Olympiad-Classroom Assessment Practice Sheet O-CAPS-01 : Pre … · 2020. 6. 9. · O-CAPS-01 :...

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Olympiad-Classroom Assessment Practice Sheet

O-CAPS-01 : Pre-Regional Mathematics Olympiad (PRMO) (For VIII, IX, X Studying Students)

1. It is given that ( ) ( )( )

( ) ( )( )

2 22 22 22 2

2 23 3

1 1 1 1 1

1 1

x x x x x xb cx x

+ − + − + + × = −+ −

, where x ≠ ± 1 and 10 > b > c > 4,

b and c both are integers. Find the number of possible order pairs of (b, c).

2. Let p and q be distinct naturals such that 1981 + p = q2 and 1981 + q = p2. Find the value of (1990 + pq).

3. If 4012 200655 12 21 3 3 2 7 ,t+ × − = then find the value of 4012 37

2t + .

4. Find the last digit of the sum (20022005 + 20032006 + 20072007).

5. Let a, b, c, be positive integers less than 10 such that (100a + 10b + c)2 = (a + b + c)5, what is the value of (a × b – c)?

6. If a, b and c are real number such that a2 + b2 + 2c2 = 4a – 2c + 2bc – 5, then find the value of (6a – 4b – 2c).

7. Find total number of real solutions to the equation 2 2 25 12 13 .z z z+ =

8. Sx = ( )... ...x x x x x x+ + + − − − − , find the value of ( )15.

2x yS S+

9. For any positive integer a, b such that a > b, the difference of the squares of 2a + 1 and 2b + 1 is always

divisible by k, where k is an integer and 4 < k < 10. Find the value of 3 224 k× .

10. If Ts = 1 + 9 + 92 + 93 + 94 ...+ 9100, and unit digit of TS is n, then find n + 9.

11. Evaluate the product of 1 1 3 24 3 4 3

5 6 7 6 7 5 5 7 6 5 6 7

2 22 2

+ + + − + − + −

12. If n(A∩B) = 10, n(B∩C) = 20 and n(A∩C) = 30, then find the greatest possible value of n(A∩B∩C).

Topics Covered :

Mathematics : Sets, Number Theory, Polynomials

MATHEMATICS

(For VIII, IX, X Studying Students) O-CAPS-01 : Pre-Regional Mathematics Olympiad (PRMO)


13. If n(Ax) = x – 1 and A1 ⊂ A2 ⊂ A3 ⊂ A4... ⊂ A99, then find 99

1x

x

n A=

.

14. For {1, 2, 3,...n} and each of its non-empty subsets an alternating sum is defined as follows:

Arrange the number of the set/subset in decreasing order and then, beginning with the largest, alternately add and subtract successive numbers for example, the alternating sum for {1, 2, 4, 6, 9} is 9 – 6 + 4 – 2 + 1 = 6 and for {5} it is simply 5. Find the sum of all such alternating sum for n = 4.

15. If (15y15 – 11y + 1) p(y) = 2005 y2005, then find the unit digit of sum of coefficients of p(y).

16. Find the number of distinct real numbers which satisfy the equation (x2 + 4x – 2)2 = (5x2 – 1)2.

17. Let a1 = 97 and for n > 1, let an 1n

na −

= . Calculate the value of a1a2a3a4a5a6.

18. Let an = 6n + 8n. Find the remainder upon dividing a83 by 49.

19. The integer n is the smallest positive multiple of 15 such that each digit of n is either 4 or 0. Compute 1110

n.

20. The product of four distinct positive integer a, b, c and d is 40320. The numbers also satisfy ab + a + b = 322 and bc + b + c = 398. Find the value of d.

21. Find the sum of all positive rational numbers n such that 2 84 1941n n+ + is an integer

22. If 12p + 1 is the cube of a positive integer, where p is a positive odd integer, then find 3 12 1p +

23. If N is the least positive integer that is both 24% less than one integer and 12% greater than another integer. Find 3N − .

24. If f(x) is a monic polynomial of degree four such that f(–1) = –1, f(2) = –4, f(–3) = –9 and f(4) = –16, then find f(1).

