Q.465. Remainder in a large division.

Question 465.
What is the remainder if 5^2009 + 13^2009 is divided by 18 ?

Answer 465.
Using,
For an odd positive integer n,
x^n + y^n
= (x + y) [x^(n-1) - x^(n-2)y + x^(n-2)y^2 - ... + y^(n-1)]

5^2009 + 13^2009
= (5 + 13) [5^2008 - 5^2007 * 13 + 5^2006 * 13^2 - ... + 13^2008]
= 18 * [5^2008 - 5^2007 * 13 + 5^2006 * 13^2 - ... + 13^2008]
=> 5^2009 + 13^2009 is divisible by 18 with zero remainder.

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Q.431. Root of a Complex Number - De Moivre's Theorem

Question 431.
Z^10 = -1 (Z is a complex number).
Prove that the answers Z1, Z2, Z3....Z10 are Geometrical progression.

Answer 431.
Z^10 = -1
=> Z
= (-1)^(1/10)
= (cosπ + isinπ)^(1/10)
= [cos(2k+1)π + i sin(2k+1)π]^(1/10)
= cos (2k+1) (π/10) + i sin(2k+1) (π/10)
k = 10, 1, 2, 3, 4, .... 9 give the values of Z1, Z2, Z3, ... Z10
For them to be in G.P., we need to prove that the ratio of two successive roots is the same

Let Zn and Zn+1 be two successive roots
=> Zn = cos(2n-1) (π/10) + i sin(2n-1) (π/10)
and Zn+1 = cos(2n+1) (π/10) + i sin(2n+1) (π/10)
=> Zn+1 / Zn
= [cos(2n+1) (π/10) + i sin(2n+1) (π/10)] / [cos(2n-1) (π/10) + i sin(2n-1) (π/10)]
= cos(π/5) + isin(π/5)
As this is independent of n, the ratio is the same for any two successive terms
=> Z1, Z2, Z3, ... Z10 are in G.P.

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Q.429. Addition/subtraction of numbers with base 7.

Question 429.
Suppose 3 four digits numbers with base seven were added and the 3rd four digit number was hidden, that I could not see it but the resulting number with base seven that was obtained when the 3 four digit numbers were added, was given. Find the 3rd four digit number (with base seven)  using the information given.
4321+1234+****=12341.

Answer 429.
4321 base 7
= 1*7^0 + 2*7^1 + 3*7^2 + 4*7^3
= 1 + 14 + 147 + 1372
= 1534

1234 base 7
= 4*7^0 + 3*7^1 + 2*7^2 + 1*7^3
= 4 + 21 + 98 + 343
= 466

12341 base 7
= 1*7^0 + 4*7^1 + 3*7^2 + 2*7^3 + 1*7^4
= 1 + 28 + 147 + 686 + 2401
= 3263

=> missing number in base 10 is
3263 - 1534 - 466
= 1263
Noting that 7^0 = 1, 7^1 = 7, 7^2 = 49, 7^3 = 343, 7^4 = 2401
1263
= 3*7^3 + 4*7^2 + 5*7^1 + 3*7^0
= 3453 (base 7) <=== Answer.

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Q.425. Arithmetic Progression

Question 425.
If 1/(a + b), 1/(b + c) and 1/(c + a) are in A.P., prove that
b^2, a^2 and c^2 are in A.P.

Answer 425.
If 1/(a+b), 1/(b+c) and 1/(c+a) are in A.P., then
1/(a+b) + 1/(c+a) = 2/(b+c)
=> (b+c)(c+a) + (a+b)(b+c) = 2(a+b)(c+a)
=> bc + c^2 + ab + ac + ab + b^2 + ac + bc
           = 2 (ac + bc + a^2 + ab)
=> b^2 + c^2 + 2(bc + ab + ac) = 2a^2 + 2(ab + bc + ca)
=> b^2 + c^2 = 2a^2
=> b^2, a^2 and c^2 are in A.P.

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Q.399. Sum of a finite series.

Question 399.
Find the sum of 1/(1*3) + 1/(3*5)+.....+1/(99*101).

