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x³ + y⁴ = z¹¹ 

Let x = 0 => y⁴ = z¹¹ => y = z^11/4 (many values of z of the form (integer)⁴ will give integral values of y. so infinite sets.

Explanation:

Using | a – b | ≥ | c |, we obtain |a – b|² ≥ |c|², which is equivalent to (a – b – c)(a – b + c) ≥ 0. Similarly, (b – c – a) (b – c + a) ≥ 0 and (c – a – b) (c – a + b) ≥ 0 .Multiplying these inequalities, We get 

- (a + b – c)² (b + c – a)² (c +a-b)² ≥ 0.

This forces that LHS is equal to zero. Hence it follows that either a + b = c, or b + c = a, or c + a = b

(166)_a × (56)_a = (8590)_a

=> [a³ + 6a² + 6] [5a + 6] = 8a³ + 9a² + 5a

=> 3a³ – 31a² – 57a – 36 = 0

=> (a – 12) (3a² + 5a + 3) = 0

a = 12, only possible value of a.

Explanation:

Suppose r pages of the book are torn off. Note that the page numbers on both the sides of a page are of the form 2k - 1 and 2k, and their sum is 4k - 1. The sum of the numbers on the torn pages must be of the form

4k₁ – 1 + 4k₂ – 1 +………….+4k_r – 1 = 4(k₁ + k₂ +…….+ k_r) – r

The sum of the numbers of all the pages in the untorn book is

1 + 2 + 3 +………+ 100 = 5050.

Hence the sum of the numbers on the torn pages is

5050 – 4949 = 101.

We therefore have

4(k₁ + k₂ +…….+ k_r) – r = 101.

This shows that r ≡ 3 (mod 4). Thus r = 4l + 3 for some l ≥ 0.

Suppose r ≥ 7, and suppose k₁< k₂ < k₃ <………<.k_r . Then we see that

4(k₁ + k₂ +……….+ k_r) – r ≥ 4 (k₁ + k₂ +……..….+ k₇) – 7 ≥ 4 (1 + 2 +………..+ 7) - 7

=4 x 28 - 7 = 105 > 101.

Hence r = 3. This leads to k₁ + k₂ + k₃ = 26 and one can choose distinct positive integers k₁, k₂, k₃ in several ways.

n^3pq – n = 0 (mod 3)

n^3pq – n = 0 (mod p)

n^3pq – n = 0 (mod q)

∴ We need to fulfill following conditions :

(i) (3 – 1) | (pq – 1) ⇒ pq is odd

(ii) (p – 1) | (3q – 1)

Now 3 must not divide (p – 1) as it doesn’t divide (3q – 1)

∴ p – 1 = 3k + 1 or 3k + 2, for some integer k.

∴ ⇒ p = 3k + 2 or 3k + 3

But p ≠ 3k + 3 (as its prime)

∴ p = 3k + 2 clearly p > 3 and k = odd = 2λ +

1 (say)

⇒ p = 6λ + 5

(ii) (q – 1) | (3p – 1) so q will also be 5 (mod 6)

by trial least values for p & q are 17 and 11.

∴ p + q = 28

Explanation:

Let the sides be x, x, y, where x, y are positive integers. Since we are looking for obtuse-angled triangle, y > x. Moreover ,2x + y = 2008 shows that y is even. But y < x + x, by triangle inequality. Thus y < 1004.Thus the possible triples (y, x, x) = (1002, 503, 503), (1000, 504, 504), (998, 505, 505), and so on. The general form is (y, x, x) = (1004 - 2k, 502 + k, 502 + k) , where k = 1, 2, 3,…., 501.

But the condition that the triangle is obtuse leads to

(1004 - 2k)² > 2(502 + k)² 

This simplifies to

502² + k² - 6(502)k > 0. 

Solving this quadratic inequality for k, we see that 

k <502 (3 - 2√2), or k > 502 (3 + 2√2),

Since k ≤ 501, we can rule out the second possibility. Thus k < 502(3 - 2√2), which is approximately 86.1432. We conclude that k ≤ 86. Thus we get 86 triangles

(y, x, x) = (1004 - 2k, 502 + k, 502 + k), k = 1, 2, 3, ………, 86.

The last obtuse triangle in this list is : (832, 588, 588). (It is easy to check that 832² - 588² - 588² = 736 > 0, whereas 830² - 589² - 589² = -4982 < 0).