25. If a, b, c, d, e are distinct integers such that (8 – a)(8 – b)(8 – c)(8 – d)(8 – e) = 12, then find a + b + c + d + e.

26. If the sum of the zeroes, product of the zeroes and the sum of the coefficients of the polynomial f(x) = ax2 + bx + c are equal and the sum of coefficients is 2, then find a.

27. Find the largest single-digit number by which the expression n3 – n is divisible for all possible integral values of n.

28. If f(x) = ax2 + bx + c is divided by x, x – 2 and x + 3, the remainders comes out to be 7, 9 and 49 respectively, then find (3a + 5b + 2c)2.

29. If a and b are positive integers such that a and b are not multiples of 3. Find the remainder when a2 + b2 is divided by 3.

30. If the numbers of solutions in positive integers of 2x + 3y = 763 is N, then find 3 2.N −


ANSWERS



1. (04) (E)

2. (10) (M)

3. (19) (E)

4. (04) (M)

5. (05) (D)

6. (18) (E)

7. (02) (E)

8. (15) (M)

9. (96) (M)

10. (10) (E)

11. (26) (D)

12. (10) (E)

13. (98) (E)

14. (32) (D)

15. (01) (M)

16. (03) (M)

17. (48) (E)

18. (35) (M)

19. (04) (E)

20. (07) (M)

21. (46) (D)

22. (13) (D)

23. (23) (M)

24. (23) (M)

25. (37) (E)

26. (02) (E)

27. (06) (E)

28. (04) (E)

29. (02) (E)

30. (05) (M)

Question Level Question Number

Easy (E) - 14 1, 3, 6, 7, 10, 12, 13, 17, 19, 25, 26, 27, 28, 29

Moderate (M) - 11 2, 4, 8, 9, 15, 16, 18, 20, 23, 24, 30

Difficult (D) - 05 5, 11, 14, 21, 22


ANSWERS & SOLUTIONS



1. Answer (04)

On solving

( ) ( )( )

( ) ( )( )

22 22 22 2

2 23 3

1 1 1 1 1

1 1

x x x x x xb cx x

+ − + − + + × = − + −

We get, 1 1, 1b c

b c= − =

−

⇒ (b, c) = (9, 8), (8, 7), (7, 6) (6, 5)

2. Answer (10)

1981 + p = q2, 1981 + q = p2

⇒ p – q = q2 – p2

⇒ p2 – q2 + p – q = 0

⇒ (p – q) (p + q + 1) = 0

⇒ p = q or p + q = – 1

But p and q are distinct numbers

∴ p + q = – 1 ...(1)

Now p2 + q2 = p + q + 3962

p2 + q2 = – 1 + 3962 (from (1))

(p + q)2 – 2pq = 3961

1 – 2pq = 3961

2pq = – 3960

pq = – 1980

⇒ 1990 + pq = 10

3. Answer (19) 4012 2006 55 12 21 3 3 2 7t = + × −

( )240124012 55 12 21 3 3 2 7t = + × −

( )( )4012 55 12 21 27 28 12 21t = + + −

( )( )4012 55 12 21 55 12 21t = + −

4012 3025 3024 1t = − =

4012 37 1 37 192 2

t + += =

4. Answer (04)

Unit digit of 20022005 is 2,

Unit digit of 20032006 is 9 and

Unit digit of 20072007 is 3

Thus unit digit of (20022005 + 20032006 + 20072007) is 2 + 9 + 3 = 14

∴ Last digit will be 4.

5. Answer (05)

Let N = (100a + 10b + c)2 = (a + b + c)5

⇒ N is both a square and fifth power

⇒ N must be tenth power

(100a + 10b + c) denotes a three digit number.

⇒ N is square of three digit number

∴ 210 = (25)2 = (32)2 is not possible

310 = (35)2 = (243)2 is possible.

410 = (45)2 = (1024)2 is not possible.

