Number System - Study Mode

[#481] The four digit smallest positive number which when divided by 4, 5, 6 or 7, it leaves always the remainder as 3:
Correct Answer

(C) 1263

Explanation

Solution: The least possible number = (LCM of 4, 5, 6 and 7) + 3 = 420 + 3 = 423
The next higher number is, (420m + 3), now we put a least value of m such that (420m + 3) ≥ 1000 At m = 3, value = 420 × 3 + 3 = 1263

[#482] A string of length 221 metre is cut into two parts such that one part is $$frac{9}{4}$$ th as long as the rest of the string, then the difference between the larger piece and the shorter piece is :
Correct Answer

(C) 85 m

Explanation

Solution: Let one part of string is x. Now, x + $$frac{{9{ ext{x}}}}{4}$$ = 221 m x = 68 meter and, $$frac{{9{ ext{x}}}}{4}$$ = 153 m Difference between two parts = 153 - 68 = 85m.

[#483] The sum of 100 terms of the series 1 - 3 + 5 - 7 + 9 - 11 .......... is:
Correct Answer

(D) - 100

Explanation

Solution: 1 - 3 + 5 - 7 + 9 - 11 .......... + 197 - 199 = (- 2) + (-2) + (-2) + .......... + (-2) (50 times) = 50 × (-2) = -100

[#484] The remainder of $$frac{{{6^{36}}}}{{215}}:$$
Correct Answer

(B) 1

Explanation

Solution: $$eqalign{
& frac{{ {{6^{36}}} }}{{215}},,{ ext{can}},{ ext{be}},{ ext{written}},{ ext{as}} cr
& frac{{{{left( {{6^3}}
ight)}^{12}}}}{{215}} cr
& Or,,frac{{{{216}^{12}}}}{{215}},,[216,{ ext{on}},{ ext{divided}},{ ext{by}},215,,{ ext{gives}},{ ext{remainder}},1] cr
& frac{{{1^{12}}}}{{215}} cr
& { ext{The}},{ ext{remainder}},{ ext{will}},{ ext{be}},1 cr} $$

[#485] The remainder when (12 13 + 23 13 ) is divided by 11.
Correct Answer

(C) 2

Explanation

Solution: $$eqalign{
& frac{{{{12}^{13}}}}{{11}},{ ext{gives}},{ ext{Remainder}},1 cr
& { ext{It}},{ ext{can}},{ ext{be}},{ ext{written}},{ ext{as}}, cr
& frac{{left( {12 imes 12 imes 12 imes 12,........,13, ext{times}}
ight)}}{{11}} cr
& { ext{On}},{ ext{dividing}},{ ext{it}},{ ext{gives}},{ ext{remainder}},{ ext{1,}},{ ext{each}},{ ext{time}}. cr
& 1 imes 1 imes 1 imes 1,.........,13,{ ext{times}} cr
& { ext{So,}},{ ext{final}},{ ext{remainder}},{ ext{will}},{ ext{be}},1 cr
& frac{{{{23}^{13}}}}{{11}} Rightarrow ,{ ext{Remainder}},1 cr
& { ext{Thus}}, cr
& { ext{The}},{ ext{remainder}},{ ext{of}},frac{{left( {{{12}^{13}} + {{23}^{13}}}
ight)}}{{11}} cr
& = left( {1 + 1}
ight) cr
& = 2 cr} $$