SOLUTION: When John sorts his collection of computer card games into groups of 3, 4, 5 and 8 there is always one card left. What is the smallest number of cards John can have? Please show

Question 186391: When John sorts his collection of computer card games into groups of 3, 4, 5 and 8 there is always one card left. What is the smallest number of cards John can have?
Please show the solution so I'm able to understand.. thanks in advance. God bless you all
Found 2 solutions by checkley77, stanbon:
Answer by checkley77(12844) (Show Source): You can put this solution on YOUR website!
3,4,5 & 8
Finding the LCD is one way of solving this problem.
3*4*5*2=120
This gives 120+1=121 is the answer.
Proof:
121/3=40 r1.
120/4=30 r1.
121/5=24 r1.
121/8=15 r1.

Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
When John sorts his collection of computer card games into groups of 3, 4, 5 and 8 there is always one card left. What is the smallest number of cards John can have?
-----------
3 is divisible by 3; but 3+1 leaves a remainder of one when you divide by 3
--------
3*4 is divisible by 3, and it is divisible by 4; but 3*4+1 leaves a remainder
of 1 when you divide by 3 or when you divide by 4
------------
Similarly 3*4*5*8+1 will have a remainder of 1 when you divide by 3,or
4, or 5, or 8
--------------------
But the 4 and the 8 share a factor of 4, so we don't need the 4 as
a separate factor to achieve divisibility by 4.
-------------------
So the smallest number that satisfies the condition is
3*5*8+1 = 121
=======================
Cheers,
Stan H.
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