Answer 399.
1/(1*3) + 1/(3*5) + ..... + 1/(99*101)
= (1/2) * [(3-1)/(1*3) + (5-3)/(5*3) + (7-5)/(7*5) + (9-7)/(9*7) + + (101-99)/(99*101)]
= (1/2) * [1 - 1/3 + 1/3 - 1/5 + 1/5 - 1/7 + ... + 1/99 - 1/101]
= (1/2) * (1 - 1/101)
= (1/2) * (100)/(101)
= (50)/(101).

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Q.385. Polynomial equation with coefficients in A.P.

Question 385.
Find a necessary and sufficient condition on a, b, c
such that the roots of x³ + ax² + bx + c = 0 are in arithmetic progression.

Answer 385.
Let the roots be m-d, m and m+d
=> m-d + m + m+d = - a
=> m = - a/3
m(m-d) + m(m+d) + (m-d)(m+d) = b
=> 3m^2 - d^2 = b
=> d^2 = a^2/3 - b
=> roots are
-a/3 - √(a^2/3-b), - a/3 and -a/3 + √(a^2/3-b)
=> product of the roots
(-a/3 - √(a^2/3 - b) * (-a/3) * (-a/3 + √(a^2/3 - b) = - c
=> - (a/3) * (a^2/9 - a^2/3 + b) = - c
=> (a/3) (b - 2a^2/9) = c
=> c = (a/27) (9b - 2a^2)
This is the necessary and sufficient condition for the roots to be in A.P.
Sufficient because the roots of the equation with the above value of c are the ones found as above for which refer to the following Wolfram Alpha link:
http://www.wolframalpha.com/input/?i=x%C2%B3+%2B+ax%C2%B2+%2B+bx+%2B+%28a%2F27%29%289b-2a%5E2%29+%3D+0

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Q.366. Compound interest - equatted monthly intalments of loan

Question 366.
A woman borrows $50 000 in order to buy a house. Compound interest at the rate of 12% per annum in charge on the loan. She agrees to pay back the loan in 25 equal instalments at yearly intervals, the first repayment being made exactly one year after the loan is taken out. Calculate the value of each instalment.

Answer 366.
Let the equal yearly instalments = $ x

Loan outstanding at the end of year 1 = $ 50000 * (1.12) - x
Loan outstanding at the end of year 2
= $ [50000 * (1.12) - x]*(1.12) - x
= $ 50000 * (1.12)^2 - x * (1 + 1.12)

Loan outstanding at the end of year 3
= $ [50000 * (1.12)^2 - x * (1 + 1.12)] * (1.12) - x
= $ 50000 * (1.12)^3 - x [1 + 1.12 + (1.12)^2]

Loan outstanding at the end of year 25
= $ 50000 * (1.12)^25 - x [1 + (1.12) + ... + (1.12)^24]

Loan outstanding at the end of year 25 should be zero
=> x * [1 + 1.12 + (1.12)^2 + ... + (1.12)^24] = 50000 * (1.12)^25
=> x * [(1.12)^25 - 1] / (1.12 - 1) = 850003.220332
=> x * (17.00006 - 1) = (0.12) * (850003.220332)
=> 16.00006 x = 102000.386
=> x = 6375
=> yearly equal instalments
= $ 6375 each year.

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Q.354. Roots of a polynomial which are in G.P.

Question 354.
Solve the equation x^3 - 19x^2 + 114x - 216 = 0 given that the roots are in G.P.

Answer 354.
Let the roots in G.P. be
a/r, a, ar
=>
a/r + a + ar = 19 and
(a/r) * a * (ar) = 216
=> a^3 = 216 => a = 6
Plugging in the first eqn.,
6/r + 6 + 6r = 19
=> 6r^2 - 13r + 6 = 0
=> 6r^2 - 4r - 9r + 6 = 0
=> 2r (3r - 2) - 3 (3r - 2) = 0
=> (2r - 3) (3r - 2) = 0
=> r = 3/2 or 2/3
=> the roots are
a/r = 6/(3/2) = 4
a = 6 and ar = 6 *(3/2) = 9
Answer: 4, 6 and 9.

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Q.347. nth term of a series

Question 347.
What is the nth term of 0,  2,   5,   9,  14,  .... ?

