Find-the-last-three-digits-49-303-3993-202-39-606- Tinku Tara June 4, 2023 Algebra 0 Comments FacebookTweetPin Question Number 151418 by mathdanisur last updated on 20/Aug/21 Findthelastthreedigits:49303⋅3993202⋅39606 Answered by Olaf_Thorendsen last updated on 21/Aug/21 x=49303.3993202.39606x=(72)303.(3.113)202(3.13)606x=(7.11.13)606.3808x=(1001)606.3808ByFermatlittletheorem3400≡1[1000]and1001≡1[1000]⇒x≡(1)606.38[1000](38=6561)x≡561[1000]Hencethelast3digitsare561. Terms of Service Privacy Policy Contact: info@tinkutara.com FacebookTweetPin Post navigation Previous Previous post: 0-ln-x-arctg-x-x-x-2-1-dx-Next Next post: To-paint-the-side-of-a-building-painter-normally-hoists-himself-up-by-pulling-on-the-rope-A-as-in-figure-The-painter-and-platform-together-weigh-200-N-The-rope-B-can-withstand-300-N-Find-the-maxim Leave a Reply Cancel replyYour email address will not be published. Required fields are marked *Comment * Name * Save my name, email, and website in this browser for the next time I comment.
![x = 49^(303) .3993^(202) .39^(606) x = (7^2 )^(303) .(3.11^3 )^(202) (3.13)^(606) x = (7.11.13)^(606) .3^(808) x = (1001)^(606) .3^(808) By Fermat little theorem 3^(400) ≡ 1 [1000] and 1001 ≡ 1 [1000] ⇒ x ≡ (1)^(606) .3^8 [1000] (3^8 = 6561) x ≡ 561 [1000] Hence the last 3 digits are 561.](https://www.tinkutara.com/question/Q151424.png)