# A Problem

elopawn
Posers and Puzzles 23 Apr '05 03:27
1. 23 Apr '05 03:27
Find a number divisible by 41 whose digits are all nines. Without doing the division find the remainder when
90 000 800 007 000 060 000 500 004 000 030 000 200 001 is divided by 41.
2. 25 Apr '05 05:38
Originally posted by elopawn
Find a number divisible by 41 whose digits are all nines. Without doing the division find the remainder when
90 000 800 007 000 060 000 500 004 000 030 000 200 001 is divided by 41.
oh no..i have only ten fingers
3. 25 Apr '05 07:032 edits
ðŸ˜µðŸ˜µðŸ˜µðŸ˜µ

i think i got it
4. 25 Apr '05 12:161 edit
Originally posted by Jusuh
oh no..i have only ten fingers
Need a clue?

Think about dividing 100, 1000, 10000, 100000 by 41 and the remainders of them. Eg they are 18, 16, 37 and 1.

How does this relate to the question?
5. 25 Apr '05 17:57
Originally posted by elopawn
Find a number divisible by 41 whose digits are all nines. .
FLT

6. 25 Apr '05 21:35
oh, i got a headache.ðŸ™„
7. 26 Apr '05 05:24
9999999999
243902439
8. 26 Apr '05 07:55
Originally posted by THUDandBLUNDER
FLT

What's FLT stand for?

Anyway I'll post the answer if no one gets it in a day.
9. 26 Apr '05 09:522 edits
Originally posted by elopawn
What's FLT stand for?
Fermat's Little Theorem:

http://mathworld.wolfram.com/FermatsLittleTheorem.html
10. 26 Apr '05 10:311 edit
Fermat's Little Throrem? The problem does not relate to it.
I'll post the answer now if you want it, it actually after you see it is rather simple.
11. 26 Apr '05 10:401 edit
Ok so we know after dividing bu 100, 1000, 10000, 100000 by 41 we get-- 18, 16, 37, 1

Thus 99999 is divisible by 41.

Let x=90 000 800 007 000 060 000 500 004 000 030 000 200 001

=9* 10power40 + 8* 10power35 +7* 10power30 + 6* 10power25 + 5* 10power20 + 4* 10power15 + 3* 10power10 + 2* 10power5 + 1

From the above, 10power5 leaves a remainder of 1 when divided by 41.

Thus the remainders of each term of the sum are 9,8,7,6,5,4,3,2,1 !!

The sum of the remainders is 45 and so the remainder when x is divided by 41 is 4 !!

12. 26 Apr '05 11:462 edits
Originally posted by phgao
Fermat's Little Throrem? The problem does not relate to it.
a^(p-1) - 1 = 0 (mod p) where a is a positive integer and p is a prime.

In your problem, a = 10 and p = 41

Therefore the number 10^40 - 1 is divisble by 41

As you didn't ask for the smallest such number, this answers your first question rather more quickly than your trial-and-error method.
13. 26 Apr '05 13:19
Originally posted by THUDandBLUNDER
a^(p-1) - 1 = 0 (mod p) where a is a positive integer and p is a prime.

In your problem, a = 10 and p = 41

Therefore the number 10^40 - 1 is divisble by 41

As you didn't ask for the smallest such number, this answers your first question rather more quickly than your trial-and-error method.
Oh thanks for that insight T&B! I appreciate it.
14. 26 Apr '05 15:42
Originally posted by phgao
Oh thanks for that insight T&B! I appreciate it.
Sorry, I thought I was replying to the original poster (elopawn).