show Source):
You can put this solution on YOUR website! .
I understand the problem in THIS WAY :
There is a set of cards.
If they are distributed equally between 2 persons, 1 card is leftover.
If they are distributed equally between 3 persons, 2 cards are leftover.
If they are distributed equally between 4 persons, 3 cards are leftover.
If they are distributed equally between 5 persons, 4 cards are leftover.
If they are distributed equally between 6 persons, 5 cards are leftover.
If they are distributed equally between 7 persons, NO cards are leftover.
Find the minimum amount of cards to satisfy the above conditions.
SOLUTION
Let N be the number under the question.
Add (mentally) 1 card to the set of cards. In other words, consider the number N+1.
Then it is divisible by 2; by 3; by 4; by 5 and by 6.
The minimal such a number is 60.
Check then, if N-1 = (60-1) = 59 is divisble by 7. ----- It is not divisible.
Next number to try is 60*2 = 120.
Check if 120-1 = 119 is divisible by 7. ----- You are LUCKY: it is (!).
So the answer to the problem is N = 119.
( ! ) S O L V E D ( ! )
----------------
It is a standard problem of a school Math circle level (typically, for children starting from 5-th or 6-th grade).
See the lesson
- The number that leaves a remainder 1 when divided by 2, by 3, by 4, by 5 and so on until 9
in this site.
Find there a lot of similar problems, solved and explained to you.
When children attend such Math circles and solve there such problems under guidance of devoted and enthusiastic teachers,
then there is a chance that they will love Math (and will learn on how to think mathematically) when they become adults . . .
Then there is also a chance that they will pass this enthusiasm and the way of thinking to their own children
and to next generations, in general . . .
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