perplexus.info :: Just Math : Quite a coincidence

Quite a coincidence (Posted on 2018-01-19)

"It's unbelievable!" exclaimed Jerry facing his three friends Adam, Dan and Betty.
"I've asked you to tell me independently each a 4-digit number and after a while, I'm happy to announce that if any of you will divide their number by mine you will end up with the same remainder! "
Now that you know it I'm sure that you will be able jointly to figure the value of my number...- of course it will be the largest of the qualifying candidate answers...

Adam: It could not be true for any three non-related 4-digit numbers!
Betty: You were extremely lucky to find such a special number!
Dan: And now we will be able to calculate the value of your number!

Indeed, Adam(2479), Betty(6181), and Dan(8649), after a not-so-long brainstorming session successfully restored Jerry's number.

a. What was it?
b. d4 bonus question:
What's the probability of Jerry "success" with 3 RANDOMLY CHOSEN
three 4-digits numbers.
Provide your estimate, listing your assumptions and reasoning.

No Solution Yet Submitted by Ady TZIDON

Rating: 3.0000 (1 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)

Part B (spoiler)

| Comment 3 of 6 |

If all the remainders were equally likely, then the probability that the 2nd and the third remainder match the first = (1/1234)^2, or one time in 1,523,756

This is a pretty good estimate, but an exact calculation is not very difficult.

1000 has a remainder of 1000.

9999 has a remainder of 127

Of the 9000 possible numbers,

8 have a remainder from 0 to 127

7 have a remainder from 128 to 999

8 have a remainder from 1000 to 1233.

So 362 come up 8 times out of 9000

and 872 come up 7 times out of 9000

The exact probability = 362*(8/9000)^3 + 872*(7/9000)^3 =

484440/729,000,000,000 which is about 1 time out of every 1,504,830

Posted by Steve Herman on 2018-01-19 10:13:57

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