Number System - Study Mode

[#216] Find the HCF of (3 125 -1) and (3 35 -1).
Correct Answer

(B) 3 5 - 1

Explanation

Solution: The solution of this question is based on the rule, The HCF of (a m - 1) and (a n - 1) is given by (a HCF of m, n - 1) Thus for this question the answer is (3 5 - 1) Since, 5 is the HCF of 35 and 125

[#217] The last digit of the number obtained by multiplying the numbers 81 × 82 × 83 × 84 × 86 × 87 × 88 × 89 will be
Correct Answer

(B) 6

Explanation

Solution: The last digit of multiplication depends on the unit digit of (81 × 82 × 83 × 84 × 86 × 87 × 88 × 89) which is given by the remainder obtained on dividing it by 10. $$eqalign{
& frac{{left( {81 imes 82 imes 83 imes 84 imes 86 imes 87 imes 88 imes 89}
ight)}}{{10}} cr
& { ext{We}},{ ext{take}},{ ext{individual}},{ ext{remainder}},{ ext{of}},{ ext{each}},{ ext{digit,}} cr
& frac{{left( {1 imes 2 imes 3 imes 4 imes 6 imes 7 imes 8 imes 9}
ight)}}{{10}} cr
& { ext{Numbers}},{ ext{multiplied}}, cr
& frac{{left( {24 imes 42 imes 72}
ight)}}{{10}} cr
& { ext{Individual}},{ ext{Remainder}},{ ext{has}},{ ext{been}},{ ext{taken}}, cr
& frac{{left( {4 imes 2 imes 2}
ight)}}{{10}} cr
& Or,,frac{{left( {16}
ight)}}{{10}} cr
& Or,,6 cr
& { ext{Remainder}} = 6 cr
& { ext{So,}},{ ext{the}},{ ext{last}},{ ext{digit}},{ ext{will}},{ ext{be}},6 cr} $$

[#218] 10531 + 4813 - 728 = ? × 87
Correct Answer

(A) 168

Explanation

Solution: Let 10531 + 4813 - 728 = x × 87 Then, (15344 - 728) = 87 × x x = $$frac{14616}{87}$$ = 168

[#219] n being any odd number greater than 1, n 65 - n is always divisible by :
Correct Answer

(C) 24

Explanation

Solution: $$eqalign{
& Leftrightarrow {n^{65}} - n cr
& = nleft( {{n^{64}} - 1}
ight) cr
& = nleft( {{n^{32}} - 1}
ight)left( {{n^{32}} + 1}
ight) cr
& = nleft( {{n^{16}} - 1}
ight)left( {{n^{16}} + 1}
ight)left( {{n^{32}} + 1}
ight) cr} $$ $$ = nleft( {{n^8} - 1}
ight)left( {{n^8} + 1}
ight)left( {{n^{16}} + 1}
ight)$$ xa0 xa0 $$left( {{n^{32}} + 1}
ight)$$ $$ = nleft( {{n^4} - 1}
ight)left( {{n^4} + 1}
ight)left( {{n^8} + 1}
ight)$$ xa0 xa0 $$left( {{n^{16}} + 1}
ight)$$ xa0$$left( {{n^{32}} + 1}
ight)$$ $$ = nleft( {{n^2} - 1}
ight)left( {{n^2} + 1}
ight)left( {{n^4} + 1}
ight)$$ xa0 xa0 $$left( {{n^8} + 1}
ight)$$ xa0$$left( {{n^{16}} + 1}
ight)$$ xa0$$left( {{n^{32}} + 1}
ight)$$ $$ = left( {n - 1}
ight)nleft( {n + 1}
ight)left( {{n^2} + 1}
ight)left( {{n^4} + 1}
ight)$$ xa0 xa0 xa0 $$left( {{n^8} + 1}
ight)$$ xa0$$left( {{n^{16}} + 1}
ight)$$ xa0$$left( {{n^{64}} + 1}
ight)$$ xa0$$left( {{n^{32}} + 1}
ight)$$ Clearly, (n - 1), n and (n + 1) are three consecutive numbers and they have to be multiples of 2, 3 and 4 as n is odd. Thus, the given number is definitely a multiple of 24

[#220] 414 × ? × 7 = 127512
Correct Answer

(C) 44

Explanation

Solution: Let 414 × x × 7 = 127512 Then, $$eqalign{
& x = frac{{127512}}{{414 imes 7}} cr
& ,,,,,, = frac{{18216}}{{414}} cr
& ,,,,,, = frac{{2024}}{{46}} cr
& ,,,,,, = frac{{1012}}{{23}} cr
& ,,,,,, = 44 cr} $$