∴ N = (243)2 = (2 + 4 + 3)5

⇒ a = 2, b = 4, and c = 3

∴ (a × b – c) = 2 × 4 – 3 = 5

O-CAPS-01 : Pre-Regional Mathematics Olympiad (PRMO) - Ans & Sol (For VIII, IX, X Studying Students)


6. Answer (18)

a2 + b2 + 2c2 = 4a – 2c + 2bc – 5

a2 + b2 + 2c2 – 4a + 2c – 2bc + 5 = 0

(a2 – 4a + 4) + (b2 + c2 – 2bc) + (c2 + 1 + 2c) = 0

(a – 2)2 + (b – c)2 + (c + 1)2 = 0

⇒ a = 2, b = c and c = – 1

∴ 6a – 4b – 2c = 12 + 4 + 2 = 18

7. Answer (02)

2 2 25 12 13z z z+ = is satisfied only when

z2 = 2, since 52 + 122 = 132, 25 + 144 = 169.

∴ z2 – 2 = 0

( )( )2 2 0z z+ − =

2, 2z = −

8. Answer (15)

... ...xS x x x x x x= + + + − − − −

... and ...a x x x b x x x= + + = − − −

anda x a b x b= + = −

2 2 and a x a b x b= + = −

a2 – b2 = a + b

(a + b) (a – b – 1) = 0, a – b = 1

⇒ Sx = 1

Similarly

Sy = 1

∴ ( )15 15 2 152 2x yS S+ ×

= =

9. Answer (96)

(2a + 1)2 – (2b + 1)2 = 4a2 + 4a + 1 – 4b2 – 4b – 1

= 4(a2 + a – b2 – b)

= 4(a (a + 1) – b (b + 1))

As we know product of two consecutive numbers are always divisible by 2

∴ 4[a(a + 1) – b(b + 1)] is always divisible by 4 × 2 i.e. 8.

∴ K = 8

∴ 3 224 8 96× =

10. Answer (10) Unit digit of 1 + 9 = 0 Unit digit of 92 + 93 = 0 Unit digit of 998 + 9999 = 0 ∴ Unit digit of 1 + 9 + 92 ... 9100 = unit digit of

9100 = 1 ⇒ n + 9 = 10 11. Answer (26)

( )( )( )( )1 2 1 33 3 4 4

5 6 7 6 7 5 5 7 6 5 6 7

2+ + +

+ + + − + − + −

( ) ( ) ( ) ( )2 2 2 2

26 7 5 5 6 7

2

+ − − −

( )( )1 13 2 42 5 5 13 2 42

4= × + − − +

( )( )1 2 42 8 2 42 8

4= × + −

( ) ( )

2 21 12 42 8 104 264 4

= × − = × =

12. Answer (10) The greatest possible value of n (A∩B∩C) is the

least value amongst the value of n (A∩B), n (B∩C) and n (A∩C).

13. Answer (98) 99

1 2 3 4 991

...xx

A A A A A A=

= ∪ ∪ ∪ ∪

Since 1 2 3 4 99...A A A A A⊂ ⊂ ⊂ ⊂

∴ 99

991

xx

A A=

=

⇒ ( )99

991

99 1 98xx

n A n A=

= = − =

14. Answer (32) Denote the desired total of alternating sums of an

n elements set with Sn. We are looking S4. Note that all alternating sums of an n element set are also alternating sums of an (n + 1) elements set. However, when we go from n to (n + 1) element set for each subset with the new element we are adding the new element and subtracting one of the alternating sum of the n element set. There are 2n subset of an (n + 1) element set that includes the new element, giving us following relationship

Sn + 1 = Sn + 2n (n + 1) – Sn = 2n (n + 1) For n = 3, this becomes S4 = 23 (3 + 1) = 8 × 4 = 32

(For VIII, IX, X Studying Students) O-CAPS-01 : Pre-Regional Mathematics Olympiad (PRMO) - Ans & Sol


15. Answer (01) Sum of coefficients of p(y) can be obtained by

getting the value of p(1). (15y15 – 11y + 1) p(y) = 2005y2005 (15 – 11 + 1) p(1) = 2005 2005(1) 401

