Tag Archives: Pre RMO 2014

part 3 of 3 — Solutions to Pre RMO Oct 2014

Question Set A.

Question 20. 

What is the number of ordered pairs (A,B) where A and B are subsets of {1,2,3,4,5} such that neither A \subseteq B nor

B \subseteq A?

Solution. Just list down A and B explicitly. Note that A and B are disjoint and that (A,B) is an ordered pair.

Question 13. For how many natural numbers n between 1 and 2014 (both inclusive) is \frac {8n}{9999-n} an integer?

Solution. Firstly, note that 9999-n is even. Hence, n is odd.

Also, 8n \geq 9999-n to yield an exact integer, so n \geq 1111.

Now, do the one of the core tricks for problem solving in number theory. Plug and play with numbers 🙂

Put n=1111. This works.

Next, note that (9999-n)k=8n for some positive integer k. Hence, we get

9999k=(k+8)n From this we observe that n=1111 is the only possible solution. Hence, the answer is 1.

Note : Question 16: HW to be posted on the blog 🙂

More later,

Nalin Pithwa

Part 2 of 3 — solutions to Pre RMO Oct 2014

pre rmo  oct 2014 no 12Question Paper Set A:

11) For natural numbers x and y, let (x,y) denote the greatest common divisor of x and y. How many pairs of natural numbers x and y with x \leq y satisfy the equation xy= x+y+ (x,y)?

Solution:

Here, if the condition, x \leq y were not there, the answer would be infinitely many. But, so, just substitute some small numbers and check what is happening. In fact, in any number theory problem, first we should play with small numbers in our head.

Upon substitution, you will find that only (2,3), (2,4), (3,3) satisfy the equation. Hence, the answer is 3.

12) Let ABCD be a convex quadrilateral with

\angle DAB = \angle BDC = 90 \deg. Let the incircles of triangles ABD and BCD touch  BD at P and Q respectively, with P lying in between B and Q. If

AD=999 and PQ=200 then what is the sum of  the radii of the incircles of triangles ABD and BDC?

Solution: Please download the jpg fig attached.

The main properties to  be used are Pythagoras’ theorem, that the angle bisectors of the vertices of a triangle meet at its incenter, the \tan 2A formula of a triangle.

Let C_{1} and C_{2} be the incenter of \Delta ABD and

\Delta BCD respectively. Then, as shown in  the attached figure:

Since, \Delta APD is a right angled isosceles triangle, DP=DQ=999, \alpha 22.5 \deg, so

\tan 45 \deg = \frac {\tan 22.5 \deg}{1- \tan^{2} 22.5 \deg}= r_{1}/999. Similarly, find the other radius and sum up the two.

Question 18.

Let f be a one-to-one function from the set of natural numbers to itself such that f(mn)=f(m)f(n) for all natural numbers m and n. What is the least possible value of f(999)?

Probable Solution. 

Use the Euler \phi function. But, the answer this gives is different from the answer key.

Please send your comments, suggestions to this.

More later,

Nalin Pithwa

 

 

 

Part I : Solutions to Pre-RMO (Regional Mathematical Olympiad) Oct 2014

Question Paper Set A:

Qs 1) A natural number is such that k^{2}<2014<(k+1)^{2}. What is the largest prime factor of k?

Solution 1:

HInt: Think of perfect square numbers near 2014.

Hence, 44^{2}<2014<(44+1)^{2}. And, the largest prime factor of 44 is 11. Hence, ans is 11.

Qs 2) The first term of a sequence is 2014. Each succeeding term is the sum of cubes of the digits of the previous term. What is the 2014th term of the sequence?

Solution 2:

Hint: Try the simplest solution…just keep calculating…:-)

a_{1}=2014

a_{2}=2^{3}+1^{3}+4^{3}=73

a_{3}=7^{3}+3^{3}=343+27=370

Clearly, a_{2014}=370, which is the answer.

Qs 3) Let ABCD be a convex quadrilateral with perpendicular diagonals. If AB=20, BC=70, and CD=90, then what is the value of DA?

Solution 3. let the diagonals meet at 0. Let OB=a,OC=b, OD=d and OA=c. Then, we can apply Pythagoras’s theorem to the 4 right angled triangles AOB, BOC, COD, DOA. We get

a^{2}+b^{2}=70^{2}

b^{2}+d^{2}=90^{2}

d^{2}+c^{2}=DA^{2}

a^{2}+c^{2}=20^{2}

Hence, we get, a^{2}+b^{2}+c^{2}+d^{2}=90^{2}+20^{2}. In the RHS, substitute for a^{2}+b^{2}=70^{2} and you will get

DA^{2}=c^{2}+d^{2}.

