Math Riddle for Fun!

[Debater Slayer]

You have a standard deck of 52 cards, but an unknown number of cards is lost. Using the new number of cards afterward, when you deal to four players, three cards remain. When you deal to three players, two cards remain. When you deal to five players, two cards remain.

What is the smallest number of cards that could have been lost?

Please use spoiler tags for answers!

 

[Altfish]

Spoiler

5

 

[Evangelicalhumanist]

Spoiler

5

 

[Debater Slayer]

Spoiler

5

Spoiler

5

How did you arrive at these answers?

 

[Altfish]

How did you arrive at these answers?

Spoiler

Taking the first clue, if there are 3 cards left over; then the answer must be a multiple of 4, plus 3.
so, it could be 48+3; 44+3; 40+3, etc.
I then tried those possible answers for the second clue and then the third clue, only one number worked.
Finally I deducted that number from 52 to get my answer.

 

[Debater Slayer]

Spoiler

Taking the first clue, if there are 3 cards left over; then the answer must be a multiple of 4, plus 3.
so, it could be 48+3; 44+3; 40+3, etc.
I then tried those possible answers for the second clue and then the third clue, only one number worked.
Finally I deducted that number from 52 to get my answer.

I love it! Thanks for elaborating.

 

[Evangelicalhumanist]

Spoiler

Taking the first clue, if there are 3 cards left over; then the answer must be a multiple of 4, plus 3.
so, it could be 48+3; 44+3; 40+3, etc.
I then tried those possible answers for the second clue and then the third clue, only one number worked.
Finally I deducted that number from 52 to get my answer.

Ditto.