5p = =

∴ Unit digit of sum of coefficient p(y) is 1 16. Answer (03) (x2 + 4x – 2)2 = (5x2 – 1)2 (x2 + 4x – 2)2 – (5x2 – 1)2 = 0 (x2 + 4x – 2 + 5x2 – 1) (x2 + 4x – 2 – 5x2 + 1) = 0 (6x2 + 4x – 3) (– 4x2 + 4x – 1) = 0 ⇒ 4x2 – 4x + 1 = 0 6x2 + 4x – 3 = 0 D = 16 – 16 = 0 D = 16 + 72 = 88 Roots are equal and D > 0

real. Roots are distinct and real

∴ Distinct numbers satisfying equation is 3. 17. Answer (48)

Since 1

nn

naa −

= , an. an–1 = n

For n = 2, 4 and 6, we get a2 a1 = 2, a4a3 = 4 and a6a5 = 6 ⇒ a1a2a3a4a5a6 = 2.4.6 = 48 18. Answer (35) a83 = 683 + 883 a83 = (7 – 1)83 + (7 + 1)83 a83 = 2(783 + 3403 × 781 ... + 83 × 7) On dividing a83 by 49 we find that all terms are

divisible except last term 2 × 83 × 7, on dividing 2 × 83 × 7 = 1162 by 49 we get 35 as remainder.

1162 = 49 × 23 + 35 19. Answer (04) According to question n should be multiple of 5 and 3. Any multiple of 5 end with 0 or 5. ∴ n should ends with ‘0’. Also sum of digits of n should be divisible by 3,

hence 4 + 4 + 4 = 12 is only divisible by 3 as we can use only 4 or 0 as digits in ‘n’

∴ n = 4440

⇒ 4440 4

1110 1110n

= =

20. Answer (07)

ab + a + b = 322

⇒ (a + 1) (b + 1) = 323 = 17 × 19 ...(i)

and bc + b + c = 398

⇒ (b + 1) (c + 1) = 399 = 19 × 21 ...(ii)

From (i) and (ii), we get

a + 1 = 17 b + 1 = 19 c + 1 = 21

a = 16 b = 18 c = 20

⇒ abcd = 40320

⇒ 40320 7dabc

= =

21. Answer (46)

n has to be integer

(n + 42)2 < n2 + 84n + 1941 < (n + 45)2

n2 + 84n + 1941 is a perfect square

n2 + 84n + 1941 = (n + 43)2 ...(i)

n2 + 84n + 1941 = (n + 44)2 ...(ii)

On solving (i) and (ii), we get 5 , 46

4n =

∴ n = 46

22. Answer (13)

Since 12p + 1 is odd

Let 12p + 1 = (2a + 1)3

12p + 1 = 8a3 + 12a2 + 6a + 1

6p = a(4a2 + 6a + 3)

We know that p is a positive odd integer since 4a2 + 6a + 3 is an odd number

∴ a = 6

p = 4(6)2 + 6 × 6 + 3

= 183

∴ 33 12 1 12 183 1p + = × +

3 2197=

= 13

23. Answer (23)

If N is 24% less than one integer,

Let k, then 76 38

100 50N k k= =

When N is 12% greater than another integer say m,



112 28100 25

N m m= =

∴ k is divisible by 50 and m is divisible by 25

∴ 38 2850 25

k m=

∴ 38k = 56m 56

38k m=

2819

m=

∴ m = 19 × 25 = 475, k = 700

N = 532

∴ 3 532 3N − = −

= 23

24. Answer (23)

f(x) + x2 = 0

f(–1) = –1

f(2) = –4

f(–3) = –9

f(4) = –16

Since f(x) is a polynomial of degree 4

f(x) + x2 = (x + 1)(x – 2)(x + 3)(x – 4)

f(1) + 12 = (2)(–1)(4)(–3)

⇒ f(1) = 23

23

25. Answer (37)

The prime factorisation of 12 is 22 × 3.

∴ The 5 distinct integer factors must have some negative numbers in them.

