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?

 

 

 

One Comment

  1. Abc
    Posted August 16, 2017 at 10:23 am | Permalink | Reply

    This was very helpful for practising some questions for olympiads

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