Number System - Study Mode

[#321] The unit digit in the sum of $${left( {124}
ight)^{372}},$$xa0 $$ + ,{left( {124}
ight)^{373}}$$ xa0= ?
Correct Answer

(D) 0

Explanation

Solution: $$eqalign{
& {left( {124}
ight)^{372}},,,,,,,,,,,,,{left( {124}
ight)^{373}} cr
& ,,,mathop {{4^{372}}}limits^ downarrow ,,,,,,,,,,,,,,,,,,,,,mathop {{4^{373}}}limits^ downarrow cr} $$ When 4 has odd power then unit digit is 4, when 4 has even power then unit digit is 6 4 1 → 4 → 4 4 2 → 16 → 6 4 3 → 64 → 4 4 4 → 256 → 6 4 372 4 373 ↓ ↓ 6 + 4 = 10 = 0 Last (unit) digit = 0

[#322] The number of integers in between 100 and 600, which are divisible by 4 and 6 both, is :
Correct Answer

(C) 41

Explanation

Solution: Take L.C.M. of and 6 = 12 and numbers should be divisible by 12. First number (after 100) = 108 Last of integers $$eqalign{
& = frac{{{ ext{ Last term }} - { ext{ First term}}}}{{{ ext{Difference}}}} + 1 cr
& = frac{{588 - 108}}{{12}} + 1 cr
& = frac{{480}}{{12}} + 1 cr
& = 40 + 1 cr
& = 41 cr} $$

[#323] Find the least number of five digits which is divisible by 333 ?
Correct Answer

(B) 10323

Explanation

Solution: Five digit least value = 10000 10000 is divided by 333 therefore, remainder = 10 Number = 10000 + 333 - 10 = 10323

[#324] One of the factors of (8 2k + 5 2k ), where k is an odd number, is:
Correct Answer

(C) 89

Explanation

Solution: 8 2k + 5 2k If K is odd number Then, K = 1, 3, 5 _ _ _ _ 8 2 × 1 + 5 2 × 1 = 64 + 25 = 89

[#325] The numerator of a fraction is multiple of two numbers. One of the numbers is greater than the other by 2. The greater number is smaller than the denominator by 4. If the denominator 7 + C (C > - 7) is a constant, then the minimum value of the fraction is:
Correct Answer

(D) -1/5

Explanation

Solution: According to the given condition Denominator = 7 + C [C > -7] [x08egin{array}{*{20}{c}}
{{ ext{Denominator}} = }&{{ ext{Greator number}}}&{}&{{ ext{Smaller number}}} \
{}&{left( {7 + C - 4}
ight)}& imes &{left( {7 + C - 4 - 2}
ight)}
end{array}] Equation $$ = frac{{left( {3 + C}
ight)left( {1 + C}
ight)}}{{7 + C}}$$ By putting value according to option C = -2 for minimum value = -1/5