There are two 2’s in the prime factorization, one of them must be negative and other positive

Distinct integer factors must be

–2, 2, 1, –1, 3

Taking a = 5

b = 6

c = 7

d = 9

e = 10

26. Answer (02)

Sum of roots b

c−

=

Product of roots ca

=

b cc a

−=

⇒ b = –c Sum of coefficients = a + b + c

= a + b – b ( b = –c)

= a ∴ a = 2 27. Answer (06) n3 – n = (n – 1)(n)(n + 1) for all integral values of n Since in a pair of consecutive integers, there is a

multiple of two and in a triplet of consecutive integers, there is a multiple of three

∴ n3 – n is divisible by 6 28. Answer (04) f(0) = 7, f(2) = 9 and f(–3) = 49 ∴ 3a + 5b + 2c = (–2)2 = 4 29. Answer (02) a and b are either 3k + 1 or 3k + 2 Let a = 3k + 1, b = 3k + 2 a2 + b2 = 18k2 + 18k + 5 = 3(6k2 + 6k + 1) + 2 ∴ The remainder is 2 30. Answer (05) 2x + 3y = 763

⇒ 763 32

yx −=

Since x is a positive integer ∴ 763 – 3y must be a positive even integer, so

that y must be a positive odd integer such that 3y ≤ 763.

There are 254 multiples of 3 less than 763, half of which are odd 2 half are even.

∴ N = 127 3 32 127 2N − = −

= 5




1. If a > b > c > d are the elements of the set and the pairwise sum of distinct elements of the set, in no

particular order are 185, 200, 211, 222, x and y. Find the sum of digits of greatest possible value of x + y.

2. If 1 3 ,

2aba b

+<

+ where a and b are positive integers and the maximum possible value of

3 3

3 31a b

a b+

+ is

xy

(x and y are relatively prime positive integers), then find x + y.

3. How many ordered pair of integers (a, b) with 0 ≤ a, b ≤ 100 satisfy ab = (a – 22)(b + 15)?

4. The number 27000001 have exactly four prime factors. Find the sum of two of the least factors.

5. If |x – 1| = |x – 2| + |x – 4|, then find the sum of all the solutions of the given equation.

6. If the number of subsets of a set with 2006 elements having an even number of elements is 2k, then find the sum of digits of k.

7. How many pairs of positive integers (a, b) are there such that hcf(a, b) = 1 and 149

a bb a

+ is an integer?

8. If m2003 ends with 3, then find unit digit of m.

9. If α is the non-real root of x2 + ax + b = 0 and α3 = 343, then find (a + b) if a and b are real.

10. Find the number of integers x satisfying 11 ≤ x ≤ 100 such that x2 + x – 110 is divisible by 19.

11. Find the value of 5xy, if

x2 + 10y2 + 1 ≤ 2y (3x –1).

12. What is the number of distinct primes which divide 14!+ 15! + 16!?

13. What is the smallest positive integral value of λ so that the equation x2 – (λ + 2)x + 2074 = 0 has integral roots?

14. Let f(x) be a polynomial of 99 degree satisfying f(k) = k, k = 1, 2, 3......99 and f(0) = 1, then find the value of f(–1).

15. Find the maximum value of λ such that 18λ divides 28!.

Topics Covered :

Mathematics : Sets, Number Theory, Polynomials

MATHEMATICS

(For VIII, IX, X Studying Students) O-CAPS-02 : Pre-Regional Mathematics Olympiad (PRMO)


16. If set A = 3 26 2: is an integer and ' ' itself is an integern nn n

n + +

, then find λ, where λ is total number of

distinct subsets of A.

17. If α and β are real numbers, satisfying α + β = k and αβ = k, where ‘k’ is a positive integer, then find the smallest value of k.

18. Find the sum of distinct remainders obtained when square of any integer is divided by 7.

19. If 3 52

x y++ = , where x and y are natural numbers, then find x + y.

20. If a – b = 3 and b – c = 5, then find the value of 2 2 2 – – –a b c ab bc ca+ + .

21. Find the unit digit of 777777 × 444444.

22. If x is the smallest natural number when multiplied by 15 and divided by 63 gives 21 as remainder, then find 2x.

23. How many real values of ‘a’ are there for which the cubic equation x3 – 3ax2 + 3ax – a = 0 has all real roots, one of which is ‘a’ itself?

24. How many non-negative integral pairs (x, y) are there for which (xy – 7)2 = x2 + y2?

25. If the maximum value for n such that n3 +100 is divisible by n + 10 is x, where x is any real number, then

find 10x

.

26. If x is any integer that satisfy the inequality 2 2 22 – 5 4 3 2 3 5 6 –x x x x x x+ + + − ≥ − , then find the

value of x2.

