SOME PROBLEMS

1)Find an integer whose decimal representation consists entirely of 1's and which is
divisible by 1987.
Solution:
Consider the numbers 1, 11, 111, 1111, ......., 111.....1111(1988 1's) .
The first one consist of 1 one, second one consists of 2 ones, third one consists of 3 ones
and so on. Since they are 1988 in total at least two of them have the same remainder when
divided by 1987. (remember that when any integer is divided by n the possible remainders
are 0, 1, 2, ........, n -1 . Now apply pigeon Hole Principle!! ).
Let those two numbers consists of exactly m 1's and n 1's respectively and without loss of
generality we assume that m > n . Then their difference which is divisible by 1987 (why!!!)
is equal to 1111.....100....00 (m- n 1's & n 0's) . We can cut out all the trailing zeroes
because these does not affect the divisibility by 1987 since neither 2 nor 5 is a factor of
1987 ( 10n is relatively prime to 1987 thus 1987 will divide 1111.....1 (m- n 1's)
Since 1111.....100....00 (m- n 1's & n 0's) =1111.....1 (m- n 1's)*10ⁿ ).
Thus we get the number 1111.....1 (m- n 1's) with the asked conditions.

2)The number of integers between 1 and 1000 (both inclusive) which are neither
perfect squares nor perfect cubes is?
Soln: We will approach the problem by finding the complement!
Number of perfect squares between 1 and 1000 are 31 (why!)
Number of perfect cubes between 1 and 1000 are 10 (why!)
Adding we get 41 and it seem the answer to be 1000-41=959 which is absolutely
wrong till we observe this!
1=1^2=1^3
64=4^3=8^2
729=9^3=27^2
so the correct answer is 959+3=962 (Remember if you have 10 rupee and if you count
it twice it will not become 20)!!!!!!

Questions:

1)1. A shop stores x kg of rice. The 1st customer buys half this amount plus half a kg of rice. The second customer buys half the remaining amount plus half a kg of rice. Then the third customer  also buys half the remaining amount plus half a kg of rice. There-after, no rice is left in the shop. Which of the following best describesthe value of x?
(1) 2 <= x<= 6
(2) 5 <= x<= 8
(3) 9 <= x<= 12
(4) 11 <= x <= 14
(5) 13 <= x<= 18  ('<=' means less than or equal to)

2)Directions for Questions 2 and 3: Let f(x) = ax² + bx + c, where
a, b and c are certain constants and a 6= 0. It is known that
f(5) = -3f(2) and that 3 is a root of f(x) = 0.
2. What is the other root of f(x) = 0?
(1) -7
(2) - 4
(3) 2
(4) 6
(5) Cannot be determined
3.What is the value of a + b + c?
(1) 9
(2) 14
(3) 13
(4) 37
(5) Cannot be determined

4) The number of common terms in the two sequences 17, 21,
25,..........., 417 and 16, 21, 26,.........., 466 is
(1) 78
(2) 19
(3) 20
(4) 77
(5) 22

5) Let f(x) be a function satisfying f(x)f(y) = f(xy) for all real x,
y. If f(2) = 4, then what is the value of f(1/2)
(1) 0
(2) 1/4
(3) 1/2
(4) 1
(5) cannot be determined

6)Consider obtuse-angled triangles with sides 8 cm, 15 cm and
x cm. If x is an integer, then how many such triangles exist?
(1) 5
(2) 21
(3) 10
(4) 15
(5) 14

7)How many integers, greater than 999 but not greater than 4000,
can be formed with the digits 0, 1, 2, 3 and 4, if repetition of digits
is allowed?
(1) 499
(2) 500
(3) 375
(4) 376
(5) 501

8)What is the number of distinct terms in the expansion of
(a + b + c)²⁰?
(1) 231
(2) 253
(3) 242
(4) 210
(5) 228