Question 4: In a triangle with integer side lengths, one side is three times as long as a second side and the length of the third side is 17. What is the greatest possible perimeter of the triangle?

Solution 4:AB=c;AC=b; BC=a; b=3a; c=17

Hint: use triangle inequalities:

a+b>c; b+c>a; c+a>b. Hence, 4a>17 and

a>17/4.. Also, a+17>3a and hence, a<17/2. So,

17/4<a<17/2. So, we get a=8, b=3a=24

Ans. 8+24+17=49.

Question 5. if real numbers a,b,c,d,e satisfy

a+1=b+2=c+3=d+4=c+5=a+b+c+d+e+3, what is the value of

a^{2}+b^{2}+c^{3}+d^{4}+e^{5} ?

Solution 5. b=a-1, c=b-1=a-2, d=c-1=a-3,

e=d-1=a-4 so we get

a+1=a+a-1+a-2+a-3+a-4+3. Hence, a=2.

Ans. 10.

Question 6: What is the smallest possible natural number n for which the equation x^{2}-ax+2014=0 has integer roots?

Solution 6:

Hint: Use the discriminant formula.

x=(n \pm \sqrt (n^{2}-8056))/2.

So, just think of plugging in some values 🙂 and note that n has to be smallest.

Also, n has to be even and also \sqrt (n^{2}-8056) has to be even and a perfect square root.

Ans. 91

Question 7: If x^{(x)^{4}}=4, what is the value of

x^{(x)^{2}}+x^{(x)^{8}} ?

Solution 7:

Warning: Do not use logarithms as there is no formula for log(M+N).

Hint: Try to input some numbers so that the original equation is satisfied.

x=\sqrt (2).Ans.

Question 8: Let S be set of real numbers with mean M. If the means of the sets S \cup {15} and S \cup {15,1} are M+2 and M+1 respectively, then how many elements does S have?

Solution: Use the definition of mean or average value followed by a little manipulation.

Ans. 4

Question 9: Natural numbers k,l,p,q are such that if a and b are roots of

x^{2}-kx+l=0, then a+(1/b) and b+(1/a) are the roots of x^{2}-px+q=0. What is the sum of all possible values of q?

Solution 9: Hint: Use the relationships between roots and coefficients:

a+b=k — equation I

ab=l — equation II

a+(1/b)+b+(1/a)=p — equation III

(a+1/b)(b+1/a)=q — equation IV

From equation IV, we get ab+1+1+(1/ab)=q, that is, l+2+(1/l)=q But q is a natural number. Hence, 1/l is also a natural number and the only way this could be possible is l=1.

Hence, ans q=4.

Question 10 In a triangle ABC, X and Y are points on the segments AB and AC respectively, such that AX:XB=1:2 and AY:YC=2:1. If the area of the triangle AXY is 10, then what is the area of the triangle ABC?

Solution 10: Let AX=k, BX=2k, AY=2m, CY=m where k and m are constants of proportionality.

Use: From trigonometry, (1/2)bc \sin A = area of triangle ABC .

Here, (1/2)k.2m.\sin A=10 and hence, mk \sin A=10.

But, area of triangle ABC is (1/2)(3k)(3m) \sin A=(9/2) \times 10=45. Ans.

Question 11: For natural numbers x and y, let (x,y) denote the greatest common divisor of x and y. How many pairs of natural numbers x and y with

x \leq y satisfy the equation xy=x+y+(x,y)?

To be discussed in the next blog.

Question 12:Let ABCD be a convex quadrilaterall with \angle DAB = \angle BDC = 90 \deg. Let the incircles of triangles ABD and BCD touch BD at P and Q, respectively, with P lying in between B and Q. If AD=999 and PQ=200, then what is the sum of the radii of the incircles of triangles ABD and BDC?

To be discussed in the next blog.

Question 13:For how many natural numbers n between 1 and 2014 (both inclusive) is 8n/(9999-n) an integer?

To be discussed in the next blog.

Question 14: One morning, each member of Manjul’s family drank an 8-ounce mixture of coffee and milk. The amounts of coffee and milk varied from cup to cup, but were never zero. Manjul drank 1/7 th of the total amount  of milk and 2/17 th of  the total amount of coffee. How many people are there in Manjul’s family?

