Integer Answer Type Questions for JEE: Mathematical induction & Binomial Theorem

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 ) = 5- 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 m– 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)3m + 1 = (23)5m + 1.22 = (7 + 1)5m + 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

The document Integer Answer Type Questions for JEE: Mathematical induction & Binomial Theorem | Chapter-wise Tests for JEE Main & Advanced is a part of the JEE Course Chapter-wise Tests for JEE Main & Advanced.
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## Chapter-wise Tests for JEE Main & Advanced

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## Chapter-wise Tests for JEE Main & Advanced

447 docs|930 tests

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