2)
Let 100a +10b+ c =A
a + 100b + 10x = Ar
10a + b + 100 c = Ar2
Add all 111 (a + b + c) = A (1 + r + r²)
37 x 3 (a + b + c) = A (1 + r + r²)
A is 3 digit number & 3 < a + b + c < 27
A = 37 (a + b + c) & r² + r + 1 = 3
r = 1 or r = -2
r = 1 All numbers are same
a = b = c
9 number
This implies (ab)² − 4(a + b) ≤ (ab − 2)² = (ab)² − 4ab + 4, from which we obtain,
ab ≤ a + b + 1. (1)
After swapping a and b if necessary, we may assume a ≤ b. If a ≥ 3, we get
ab ≥ 3b ≥ a + b + b > a + b + 1 in contradiction to (1). Hence a = 2 or a = 1.
If a = 1, we have b²−4(b+1) = x², which is equivalent to (b−2−x)(b−2+x) =8. Because (b − 2 − x) + (b − 2 + x) = 2b − 4 is even and b − 2 − x ≤ b − 2 + x,the only possibility is b−2−x = 2 and b−2+x = 4. This yields (a, b) = (1,
5) as the only possible solution with 1 =
a ≤ b.If a = 2, we have 4b² − 4(b + 2) = x², which is equivalent to (2b − 1 − x)(2b −1 + x) = 9. Here we have two possibilities. Either 2b − 1 − x = 2b − 1 + x = 3 or 2b − 1 − x = 1, 2b − 1 + x = 9. In the first case we obtain b = 2 and in the second b = 3. So we have shown that (a, b) = (2, 2) and (a, b) = (2, 3) are theonly possible solutions with 2 = a ≤ b.A simple calculation verifies that the five pairs (1, 5),(5, 1),(2, 2),(2, 3) and (3, 2) indeed satisfy the requirements of the problem.
4.Let m be the g.c.d. of E(1), E(2), E(3), . . . , E(2009).
Since m|E(1) = 2 · 3 · . . . · 11, it follows that any prime divisor of m is less than or equal to 11. Let p be a prime number such that p|m. Since p ≤ 11 < 2009,it follows that m|E(p) = p(p + 1)(2p + 1)(3p + 1)· · ·(10p + 1). Remark that p + 1, 2p + 1, 3p + 1, . . . , 10p + 1 are relatively prime to p, so E(p) (and thus m) is divisible by p but not by p². We have thus proved that m is not divisible
by the square of any prime number.Since m|E(1) = 2 · 3 · . . . · 11, it follows that m divides the product of all prime numbers less than or equal to 11, that is, m|2310.To show that m = 2310 it is enough to prove that for all n ≥ 1, the number E(n) is divisible by 2310.Let n ≥ 1. Then, one of the numbers n or n + 1 is divisible by 2, so 2|E(n).
Similarly, one of the numbers n, n+1, 2n+1 is divisible by 3 so 3|E(n). Then,one of the numbers n, n+ 1, 2n+ 1, 3n+ 1, 4n+ 1 is divisible by 5 which yields 5|E(n). In the same manner we obtain 7|E(n) and 11|E(n). Therefore E(n) is a multiple of 2 · 3 · 5 · 7 · 11 = 2310 and so, the g.c.d. is 2310.
6.Let n = abc
where n = a×100 + b × 10 + c a three digit number
a + b + c = S(n); Here note that S(n) ≤ 27
Since S(S(n)) = 2,
It means sum of digits of S(n) is 2
Now, S(n) can be, S(n) = 2, 11, 20 only.
9) answer 256
solution-
Number of prime factors of 27! are 9
No of ways of arranging them as numerator and denominator are
2⁹
among them only half of them are greater than 1
so required number of such fractions are 2⁹/2=2⁸=256
LEVEL 2
1) As a Olympiad aspirant you should be able to solve this question yourself
2)Pretty clearly, if we choose each sock-shoe pair individually, they will automatically align themselves.
16!/2^8
3)
4)The quadrilateral that we want to prove is cyclic is a ratio-2 dilation of the one
determined by the projections of P onto the sides, and thus it suffices to prove that the
latter is cyclic. Let X,Y,Z,W be the projections of P onto the sides AB, BC, CD, and
AD, respectively (Figure 1.2.). The quadrilaterals AXPW, BYPX, CZPY, and DWPZ
are cyclic, since they all have a pair of opposite right angles.
Considering angles formed by a side and a diagonal, we get WAP = WXP,
PXY = PBY, YZP = YCP, and PZW = PDW. In the triangles APD and BPC
we have PAD+ PDA = 90◦ and PBC+ PCB = 90◦. Hence
WXY+ WZY = WXP+ PXY+ YZP+ PZW
= WAP+ PDW+ PBY+ YCP
= 90◦ +90◦ = 180◦
which shows that the quadrilateral XYZW is cyclic, and the problem is solved.
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