Explanation:

We write

a² - 3a - 19 = a² - 3a - 70 + 51 = (a - 10)(a + 7) + 51

Suppose 289 divides a² - 3a - 19 for some integer a. Then 17 divides it and hence 17 divides (a - 10)(a + 7). Since 17 is a prime, it must divides (a - 10) or (a + 7). But (a + 7) - (a - 10) = 17. Hence whenever 17 divides one of (a - 10) and (a + 7), it must divide the older also. Thus 17² = 289 divides (a - 10)(a + 7). It follows that 289 divides 51, which is impossible. Thus, there is no integer a for which 289 divides a² - 3a - 19.

Suppose a = b. Then we get one equation a² = ac + 1. This reduces to a(a – c) = 1. Therefore a = 1, a - c = - 1. Thus we get (a, b, c)=(1, 1, 0)and (-1, -1, 0).

If a b, subtracting the second relation from the first we get

a2 - b2 = c(b - a).

Thus a² + b² + ab = 1.  

Multiplication by 2 gives (a + b)² + a² + b² = 2.

Thus (a, b) = (1, -1), (-1, 1), (1, 0), (-1, 0), (0, 1), (0, - 1).  

We get respectively = 0, 0, - 1,1, - 1, 1. 

Thus we get the triples:

(a, b, c)

= (1, 1, 0), (- 1, - 1, 0), (1, - 1, 0), (- 1, 1, 0), (1, 0, - 1), (- 1, 0, 1), (0, 1, - 1), (0, - 1, 1).

If a prime p  divides a, then p|b⁴ and hence p|b. This implies that p|c⁴ and hence |c. Thus every prime divides a also divides b and c. By symmetry, this is true for b & c as well. We conclude that a, b, c have the same set of prime divisors.

Let p^x ||a, p^y|| b and p^z||c. (Here we write px || a to mean px ||a and (|px + 1|)/a we may assume min {x, y, z} = x. Now b|c4 implies that y ≤ 4z : c| a⁴ implies that z ≤ 4x. we obtain

y ≤ z ≤ 16x

Thus x + y + z ≤ x + 4x + 16x = 21x  Hence the maximum power of p that dividesabc is x + y + z ≤ 21x. Since x is the minimum among x, y, z, px divides a, b, c. hence p^x divides a + b + c. This implies that p^21x divides (a + b + c)^21. since x + y + z ≤ 21x, it follows that p^{x + y + z} divides (a + b + c)²¹. this is true of any prime p dividing, c. hence abc divides (a + b + c)²¹.

Since all five numbers are 2 digit

∴ Number should have 2⁴ or 3⁴ included no other

possibility and common ratio should be 3/2 or 2/3.

∴ Numbers are 16, 24, 36, 54, 81.

Suppose that A' is a point on A such that ∠A'PC = ∠BPD. Then the segments OA' and OB subtends same angle in the respective minor arcs, so OA' = OB. This shows that A lies on Γ and hence A' =A.

This proves that ∠APC = ∠BPD.

Explanation:

We divide the even 4-digit numbers having non-zero digits into 4 classes: those ending in 2,4,6,8.

(A) Suppose a 4-digit number ends in 2. Then the second right digit must be odd in order to be divisible by 4. Thus the last 2 digits must be of the form 12, 32, 52, 72 or 92. If a number ends in 12, 52, or 92, then the previous digit must be even in order not to be divisible by 8 and w have 4 admissible even digits. Now the left most digit of such a 4-digit number can be any non-zero digit and there are 9 such ways, and we get 9x4x3 = 108 such numbers. If a number ends in 32 or 72, then the previous digit must be odd in order not to be divisible by 8 and we have 5 admissible odd digits. Here again the left most digit of such a 4-digit number can be any non-zero digit and there are 9 such ways,and we get 9x5x2=90 such numbers. Thus the number of 4- digit numbers having non-zero digits, ending in 2,divisible by 4 but not by 8 is 108 + 90 = 198.

 (B) If the number ends in 4, then the previous digit must be even for divisibility by 4. Thus the last two digits must be of the form 24, 44, 54, 84. If we take numbers ending with 24 and 64, then the previous digit must be odd for non-divisibility by 8 and the left most digit can be any non-zero digit. Here we get 952=90 such numbers. If the last two digits are of the form 44 ant 84 ,then previous digit must be even for non-divisibility by 8. And the left most digit Can be take 9 possible values. We thus get 9 x 4 x 2 = 72 numbers. Thus the admissible numbers ending in 4 is 90+72=162.

(C) If a number ends with 6 , then the last two digits must be of the form 16, 36, 56, 76, 96. For numbers ending with 16, 56, 76, the previous digit must be odd. For numbers ending with 36, 76, the previous digit must be even. Thus we get here (9 x 5 x 3)(9 x 4 x 2) = 135 + 72 = 207 numbers.