27. How many solutions of the equation x4 – x2 + 9 = 3x3 + x are negative?

28. If x, y are natural numbers satisfying equation x2 + y2 – 45x – 45y + 2xy – 46 = 0, then find the value of x + y.

29. What is the sum of the three positive integers a, b and c satisfying 1 39

3 7a

bc

+ =+

?

30. If ‘p’ is the root of x4 + x2 – 1 = 0, then find the value of ( )10086 42p p+ .


ANSWERS



1. (13) (M)

2. (36) (D)

3. (04) (M)

4. (50) (M)

5. (08) (E)

6. (07) (E)

7. (04) (D)

8. (07) (M)

9. (56) (E)

10. (08) (M)

11. (15) (M)

12. (06) (E)

13. (93) (M)

14. (99) (M)

15. (06) (M)

16. (16) (M)

17. (04) (M)

18. (07) (E)

19. (12) (E)

20. (07) (E)

21. (02) (E)

22. (28) (M)

23. (02) (D)

24. (04) (D)

25. (89) (M)

26. (09) (M)

27. (00) (E)

28. (46) (M)

29. (10) (M)

30. (01) (M)

Question Level Question Number

Easy (E) - 09 5, 6, 9, 12, 18, 19, 20, 21, 27

Moderate (M) - 17 1, 3, 4, 8, 10, 11, 13, 14, 15, 16, 17, 22, 25, 26, 28, 29, 30

Difficult (D) - 04 2, 7, 23, 24


ANSWERS & SOLUTIONS



1. Answer (13) a > b > c > d a + b > a + c > b + c > a + d > b + d > c + d a + b = x, a + c = y, b + c = 222, a + d = 211,

b + d = 200, c + d = 185 x + y = (a + b) + (a + c) = 2((a + d) + (b + c)) – ((c + d) + (b + d)) = 2(211 + 222) – (185 + 200) = 866 – 385 = 481 ∴ Sum of digits = 4 + 8 + 1 = 13 2. Answer (36) 1 3

2aba b

+<

+

⇒ 2ab + 2 < 3a + 3b ⇒ 4ab – 6a – 6b + 4 < 0 ⇒ (2a – 3)(2b – 3) < 5 (2a – 3)(2b – 3) = 1, 3 , –1, –3, – 5, –7... ⇒ (2a – 3, 2b – 3) = (1, 1), (1, 3), (3, 1) (1, –1),

(3, –1) (a, b) = (2, 2)(2, 3), (3, 2), (k, – (5, –1), (7, –1)... 3 3

3 31 65 31, ,1

16 5a ba b

+=

+

For maximum value, 3 3

3 31 31

5a b x

ya b+

= =+

∴ x + y = 31 + 5 = 36

3. Answer (04) ab = (a – 22)(b + 15), 0 ≤ a, b ≤ 100

Ordered pair (a, b) = (22, 0)(44, 15), (66, 30) and (88, 45)

= 4 4. Answer (50) Let x = 10 27000001 = 27 × 106 + 1 = 27x6 + 1 = (3x2)3 + 1 = (3x2 + 1)(9x4 – 3x2 + 1) = (3x2 + 1) [(3x2 + 1)2 – 9x2] = (3x2 + 1)(3x2 + 3x + 1)(3x2 – 3x + 1) = (301)(331)(271) = 7 × 43 × 271 × 331 = Sum = 7 + 43 = 50 5. Answer (08) ∴ x = 3 and 5 are the only solution

