A young boy dropped his deck of baseball cards. Unwilling to count them again, he remembered that:
- When he put them in piles of two, there was one card left over.
- When he put them in stacks of three, there was one card left over.
- The same thing happened for stacks of 4, 5, and 6.
- However, when he put them in stacks of 7, there were no cards left over.
How many cards did the boy have? Is there only one possibility? Can you generalize a solution?
Answer (2)
so if he put them in 2 stacks and there was 1 card left over, he has an odd number of cards
when he put them in 3 stacks there was 1 card left over we can tell that if he took away one of his card and divided the stack of cards by two or three the number would be the same
so he divides it by 4 5 and 6 so the number of cards is not divisible by 2,3,4,5,6 but it is divisable by 7 so we must find a number that when you subtract one you can divide it by 2,3,4,5,6 so i just guessed a term from 7^n so 49 doesn't work because 48 doesn't divide into 5 343 doesn't work because 342 doesn's divide by 6 but 2401 (7^4) works because 2400 divides evenly by 2,3,4,5, and 6
Just a note, this number is not very logical that he would have time to count them all but hey, it's math
The boy has 301 baseball cards, and any number in the form of 420m + 301 (for integer m) is a possible number of cards he could have. The conditions reveal multiple valid solutions, with 301 being the smallest. More generally, the problem can be solved using modular arithmetic and the least common multiple concept.
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