Number System - Study Mode

[#471] A magazine publisher publishes a monthly magazine of 84 pages. One I found that in a magazine 4 pages was missing. One out of them was page number 29 it is known that the page number of the last page of the magazine is 84, (including the cover pages). The numbers printed on the missing pages were :
Correct Answer

(B) 30, 55, 56

Explanation

Solution: Since, the magazine has 84 pages it means that there are 21 sheets of paper, which are folded in the middle. The pattern of pages will be this way, Left Right 1, 2 83, 84 3, 4 81, 82 - - - - - - - - 29, 30 55, 56

[#472] The remainder when 6 6 6 6 . . ∞ is divided by 10 is :
Correct Answer

(B) 6

Explanation

Solution: $$eqalign{
& { ext{Since}}, cr
& frac{6}{{10}} o ,{ ext{remainder}},{ ext{is}},{ ext{6}}{ ext{}} cr
& frac{{{6^6}}}{{10}} o ,{ ext{remainder}},{ ext{is}},{ ext{6}}{ ext{}} cr
& frac{{{6^{{6^6}}}}}{{10}} o ,{ ext{remainder}},{ ext{is}},{ ext{6}}{ ext{}} cr
& frac{{{6^{{6^{{6^6}}}}}}}{{10}} o ,{ ext{remainder}},{ ext{is}},{ ext{6}}{ ext{}} cr
& frac{{{6^{{6^{{6^{{6^6}}}}}}}}}{{10}} o ,{ ext{remainder}},{ ext{is}},{ ext{6}},{ ext{and}},{ ext{so}},{ ext{on}}. cr
& cr
& { ext{Thus,}},{ ext{in}},{ ext{all}},{ ext{the}},{ ext{such}},{ ext{cases}}, cr
& { ext{The}},{ ext{remainder}},{ ext{will}},{ ext{always}},{ ext{be}},6 cr} $$

[#473] $$frac{{left[ {left( {888 imes 888 imes 888}
ight) - left( {222 imes 222 imes 222}
ight)}
ight]}}{{left[ {left( {888 imes 888}
ight) + left( {888 imes 222}
ight) + left( {222 imes 222}
ight)}
ight]} }=, ?$$
Correct Answer

(D) 666

Explanation

Solution: $$eqalign{
& frac{{left[ {left( {888 imes 888 imes 888}
ight) - left( {222 imes 222 imes 222}
ight)}
ight]}}{{left[ {left( {888 imes 888}
ight) + left( {888 imes 222}
ight) + left( {222 imes 222}
ight)}
ight]}} cr
& = frac{{left( {{{888}^3} - {{222}^3}}
ight)}}{{left[ {{{888}^2} + left( {888 imes 222}
ight) + {{222}^2}}
ight]}} cr
& = 888 - 222 cr
& = 666 cr} $$

[#474] The remainder of $$frac{{{{32}^{{{32}^{32}}}}}}{7}:$$
Correct Answer

(D) 4

Explanation

Solution: $$eqalign{
& {32^{{{32}^{32}}}},{ ext{means}},{32^{left( {32.32.32.32.......32, ext{times}}
ight)}} cr
& { ext{and}},{ ext{the}},{ ext{remainder}},{ ext{of}},frac{{32}}{7},{ ext{is}},4 cr
& { ext{So}}, cr
& frac{{{4^{left( {32.32.32.32.......32, ext{times}}
ight)}}}}{7} cr
& frac{{{4^{left( {2.2.2.2.2........32, ext{times}}
ight)}}}}{7} cr
& { ext{Remainder}} = 4 cr
& { ext{Since}},,frac{4}{7} o ,{ ext{Remainder}},4 cr
& frac{{{4^2}}}{7} o ,{ ext{Remainder}},2 cr
& frac{{{4^3}}}{7} o ,{ ext{Remainder}},1 cr
& frac{{{4^4}}}{7} o ,{ ext{Remainder}},4 cr} $$

[#475] The Remainder of $$frac{{ {{{888}^{222}} + {{222}^{888}}} }}{3},{ ext{is}}:$$
Correct Answer

(A) 0

Explanation

Solution: Remainder of $$frac{{ {{{888}^{222}} + {{222}^{888}}} }}{3}$$ xa0 xa0 = 0 Since, 888 and 222 both (bases) are divisible by 3