Answer 347.
It can be observed that tn - t(n-1) = n
=>
t2 - t1 = 2
t3 - t2 = 3
t4 - t3 = 4
.....
......
tn - t(n-1) = n
-----------------------
Adding,
tn - t1 = 2 + 3 + 4 + ... + n
=> tn - 0 = n(n+1)/2 - 1
=> tn = (1/2) (n^2 + n - 2).

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Q. 338. Series summation of powers of natural numbers.

Question 338.

Find Σ n^4.

Answer 338.

Knowing

1 + 2 + 3 + 4 + ... + n = (1/2)n(n+1)
1^2 + 2^2 + 3^2 + 4^2 + ... + n^2 = (1/6) n(n+1)(2n+1) and
1^3 + 2^3 + 3^3 + 4^3 + ... +n^3 = (1/4)n^2 (n+1)^2,
the sum of 4th powers of natural numbers can be found as under.

n^5 - (n-1)^5 = n^5 - (n^5 - 5n^4 + 10n^3 - 10n^2 + 5n - 1)
=> n^5 - (n-1)^5 = 5n^4 - 10n^3 + 10n^2 - 5n + 1

Plugging n = 1, 2, 3, 4, ... n

1^5 - 0^5 = 5*(1^4) - 10*(1^3) + 10*(1^2) - 5*(1) + 1
2^5 - 1^5 = 5*(2^4) - 10*(2^3) + 10*(2^2)- 5*(2) + 1
3^5 - 2^5 = 5*(3^4) - 10*(3^3) + 10*(3^2) - 5*(3) + 1
4^5 - 3^5 = 5*(4^4) - 10*(4^3) + 10*(4^2) - 5*(4) + 1
.......
.....
n^5 - (n-1)^5 = 5(n^4) - 10*(n^3) + 10*(n^2) - 5*(n) + 1
---------------------------------------…
Adding all,
n^5 = 5 Σn^4 - 10 Σn^3 + 10 Σn^2 - 5 Σn + Σ 1
=> 5 Σn^4
= n^5 + 10 Σn^3 - 10 Σn^2 + 5 Σn - n
= n^5 + (10/4)n^2 (n+1)^2 - (10/6) n (n+1) (2n+1) + (5/2) n (n+1) - n
= (1/12) [12n^5 + 30n^4 + 60n^3 + 30n^2 - 20(2n^3 + 3n^2 + n) + 30 (n^2 + n) - 12n]
= (1/12) (12n^5 + 30n^4 + 20n^3 - 2n)

= (1/6) (6n^5 + 15n^4 + 10n^3 - n)

=> Σn^4 = (1/30) (6n^5 + 15n^4 + 10n^3 - 30n^2 - 31n).

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Q.297. Geometric Progression

Question 297.
If a, b, c, d are in GP, show that (a^2 + b^2 + c^2)(b^2 + c^2 + d^2) = (ab + bc + cd)^2.

Answer 297.
Let a, b, c, d be
m, mr, mr^2, mr^3

=> (a^2 + b^2 + c^2) ( b^2 + c^2 + d^2)
= (m^2 + m^2r^2 + m^2r^4) (m^2r^2 + m^2r^4 + m^2r^6)
= m^4r^2 (1 + r^2 + r^4) (1 + r^2 + r^4)
= m^4r^2(1 + r^2 + r^4)^2
= (m^2r + m^2r^3 + m^2r^5)^2
= [m*(mr) + (mr)*(mr^2) + (mr^2)(mr^3)]^2
= (ab + bc + ca)^2.

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Q.292. Arithmetical Progression.

Question 292.
Find 4 terms of an AP whose sum is 26 and the sum of their squares is 189.

Answer 292.
Let the four terms be
(a - 3d), (a - d), (a + d) and (a + 3d)
Sum = 26
=> 4a = 26
=> a = 6.5

Sum of squares = 189
=> (a - 3d)^2 + (a - d)^2 + (a + d)^2 + (a + 3d)^2 = 189
=> 4a^2 + 20d^2 = 189
=> 20d^2 = 189 - 4a^2 = 189 - 4(6.5)^2 = 20
=> d^2 = 1
=> d = 1 or - 1

=> four terms are
3.5, 5.5, 7.5, 9.5
OR
9.5, 7.5, 5.5, 3.5.

