Subject
Quantitative Aptitude
Topic
Algebra
Find the largest positive integer n such that n3+ 100, is divisible by n + 10.
Find the largest positive integer n such that n3+ 100, is divisible by n + 10.
ক)
890খ)
920গ)
990ঘ)
940ঙ)
None of theseExplanation
We know that, a 3 + b 3 = (a + b) (a 2 –ab + b 2 ) ∴ n 3 + 10 3 = n 3 + 1000 = (n + 10) (n 2 – 10n + 100) Thus, (n + 10) divides (n 3 + 1000). Given equation, n 3 + 100 = (n 3 + 1000) – 900 Since, (n + 10) divides (n 3 + 100) and (n 3 + 1000), it must divide 900 also. So, largest value of (n + 10) = 900 ∴ n = 890Related questions
if then what would be the value of ?Multiplication and difference of two numbers are 93 and 28 respectively. Find the difference of reciprocals of the numbers.if , then find the value of In following question, two equations are given, you have to solve them and choose the correct option:
x2– 3x – 4 = 0,
y2– 4 = 0In following question, two equations are given, you have to solve them and chose the correct option:
x2+ 11x + 30 = 0,
y2+ 7y + 12 = 0In the following question, one or two equation(s) is/are given. You have to solve both the equations and find the relation between ‘l’ and ‘m’ and mark correct answer.
I. (125)1/3l – (1331)1/3= 4 × (729)1/6+ (1024)1/10II. (m)2– (15625)1/3= 0Direction: In the following question two equations numbered I and II are given. You have to solve both the equations and give answers:I: x - √121 = 0
II: y2- 121 = 0The number of solutions of x2+ 4|x| + 5 = 0 is