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.

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.

😵😵😵😵

i think i got it

Originally posted by Jusuh
oh no..i have only ten fingers

Originally posted by elopawn
Find a number divisible by 41 whose digits are all nines. .

oh, i got a headache.🙄

9999999999
243902439

Originally posted by THUDandBLUNDER
FLT

Originally posted by elopawn
What's FLT stand for?

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.

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 !!

Originally posted by phgao
Fermat's Little Throrem? The problem does not relate to it.

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.

Originally posted by phgao
Oh thanks for that insight T&B! I appreciate it.