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Q.290. Logarithm.

Question 290.
If a^(1/3) + b^3(1/3) + c^3(1/3) = 0,
show that log [(a+b+c)/3] = (1/3) (log a + log b + log c).

Answer 290.
a^(1/3) + b^(1/3) + c^(1/3) = 0
=> a^(1/3) + b^(1/3) = - c^(1/3)
 Cubing both sides,
[a^(1/3) + b^(1/3)]^3 = - c
=> [a^(1/3)]^3 + [b^(1/3)]^3 - 3[a^(1/3)] * [b^(1/3)] * [a^(1/3) * b^(1/3)] = - c
=> a + b - 3(ab)^(1/3) * (-c)^(1/3) = - c
=> a + b + c = 3(abc)^(1/3)
=> log [(a+b+c)/3] = (1/3) log(abc)
=> log [(a+b+c)/3] = (1/3) (loga + logb + logc).

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Q.289. Logarithm

Question 289.
If [logx/(a - b)] = [logy/(b - c)] = [logz/(c - a)], then prove that xyz = 1 and x^c y^a z^b = 1.

Answer 289.
Let logx / (a - b) = logy / (b - c) = logz / (c - a) = k

=> logx = k (a - b), log y = k (b - c) and logz = k (c - a)
=> logx + logy + logz = k (a - b + b - c + c - a) = 0
=> log (xyz) = 0
=> xyz = 1.

x^c y^a z^b
= Antilog log (x^c y^a z^b)
= Antilog [clogx + alogy + blogz)
= Antilog [ck(a - b) + ak(b - c) + bk(c - a)]
= Antilog 0
= 1.

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Q.253. Simple but tricky algebra question.

Question 253.
Find x^6 + 1, if (x + 1/x)^2 = 3.

Answer 253.
(x + 1/x)^2 = 3
=> x^2 + 2 + 1/x^2 = 3
=> x^4 - x^2 + 1 = 0

x^6 + 1
= (x^2)^3 + 1
= (x^2 + 1)(x^4 - x^2 + 1)
= (x^2 + 1) * 0
= 0.

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Q.247. Surds

Question 247.
Evaluate √(21 + 8√5) - √(9 + 4√5).

Answer 247.

21 + 8√5
= 21 + 2√(80)

= [√(16) + √5]^2

= (4 + √5)^2

=> √[21 + 8√5]

= (4 + √5)

9 + 4√5
= 9 + 2√(20)
= (√4 + √5)^2
= (2 + √5)^2
=> √(9 + 4√5)
= (2 + √5)

=> √(21 + 8√5) - √(9 + 4√5)
= (4 + √5) - (2 + √5)
= 2.

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Q.246. To find the equation of a quadratic curve with given conditions.

Question 245.
The quadratic function f(x) is negative for x > 9/2 and x < -1, but for no other value of x. If f(1) = 28 determine f(x) algebraically and sketch a graph of the function.

Answer 246.
Let f(x) = ax^2 + bx + c
f(1) = a + b + c = 28 ... ( 1 )
f(-1) = a - b + c = 0 ... ( 2 )
f(9/2) = 81a/4 + 9b/2 + c = 0 ... ( 3 )

Adding eqns. ( 1 ) and ( 2 ),
a + c = 14 ... ( 4 )
Multiplying eqn. ( 1 ) by 9/2 and subtracting from eqn. ( 3 ),
63a/4 - 7c/2 = -126
=> 9a/2 - c = - 36 ... ( 5)
Adding eqns. ( 4) and ( 5 ),
11a/2 = -22 => a = - 4
c = 14 - a = 14 + 4 = 18
b = a + c =18 - 4 = 14
f(x) = -4x^2 + 14x + 18 = - 2(2x - 9)(x + 1)

To sketch the graph of the parabola, f(x)
= - 4 (x^2 - 7x/2 - 9/2)
= - 4 [ (x - 7/4)^2 - 121/16 ]
=> f(x) - 121/4 = - 4 (x - 7/4)^2

This is a parabola with axis parallel to y-axis and vertex at (7/4, 121/4) which is the local maxima. The two arms of the parabola tend to the negative y-direction.

