Number System - Study Mode
[#631] 64329 is divided by a certain number, 175, 114 and 213 appears as three successive remainders. The divisor is -
Correct Answer
(C) 234
Explanation
Solution: $$eqalign{
& { ext{Number at (1) = 643 - 175}} cr
& ,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,{ ext{ = 468}} cr
& { ext{Number at (2) = 1752 - 114}} cr
& ,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,{ ext{ = 1638 }} cr
& { ext{Number at (3) = 1149 - 213 }} cr
& ,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,{ ext{ = 936}} cr
& { ext{HCF of 468, 1638, 936 = 234}} cr
& { ext{The divisor is 234}}{ ext{.}} cr} $$
[#632] Three numbers are in the ratio 1 : 2 : 3, and the sum of their cubes is 4500. The smallest number will be -
Correct Answer
(B) 5
Explanation
Solution: $$eqalign{
& x:2x:3x cr
& {x^3} + 8{x^3} + 27{x^3} = 4500 cr
& 36{x^3} = 4500 cr
& {x^3} = frac{{4500}}{{36}} = 125 cr
& x = 5 cr
& { ext{Smallest number is 5}} cr} $$
[#633] The number 2272 and 875 are divided by a 3 digit number N, giving the same remainders. The sum of the digit is :
Correct Answer
(A) 10
Explanation
Solution: Let the remainder in each case be x Then, (2272 - x) and (875 - x) are exactly divisible by three digit number Difference : = (2272 - x) - (875 - x) = 1397 Factor of 1397 = 11 × 127 Since, both 11 and 127 are prime number Three digit number is 127 Sum of digits = 1 + 2 + 7 = 10 Note: - In this type of questions the number which divide the given number and leaves no remainder is either difference of number or a factor of difference.
[#634] When 2 31 is divided by 5 the remainder is -
Correct Answer
(B) 3
Explanation
Solution: $${{ ext{2}}^{31}} div 5$$ [x08egin{gathered}
{ ext{power }},,,,,,,,,{ ext{ remainder}} hfill \
left[ x08egin{gathered}
{2^1},,, o ,,,frac{2}{5} o ,,,2 hfill \
{2^2},,, o ,,,frac{4}{5} o ,,,4 hfill \
{2^3},,, o ,,,frac{8}{5} o ,,,3 hfill \
{2^4},,, o ,,,frac{{16}}{5} o ,,1 hfill \
end{gathered}
ight]{ ext{cycle 1}} hfill \
left[ {{2^5},,, o ,,,frac{{243}}{5} o ,,2}
ight]{ ext{cycle 2}} hfill \
end{gathered} ] $$eqalign{
& { ext{Divided = }}frac{{{2^{31}}}}{5} = frac{{{2^{4 imes 7}} imes {2^3}}}{5} cr
& Rightarrow { ext{remainder = 3}} cr
& { ext{So }}{{ ext{2}}^3}{ ext{ has 3 remainder }} cr
& {{ ext{2}}^{31}}{ ext{ has 3 remainder}} cr} $$
[#635] $$left( {{3^{25}} + {3^{26}} + {3^{27}} + {3^{28}}}
ight)$$ xa0 xa0 is divisible by :
Correct Answer
(D) 30
Explanation
Solution: $$eqalign{
& left( {{3^{25}} + {3^{26}} + {3^{27}} + {3^{28}}}
ight), cr
& = {3^{25}}left( {{3^0} + {3^1} + {3^2} + {3^3}}
ight), cr
& = {3^{25}} imes 40 cr
& = {3^{24}} imes 120 cr
& { ext{now check with option }} cr
& { ext{Only 30 can divide this}}{ ext{.}} cr} $$