Solution 14: Key Idea: When two different fluids are mixed, they are in a certain proportion or ratio; this ratio of the two fluids remains the same when the mixture is poured out in different amounts.

Now, ratio of milk to coffee in Manjul’s mixture is 17/14.

Let x be the constant of proportionality. Total mixture is 8 ounce. Hence,

17x + 14x = 8 and therefore, x=8/31. But, total milk is

17x=(17/31) \times 8 and total coffee is 14x = (14/31) \times 8.

Hence, ans. 8.

Question 15: Let XOY be a triangle with \angle XOY=90 \deg. Let M and N be the midpoints of legs OX and OY, respectively. Suppose that XN=19 and YM=22. What is XY?$

Solution 15. Let XM=a=MO and ON=b=NY.

Apply Pythagoras’s theorem to right angled triangle MOY:

a^{2}+(2b)^{2}=22 \times 22.

Similarly, from right angled triangle XON, we get:

(2a)^{2}+ b^{2}=19 \times 19

Adding the above two equations, 5(a^{2}+b^{2})=845 and hence,

2(a^{2}+b^{2})=26, which is the desired answer.

Question 16: In a triangle ABC, let I denote the incenter. Let the lines AI, BI, and CI intersect the incircle at P, Q and R, respectively. If \angle BAC=40 \deg, what is the value of \angle QPR in degrees?

To be discussed in the next blog.

Question 17: For a natural number b, let N(b) denote the number of natural numbers a for which the equation x^{2}+ax+b=0 has integer roots. What is the smallest value of b for which N(b)=6?

Question 18: Let f be a one-one function from the set of natural numbers to itself such that f(mn)=f(m)f(n) for all natural numbers m and n. What is the least possible value of f(999)?

Question 19: Let x_{1}, x_{2}, x_{3}, \ldots x_{2014} be real numbers different from 1, such that x_{1}+x_{2}+x_{3}+ \ldots + x_{2014}=1 and

\frac {x_{1}}{1-x_{1}} + \frac {x_{2}}{1-x_{2}}+ \frac {x_{3}}{1-x_{3}}+ \ldots +\frac {x_{2014}}{1-x_{2014}}=1.

What is the value of \frac {x_{1}^{2}}{1-x_{1}}+\frac {x_{2}^{2}}{1-x_{2}}+\frac {x_{3}^{2}}{1-x_{3}}+\ldots + \frac {x_{2014}^{2}}{1-x_{2014}} ?

Solution. 

\frac {2x_{1}}{1-x_{1}}+\frac {2x_{2}}{1-x_{2}}+\frac {2x_{3}}{1-x_{3}}+\ldots+\frac {2x_{2014}}{1-x_{2014}}=2

Let S= \frac {x_{1}^{2}}{1-x_{1}}+\frac {x_{2}^{2}}{1-x_{2}}+\frac {x_{3}^{2}}{1-x_{3}}+\ldots+\frac {1-x_{2014}^{2}}{1-x_{2014}}

hence, S-2=\frac {x_{1}^{2}}{1-x_{1}}-\frac {2x_{1}}{1-x_{1}}+    \frac {x_{2}^{2}}{1-x_{2}}-\frac {2x_{2}}{1-x_{2}}+\ldots+\frac{x_{2014}^{2}}{1-x_{2014}}-\frac {2x_{2014}}{1-x_{2014}}

But, x_{1}+x_{2}+x_{3}+\ldots+x_{2014}=1 and also given that

\frac{x_{1}}{1-x_{1}}+\frac {x_{2}}{1-x{2}}+\frac {x_{3}}{1-x_{3}}+\ldots+\frac {x_{2014}}{1-x_{2014}}=1

(\frac {x_{1}}{1-x{1}}+1)+(\frac {x_{2}}{1-x{2}}+1)+(\frac {x_{3}}{1-x_{3}} +1)+\ldots +(\frac{x_{2014}}{1-x{2014}}+1)=2015.

Hence, S-2+2015=\frac {(1-x_{1})^{2}}{1-x_{1}} + \frac {(1-x_{2})^{2}}{1-x_{2}} +\ldots + \frac {(1-x_{2014})^{2}}{1-x_{2014}}

which  equals 1-x_{1}+1-x_{2}+\ldots +1-x_{2014}=2014 -1=2013.

Hence, S=0.

Question 20: What is the number of ordered pairs (A,B) where A and B are subsets of {1,2,\ldots 5} such that neither A \subseteq B nor

B \subseteq A?