∴ 3 + 5 = 8

6. Answer (07) Total number of subsets = 22006 Half of the subsets will have an odd number of

elements and other half will have an even number of elements

So, ( )2006 20051 2 2 22

k= =

k = 2 + 0 + 0 + 5 = 7



7. Answer (04)

2 214 9 14

9 9a b a bb a ab

++ = must be an integer

Since the denominator contains a factor of 9 9 | 9a2 + 14b2

⇒ 9 | b2 ⇒ 3 | b Since, b = 3n for some positive integer n

∴ 2 214

3a n

an+

The denominator now contains a factor of n n | a2 + 14n2 ⇒ n | a2 But since 1 = hcf(a, b) = hcf(a, 3n) = hcf(a, n) ∴ n = 1 and thus b = 3

for b = 3, 2 143

aa+

for integer a must be a factor of 14 a ∈ {1, 2, 7, 17} ∴ There are four solutions (1, 3)(2, 3)(7, 3)(14, 3) = 4 8. Answer (07) 2003 = 4(500) + 3 The unit digit of m2003 is 3 Let m ends with digit 2 then unit digit of m2003 is

same as unit digit of 22003 22003 = (24)500 × 23 = 16500 × 8, product ends with 8, not 2 Let unit digit of m is 7 then 72003 = (74)500 × 73 Since 74 ends with 1 and 73 end with 3, the

product end with 3, which is the required digit ∴ Unit digit of m is 7 9. Answer (56) α3 = 343 ⇒ α3 – 343 = 0 ⇒ (α –7)(α2 + 7α + 49) = 0 ⇒ α = 7 or α2 + 7α + 49 = 0 α ≠ 7, since α is non-real. ⇒ α2 + 7α + 49 = 0 D = (7)2 – 4(1)(49) = –147 D < 0

Since, α is also a non-real root of ax2 + bx + c

α2 + aα + b = 0 a = 7, b = 49 ∴ a + b = 56 10. Answer (08) x2 + x – 110 = (x – 10)(x + 11) is divisible by 19 either (x – 10) or (x + 11) divisible by 19 Integers x with 11 ≤ x ≤ 100 and (x – 10) divisible

by 19 are 29, 48, 67, 86 (x + 11) divisible by 19 are 27, 46, 65, 84 ∴ {27, 29, 46, 48, 65, 67, 84, 86} Required number of integers = 8 11. Answer (15) x2 + 10y2 + 1 – 6xy + 2y ≤ 0 ⇒ x2 + 9y2 – 6xy + 2y + y2 + 1 ≤ 0 ⇒ (x – 3y)2 + (y + 1)2 ≤ 0

⇒ x = 3y, y = –1 12. Answer (06) 14! (1 + 15 × 17) = 256 × 14! ⇒ Numbers are 2, 3, 5, 7, 11, 13 13. Answer (93) αβ = 2074 = 2 × 17 × 61 Positive roots can be {1, 2074}, {2, 1037},

{17, 122}, {61, 34} ⇒ λ + 2 = 95

λ = 93 14. Answer (99) Let g(x) = f(x) – x g(k) = 0 for k = 1, 2, 3.....99 g(x) is polynomial of degree 100 whose roots are 1, 2, 3......99. g(x) = c(x – 1)(x – 2)(x – 3)... (x – 99) f(x) = g(x) + x f(x) = x + c (x – 1)(x – 2) (x – 3)...(x – 99) [at x = 0] f(0) = c(–1)(–2)(–3)... (–99) 1–

99!c =

and f(–1) = 1–1–

99! (–2) (–3) ... (–100)

f(–1) = 100!–199!

+

f(–1) = 99

(For VIII, IX, X Studying Students) O-CAPS-02 : Pre-Regional Mathematics Olympiad (PRMO) - Ans & Sol


15. Answer (06)

18λ = (2 × 32) λ

228 28 28 28 252 4 8 16

H = + + + =

328 28 28 133 9 27

H = + + =

28! = 225 × 313 × p1 = (2 × 32)6 × p2

⇒ λ = 6

16. Answer (16) According to the condition

3 26 2n n

n+ + should be integer (where n is integer)

⇒ n2 + 6n + 2n

will be integer.

⇒ 2n

should be integer for different value of n.

⇒ n = – 1, – 2, 1, 2 will satisfy.

∴ λ = 24 = 16

17. Answer (04) According to condition,

α and β must be the roots of x2 – kx + k = 0

⇒ D ≥, 0 (α and β are real)

⇒ k2 – 4k ≥ 0

k (k – 4) ≥ 0

k is positive

∴ k – 4 ≥ 0

⇒ k ≥ 4

18. Answer (07) An integer n is of form 7k, 7k ± 1, 7k ± 2,

7k ± 3

⇒ n2 is of form 7k, 7k + 1, 7k + 4, 7k + 9.

