Integer Answer Type Questions for JEE: Mathematical induction & Binomial Theorem | Chapter-wise Tests for JEE Main & Advanced PDF Download

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Q.1. Let P (n) = 5ⁿ - 2ⁿ , P(n)  is divisible by 3λ where λ and n both are odd positive integers then the least value of n and λ will be

Ans. 1
P (n ) = 5ⁿ - 2ⁿ 
n = 1
∴ P (5) = 55 - 25
= 3125 - 32
= 3093
= 3 × 1031
In this case λ = 1031
Similarly we can check the result for other cases and find that the least value of λ and n is 1.

Q.2. The greatest positive integer. Which divides (n + 16) (n + 17) (n + 18) (n + 19), for all n ∈ N, is-

Ans. 24
Since product of any r consecutive integers is divisible by r! and not divisible by r + 1!.
So given product of 4 consecutive integers is divisible by 4! or 24.

Q.3. The sum of the cubes of three consecutive natural numbers is divisible by-

Ans. 9
Let three consecutive natural numbers are n, n + 1, n + 2, P(n)
= (n)³ + (n + 1)³ + (n + 2)³
P(1)=1³ + 2³ + 3³ = 36, which is divisible by 2 and 9
P(2) = (2)³ + (3)³ + (4)³ = 99, which is divisible by 9 (not by 2).
Hence P(n) is divisible 9 ∀  n ∈ N.

Q.4. The difference between an +ve integer and its cube is divisible by-

Ans. 6
Let n is a positive integer.
P(n) = n³ – n P(1) = 0,
which is divisible by for all n ∈ N P(2) = 6,
which is divisible by 6 (not by 4 and 9)

Q.5. If 10n + 3.4ⁿ⁺² + λ is exactly divisible by 9 for all n ∈ N, then the least positive integral value of λ is-

Ans. 5
Let P(n) = 10n + 3.4ⁿ⁺² + λ is divisible by 9 ∀  n ∈ N.
P(1) = 10 + 3.43 + λ = 202 + λ = 207 + (λ - 5)
Which is divisible by 9 if λ = 5

Q.6. The remainder when 5⁹⁹ is divided by 13 is

Ans. 8
5⁹⁹ = (5) (5²)⁴⁹ = 5(25)⁴⁹ = 5(26 -1)⁴⁹
= 5 x (26) x (Positive terms)5, So when it is divided by 13 it gives the remainder - 5 or (13 - 5)
i.e., 8.

Q.7. For every natural number n, n (n² - 1) is divisible by 

Ans. 6
n(n² -1) = (n -1) (n) (n + 1)
It is product of three consecutive natural numbers, so according to Langrange's theorem it is divisible by 3 ! i.e., 6.

Q.8. If m, n are any two odd positive integer with n < m then the largest positive integers which divides all the numbers of the type m² – n² is:

Ans. 8
Let m = 2k + 1
n = 2k –1 as n < m.
m² – n² = (2k + 1)² – (2k – 1)²
= 4k² + 1 + 4k – 2k² – 1 + 4k
P(k) = 8k
P(1) is divisible by 8, P(2) is divisible by 8
P(k) is divisible by 8

Q.9. For all n ∈ N, 49n + 16n - 1 is divisible by

Ans. 64
P(n) : 49ⁿ + 16n –1
P(1) : 49 + 16 – 1, divisible by 64, 16, 8, 4
P (n + 1) = 49ⁿ⁺¹ + 16n +16 – 1
= 49ⁿ.49 + 16n + 16 –1 + 49. 16n –49 –49.16n + 49
= 49(49ⁿ +16n – 1) – 48.16n + 64
= 49 (49ⁿ + 16n – 1) + 64(1 – 12n)
P(n + 1), is divisible by 64,
So, P(n) is divisible by 64

Q.10. The greatest positive integer, which divides (n + 2) (n + 3) (n + 4) (n + 5) (n + 6) for all  n∈N is

Ans. 120
Product of r consecutive integers is divisible by r!.
So given expression is divisible by 5! i.e. 120.

Q.11. If x = (7 + 4√3)²ⁿ = [x] + f,  then x(1 - f) is equal to _____.

Ans. 1
We have
∴ 0 < 7 - 4√3 < 1
⇒ 0 < (7 - 4√3)²ⁿ < 1
Let F = (7 - 4√3)²ⁿ. Then

= 2m, where m is some positive integer.
⇒ [x] + f + F = 2m
⇒ f + F = 2m - [x]
Since 0 ≤ f < 1 and 0 < F < 1, we get 0 < f + F < 2.
Also, since f + F is an integer, we must have f + F = 1. Thus,
x(1 - f) = xF = (7 + 4√3)²ⁿ (7 - 4√3)²ⁿ = (49 - 48)²ⁿ = 1²ⁿ = 1.

Q.12. The coefficient of a¹⁰b⁷c³ in the expansion of (bc + ca + ab)¹⁰ is _____.

Ans. 120
The general term in the expansion of (bc + ca + ab)¹⁰ is

where r + s + t = 10
For coefficient of a¹⁰b⁷c³ we set t + s = 10, r + t = 7, r + s = 3.
Since r + s + t = 10, we get r = 0, s = 3, t = 7.
Thus, the coefficient of a¹⁰b⁷c³ in the expansion of

Q.13. When  is divided by 7, the remainder is ______ .

Ans. 4
We have     32³² = (2⁵)³² = 2¹⁶⁰
=
= 3m + 1, where m is some positive integer.
Now,  = (2⁵)3^{m + 1} = (2³)^{5m + 1}.2² = (7 + 1)5^{m + 1}.4
But (7 + 1)^{5m + 1} = 7n + 1 for some positive integer n.
Thus, N = = 28n + 4.
Thus, shows that when N is divided by 7, the remainder is 4.

Q.14. The digits at unit’s place in the number 17¹⁹⁹⁵ + 11¹⁹⁹⁵- 7¹⁹⁹⁵ is

Ans. 1
We have
17¹⁹⁹⁵ + 11¹⁹⁹⁵ - 7¹⁹⁹⁵  = (7 + 10)¹⁹⁹⁵ + (1 + 10)¹⁹⁹⁵ - 7¹⁹⁹⁵
=
=
= a multiple of 10 + 1
Thus, the unit’s place digits is 1.

Q.15. Find the number of terms and coefficient of  x⁵ in ( 1+x+x²)⁷.

Ans. 266
Here  the  variables 1, x, x² are  not independent so the general formula is not applicable but as power of  x varies from 0 to 14, therefore total number of terms = 15.  Now x5 can be formed in three ways: 1² x⁵(x²)⁰, 1³ x³( x² )¹ or
1⁰ x¹ (x²)² , so total arrangement of coefficient =  
= 266