+0

# Modular Math

0
42
1

How many values of n less than the LCM ( other than the trivial case of n=1), are there that satisfy the following congruences: n^9 mod 7007=1, n^9 mod 7008=1 and n^9 mod 7009=1?

Jul 1, 2023

#1
0

(1, 716 , 1145 , 4852 , 5391 , 5567 , 5996 , 6106 , 6535)^9 mod 7007==1 - there are 9 values of n.

(1, 673 , 1249 , 1537 , 4225 , 4417 , 5185 , 6529 , 6625)^9 mod 7008==1 - there are 9 values of n

(1, 216 , 221 , 466 , 737 , 853 , 1111 , 1468 , 1670 , 1683 , 2014 , 2530 , 3261 , 3476 , 3527 , 4215 , 4602 , 4860 , 4994 , 5419 , 5682 , 5763 , 6069 , 6279 , 6327 , 6787 , 6886)^9 mod 7009==1 - there are 27 values of n.

The total: 9 x 9 x 27 ==2,187 values of n

Check: taking any 3 random numbers such as: 5391, 4417, 3261 and applying Chinese Remainder Theorem, we have:

344,177,337,504 m +  339,715,607,617, where m=0, 1, 2, 3....etc. When m=0, then we have: 339,715,607,617, which is < than the LCM. And:

339,715,607,617^9 mod 7007==1
339,715,607,617^9 mod 7008==1
339,715,607,617^9 mod 7009==1

Jul 2, 2023