Q.1. Let P (n) = 5^{n}  2^{n} , 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^{n } 2^{n}
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)=1^{3} + 2^{3} + 3^{3} = 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) = n^{3} – 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 + 3.4^{n+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 + 3.4^{n+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 5^{99} is divided by 13 is
Ans. 8
5^{99} = (5) (5^{2})^{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 (n^{2}  1) is divisible by
Ans. 6
n(n^{2} 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^{2 }– n^{2} is:
Ans. 8
Let m = 2k + 1
n = 2k –1 as n < m.
m^{2} – n^{2} = (2k + 1)^{2} – (2k – 1)^{2}
= 4k^{2} + 1 + 4k – 2k^{2} – 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) : 49^{n} + 16n –1
P(1) : 49 + 16 – 1, divisible by 64, 16, 8, 4
P (n + 1) = 49^{n+1} + 16n +16 – 1
= 49^{n}.49 + 16n + 16 –1 + 49. 16n –49 –49.16n + 49
= 49(49^{n} +16n – 1) – 48.16n + 64
= 49 (49^{n} + 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} = 1^{2n} = 1.
Q.12. The coefficient of a^{10}b^{7}c^{3} 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 a^{10}b^{7}c^{3} 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 a^{10}b^{7}c^{3} in the expansion of
Q.13. When is divided by 7, the remainder is ______ .
Ans. 4
We have 32^{32} = (2^{5})^{32} = 2^{160}
=
= 3m + 1, where m is some positive integer.
Now, = (2^{5})3^{m + 1} = (2^{3})^{5m + 1}.2^{2} = (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 17^{1995} + 11^{1995} 7^{1995} is
Ans. 1
We have
17^{1995} + 11^{1995 } 7^{1995 } = (7 + 10)^{1995} + (1 + 10)^{1995}  7^{1995}
=
=
= a multiple of 10 + 1
Thus, the unit’s place digits is 1.
Q.15. Find the number of terms and coefficient of x^{5} in ( 1+x+x^{2})^{7}.
Ans. 266
Here the variables 1, x, x^{2} 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: 1^{2} x^{5}(x^{2})^{0}, 1^{3} x^{3}( x^{2} )^{1} or
1^{0} x^{1} (x^{2})^{2} , so total arrangement of coefficient =
= 266
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