On dividing n2 by 7 remainders are 0, 1, 4, 2. 19. Answer (12) 5 13 5

2 2+ = +

5 1 22 2+

= ×

10 22+

=

∴ x + y = 12

20. Answer (07)

c – a = (c – b) + (b – a)

= –(b – c) – (a – b)

= –5 – 3 = –8

∴ a2 + b2 + c2 – ab – bc – ca

( ) ( ) ( )2 2 212

a b b c c a = − + − + −

[ ]1 9 25 64

2= + +

= 49

21. Answer (02)

Unit digit of 777777 = 7

Unit digit of 444444 = 6

∴ 7 × 6 = 42

∴ Required unit digit is 2.

22. Answer (28)

15x – 21 = 63n, n ∈ N

∴ 15x = 63n + 21

⇒ x = 74

5nn +

+

∴ n = 3 (x ∈ N)

∴ ( ) 3 74 35

x += +

= 14

2x = 14 × 2

= 28

23. Answer (02)

f(x) – x3 – 3ax2 – 3ax – a – 0

a3 – 3a3 + 3a2 – a = 0

⇒ 2a3 – 3a2 + a = 0

⇒ a(2a2 – 3a + 1) = 0

⇒ a = 0

2 12 – 3 1 0 1,2

a a a+ = ⇒ =

∴ but 1,2

a = at the cubic has only one red root

24. Answer (04)

(xy – 7)2 = x2 + y2

⇒ (xy)2 – 14xy + 49 – x2 + y2

⇒ 13 = [(x + y) + (xy – 6)][(x + y) – (xy – 6)]



Since 13 is a prime number, the only possible factors are ±1 and ±13.

Case-I (x + y) + (xy – 6) = 13 and (x + y) – (xy – 6) = 1

Case-II (x + y) – (xy – 6) = –13 and (x + y) + (xy – 6) = –1

∴ (7, 0), (0, 7), (3, 4), (4, 3) are the solutions.

25. Answer (89)

Let 3 2100 ( 10)( )n n n an b c+ = + + + +

( )3 2 10 ( 10 ) 10n n a n b a b c= + + + + + +

Equating coefficients,

10 + a = 0

b + 10a = 0

10b + c = 100

a = –10, b = 100 and c = –900

∴ Maximum value for n is 890.

26. Answer (09)

x2 – 5x + 4 ≥ 0

(x – 1) (x – 4) ≥ 0

x ≥ 1, x ≥ 4

x2 + 2x – 3 ≥ 0

(x + 3) (x – 1) ≥ 0

x ≤ –3 or x ≥ 1

6 – x – x2 ≥ 0

(x + 3) (x – 2) ≤ 0

–3 ≤ x ≤ 2

All roots are defined at x = –3 and x = 1.

At, x = –3, the inequality becomes

2 28 0≥

At x = 1, the inequality becomes 0 5 6≥

∴ Only x = –3 satisfies.

x2 = 9

27. Answer (00)

When we put (–x) in equation, we get always +ve in L.H.S. and –ve in R.H.S.

28. Answer (46)

x2 + y2 – 45x – 45y + 2xy – 46 = 0

⇒ x2 + y2 + 2xy – (45x + 45y) –46 = 0

⇒ (x + y)2 – 45(x + y) – 46 = 0

⇒ (x + y + 1) (x + y – 46) = 0

x + y = –1 is not possible since x and y are natural numbers.

∴ x + y = 46

29. Answer (10)

1 39 4 1 15 5 53 7 37 7 1

4 4

ab

c

+ = = + = + = ++ +

a + b + c = 5 + 1 + 4 = 10

30. Answer (01)

p4 + p2 – 1 = 0

p4 = 1 – p2 ...(i)

p6 = p2 – p4

p6 + p4 = p2

p6 + p4 = 1 – p4 [from (i)]

p6 + 2p4 = 1

⇒ (p6 + 2p4)1008 = (1)1008 = 1

Edition: 2020-21

Olympiad-Classroom Assessment Practice Sheet O-CAPS-01 : Pre … · 2020. 6. 9. · O-CAPS-01 :...

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