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Q.243. Algebraic inequality.

Question 243.
If a, b, c, d are distinct positive real numbers such that 3s = a + b + c + d, then prove that abcd > 81(s - a)(s - b)(s - c)(s - d).

Answer 243.
3s = a + b + c + d
=> a = 3s - b - c - d = (s - b) + (s - c) + (s - d)
 Now, AM ≥ GM
=> (1/3) [ (s - b) + (s - c) + (s - d) ] ≥ [ (s - b)(s - c)(s - d) ]^(1/3)
=> a ≥ 3 [ (s - b)(s - c)(s - d) ]^(1/3) ... (1)

Similarly,
b ≥ 3 [ (s - c)(s - d)(s - a) ]^(1/3) ... (2)
c ≥ 3 [ (s - d)(s - a)(s - b) ]^(1/3) ... (3)
d ≥ 3 [ (s - a)(s - b)(s - c) ]^(1/3) ... (4)

Multiplying equations (1), (2), (3) and (4),

abcd ≥ 81 [ (s - a)^3(s - b)^3(s - c)^3(s - d)^3 ]^(1/3)
=> abcd ≥ 81(s - a)(s - b)(s - c)(s - d)

As a, b, c and d are distinct positive real numbers, s - a, s - b, s - c and s - d are also distict positive numbers. Hence, only inequality holds.
=> abcd > 81 (s - a)(s - b)(s - c)(s -d).

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Q.225. Relation between roots of a polynomial

Question 225.
The equation x³ + px² + qx + r = 0 (where p, q, r are non zero) has roots α, β, γ
such that 1/α , 1/β, 1/γ are consecutive terms in an arithmetic sequence,
show that β = -3r / q.
The equation x³ - 26x² + 216x - 576 = 0, has roots α, β, γ such that 1/α , 1/β, 1/γ are consecutive terms in an arithmetic sequence. Find the values of α, β, γ.

Answer 225.
If 1/α , 1/β, 1/γ are in A.P., then
1/α + 1/γ = 2/β,
=> βγ + αβ = 2γα
=> βγ + αβ + γα = 3γα ... ( 1 )

Also,
x^3 + px^2 + qx + r = (x - α) (x - β) (x - γ)
=>
x^3 + px^2 + qx + r = x^3 - (α + β + γ)x^2 + (βγ + αβ + γα)x - αβγ
=> βγ + αβ + γα = q ... ( 2 )
and αβγ = - r ... ( 3 )

From ( 1 ) an ( 2 ),
3γα = q
=> 3αβγ = qβ

Plugging αβγ = - r from ( 3 ),
- 3r = qβ
=> β = - 3r/q.

x^3 - 26x^2 + 216x - 576 = 0
Now, β = -3r / q
=> β = - 3(-576)/216 = 8

Coefficient of x^2 is
α + β + γ = 26
=> α + γ = 26 - β = 26 - 8 = 18 ... ( i )
and αβγ = r
=> αγ = r/β = 576/8 = 72 ... ( ii )
 Solving ( i ) and ( ii ),
α = 6, β = 8 and γ = 12
OR
α = 12, β = 8 and γ = 6.

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Q.222. Algebra.

Question 222.
If x^4 + 1/x^4 = A, then what is x^5 + 1/x^5 in terms of A?
What is its positive, real value if A = 47?

Answer 222.
(x^2 + 1/x^2)^2 - 2 = A
=> x^2 + 1/x^2 = sqrt(A + 2)
=> (x + 1/x)^2 - 2 = sqrt(A + 2)
=> x + 1/x = sqrt [sqrt(A + 2) + 2]

x^5 + 1/x^5
= (x + 1/x) (x^4 - x^2 + 1 - 1/x^2 + 1/x^4)
= sqrt [sqrt(A + 2) + 2] * [A + 1 - sqrt(A + 2)]

For A = 47
x^5 + 1/x^5
= sqrt [sqrt(47 + 2) + 2] * [47 + 1 - sqrt(47 + 2)]
= 123.

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