(D) If a numbers ends with 8, then the last two digits must be of the form 28,48,68,88. For numbers ending with 28,68, the previous digit must be even. For numbers ending with 48,88, the previous digit must be odd. Thus we get (9 x 4 x 2) + (9 x 5 x 2) = 72 + 90 = 162 numbers . Thus the number of 4-digit numbers, having non-zero digits. and divisible by 4 but not by 8 is

198 + 162 + 207 + 162 = 729.

Alternative Solution: 

If we take any four consecutive even numbers and divide them by 8, we get remainders 0, 2, 4, 6 in some order. Thus there is only one number of the form 8k + 4 among them which is divisible by 4 but not by 8. Hence we take four even consecutive numbers

1000a + 100b + 10c + 2, 1000a + 100b + 10c + 4, 1000a + 100b + 10c + 6, 1000a + 100b + 10c + 8,

There is exactly one among these four which is divisible by 4 but not by 8. Now we can divide the set of all 4-digit even numbers with nonzero digits into groups of 4 such consecutive even numbers with a, b, c nonzero. And in each group, there is exactly one number which is divisible by 4 but not by 8 . The number of such groups is precisely equal to 9x9x9 =729, since we can vary a, b, c in the set {1, 2, 3, 4, 5, 6, 7, 8, 9,}.

∵ Roots of (x² + ax + 20)(a² + 17x + b) = 0 are negative integers and we required (a + b)_min 

∴ a, b each should be minimum

a² – 80 is perfect square and a is a positive integer

⇒ a_min = 9

289 – 4b is a perfect square and b is a positive integer

⇒ b_min = 16                   (using 8, 15, 17)

∴ (a + b)_min = 25

Let pens be x and notebooks be y.

∴ 11x + 13y = 1000

11x = 1000 – 13y

when y = 5 then x = 85

Now every common number will occur at the interval of 11 × 13 = 143 is RHS starting with y = 5, RHS is

935, 792, ..., 66

∴ We obtain total 7 such ordered pairs.

i = 1 => 1 + (2 + 4 + 6 + 8 + 10 – 3 – 5 – 7 – 9)

= 1 – 4 + 10 = 7

i = 2 => 0 × 2 + (3 + 5 + 7 + 9 – 4 – 6 – 8 – 10)

= – 4

i = 3 => 1 × 3 + (4 + 6 + 8 + 10 – 5 – 7 – 9)

= 3 – 3 + 10 = 10

i = 4 => 0 × 4 + (5 + 7 + 9 – 6 – 8 – 10) = – 3

i = 5 => 1 × 5 + (6 + 8 + 10 – 7 – 9) = 5 – 2 + 10

= 13

i = 6 => 0 × 6 + (7 + 9 – 8 – 10) = – 2

i = 7 => 1 × 7 + (8 + 10 – 9) = 7 – 1 + 10 = 16

i = 8 => 0 × 8 + (9 – 10) = – 1

i = 9 => 1 × 9 + (10) = 19

∴ S = (7 + 10 + 13 + 16 + 19) – (4 – 3 – 2 – 1)

= 55

He can visit all 4 cities or 3 cities or 2 cities from B, C, D, E.

If he visits all 4 cities = 4! = 24

If he visits 3 cities = ⁴C₃.3! = 24

If he visits 2 cities = ⁴C₂.2! = 12

∴ Total ways = 60

Let

2a – b = n² …(i)

a – 2b = p² …(ii)

a + b = k² …(iii)

Adding (ii) and (iii) we get

2a – b = p² – k²

p² + k² = n² (p < k as a + b < a – 2b)

For b to be smallest, k² and p² is also small and

must be multiple of 3 (as 3b = k² – p²)

For smallest b, the least value of k and p be 12 and 9 resp.

∴ Least value of b = 21

∵ Sum of the roots = 0

∴ 1 + 2 + 3 + α = 0 ⇒ α = –6

∴ Product of roots = –c = 1 × 2 × 3 × –6

⇒ c = 36

Let  A∪B = Y, B\A = M, A\B =N & X\Y = L. Then X is the  disjoint union of M, N, L & A ∩ B. Now A ∩ B = {2, 3, 5, 7, 8} is fixed. The remaining seven elements 1, 4, 6, 9 10, 11, 12 can be distributed in any of remaining sets M N, L

This can be done in 3⁷ ways. Of there if all the elements are in the set L then A = B = {2, 3, 5, 7, 8} and this case has to be omitted. Hence the total number of pairs {A, B} such that A ⊆ X, B ⊆ X, AB and A ≠ B = {2, 3, 5, 7, 8} is 3⁷ - 1.

Explanation:

We observe that

f(4 + 5) = f(4.5) = f(20) = f(16 + 4) = f(16.4) = (64) = f(8.8) 

= f(8 + 8) = f(16) = f(4 + 4) = f(8).

