Number System - Study Mode

[#476] Find the remainder when 10 + 10 2 + 10 3 + 10 4 + ........ + 10 99 is divided by 6.
Correct Answer

(A) 0

Explanation

Solution: The remainder when 10 is divided by 6 is 4. The remainder when 10 2 is divided by 6 is 4. The remainder when 10 3 is divided by 6 is 4. The remainder when 10 4 is divided by 6 is 4. Thus, remainder is always 4. So, the required remainder, $$eqalign{
& = frac{{4 + 4 + 4 + 4 + ,.....,99{ ext{ times}}}}{6} cr
& = frac{{396}}{6} cr} $$ Thus, Required remainder is 0.

[#477] The remainder of (3) 67! divided by 80 is :
Correct Answer

(B) 1

Explanation

Solution: Since, $${3^4} = 81$$ xa0gives remainder 1 on divided by 80 So, $$frac{{{3^{4n}}}}{{80}}$$ xa0gives remainder 1 Thus, $$frac{{{3^{67!}}}}{{80}}$$ xa0will also give the remainder as 1 Since, 67! = 4n for a positive integer n

[#478] The remainder of $$frac{{{{39}^{97!}}}}{{40}}$$ is :
Correct Answer

(C) 1

Explanation

Solution: Since, $$frac{{{{ ext{a}}^{ ext{n}}}}}{{{ ext{a}} + 1}}$$ xa0 gives remainder 1 when 'n' is even. Now since, 97! is an even number so remainder will be 1. Note:- Factorial of any no. is even.

[#479] The remainder of $$frac{{{2^{59}}}}{{255}}$$ is:
Correct Answer

55

Explanation

Solution: $$eqalign{
& {2^{59}},{ ext{can}},{ ext{be}},{ ext{expressed}},{ ext{as}},{2^{8n}} cr
& { ext{So}}, cr
& { ext{Remainder}},frac{{{2^{59}}}}{{255}} cr
& = { ext{Remainder}},frac{{{2^{8n}}}}{{255}} cr
& = { ext{Remainder}},frac{{{{256}^n}}}{{255}} = 1 cr
& { ext{Required}},{ ext{remainder}} = 1 cr} $$

[#480] The HCF and LCM of 2 4 , 8 2 , 16 2 , 20 3 are :
Correct Answer

(B) 2 4 , 32000

Explanation

Solution: HCF of 2 4 , 8 2 , 16 2 , 20 3 = 2 4 LCM of 2 4 , 8 2 , 16 2 , 20 3 = 2 4 = 2 8 × 125 = 32000