If f(8) = 9, then f(9 = 9) This is one string. There may be other different way changing f(8) from f(9). The important thing to be observed is the fact that the f(x + y) = f(xy) applies only when x and y are at least 4. One may string using x and y which are smaller then 4, but that is not valid. For example

f(9) = f(3.3) = f(3 + 3) = f(6) = f(4 + 2) = f(4.2) = f(8), (is filled string.)

Let boys be 4x and girls be 3x.

∵ 4x – 8 = (3x – 14)²

∴ 9x² – 88x + 204 = 0

9x² – 54x – 34x + 204 = 0

(9x – 34) (x – 6) = 0

x = 6, 34/9

∴ Total number of students = 7x = 42

Explanation:

We observe that 0 + 1 + 2 + 3 = 6. Hence the remaining two digits must account for the sum 4. This is possible with 4 = 0 + 4 = 1 + 3 = 2 + 2. Thus we see that the digits in any such 6-digit number must be from one of the collections: 

{0, 1, 2, 3, 0, 4}, {0, 1, 2, 3, 1, 3} or {0, 1, 2, 3, 2, 2}.

Consider the case in which the digits are from the collection {0, 1, 2, 3, 0, 4}. Here 0 occurs twice and 1,2,3,4 occur one each. But 0 cannot be the first digit. Hence the first digit must be one of the 1,2,3,4. Suppose we fix 1 as the first digit. Then the number of 6-digit numbers in which the remaining 5 digits are 0,0,2,3,4 is 5!/2! = 60. Same is the case with other digits: 2, 3, 4.

Thus the number of 6-digit numbers in which the digits 0,1,2,3,0,4 occur is 60 x 4 = 240.

Suppose the digits are from the collection {0,1,2,3,1,3} .The number of 6- digit numbers beginning with 1 is 5!/2! = 60.  The number of those beginning with 2 is 5!/2!2! = 30.   and the number of those beginning with 3 is 5!/2! = 60. Thus the total number, in this case, is 60 + 30 + 60 =150. Alternately, we can also, count it as follows: the number of 6-digit numbers one can obtain from the collection {0,1,2,3,1,3} with 0 also as a possible first digit is 6!/2!2! = 180; the number of 6-digit numbers one can obtain from the collection {0,1,2,3,1,3} in which 0 is the first digit is 5!/2!2! = 30. Thus the number of 6- digit numbers formed by the collection {0,1,2,3,1,3} ,such that no number has its first 0 is 180-30=150.

Finally look at the collection {0,1,2,3,2,2}. Here the number of 6-digit numbers in which 1 is first digit 5!/3! = 20; the number of those having 3 as the first digit is 5!/3! = 20;  Thus the number of admissible 6-digit number here is 20+60+20=100.This may also be obtained using the other method of counting: 6!/3! - 5!/3! = 120 - 20 = 100;

Finally the total number of 6-digit numbers in which each of the digits 0,1,2,3 appears at least once is 240 + 150 + 100 = 490. 

Note that 12346 is even, 3 and 5 divide 12345, and 7 divides 12348. 

Consider a 5 digit number n = abcde with 0 < a < b < c < d < e < 10. 

Let S = (a + c + e) – (b + d). 

Then S = a + (c – b) + (e – d) > a > 0 and S = e - (d – c) – (b – a) < e ≤ 10, 

so S is not divisible by 11 and hence n is not divisible by 11. 

Thus 11 is the smallest prime that does not divide any fivedigit number whose digits are in a strictly increasing order.

One can chose 3 objects out of 32 objects in (³²₃) ways. Among these choices all would be together in 32 cases; exactly two will be together in 32 x 28 cases. Thus three object can be chosen such that no two adjacent in (³²₃) - 32 - (32 x 28) ways. Among these, further, two objects will be diametrically opposite in 16 ways and third would be on either semicircle in a non adjacent portion in 32 – 6 = 26 ways. Thus required number is

(³²₃) - 32 - (32 x 28) - (16 x 26) = 3616 

Number of line segment parallel to co-ordinate axis of 1 unit length equals to 8 × 9 × 2.

Number of line segment parallel to co-ordinate axis of 2 unit length equals to 7 × 9 × 2.

Number of line segment parallel to co-ordinate axis of 3 unit length equals to 6 × 9 × 2. and soon…..

Number of line segment parallel to co-ordinate axis of 8 unit length equals to 1 × 9 × 2.

Number of line segment not parallel to co-ordinate axis of 5 unit length equals to 5 × 8 × 2 × 2.

Number of line segment not parallel to co-ordinate axis of 10 unit length equals to 7 × 3 × 2 × 2.

Total line segment equals to 780.