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Sum of digits Problem | PRMO 2016 | Question 6

Try this beautiful problem from Number system based on sum of digits.

Sum of digits | PRMO | Problem 6


Find the sum of digits in decimal form of the number \((9999….9)^3\) (There are 12 nines)

  • $200$
  • $216$
  • $230$

Key Concepts


Number system

Digits

counting

Check the Answer


Answer:$216$

PRMO-2016, Problem 6

Pre College Mathematics

Try with Hints


we don’t know what will be the expression of \((9999….9)^3\). so we observe….

\(9^3\)=\(729\)

\((99)^3\)=\(970299\)

\((999)^3\)=\(997002999\)

……………..

……………

we observe that,There is a pattern such that…

In \((99)^3\)=\(970299\) there are 1-nine,1-seven,1-zero,1-two,2-nines & \((999)^3\)=\(997002999\) there are 2- nines,1-seven,2-zeros,1-two,3-nines….so in this way…..\((999….9)^3\) will be 11-nines,1-seven,11-zeros,1-two,12-nines……….

Therefore \((999….9)^3\)=\((99999999999) 7 (00000000000) 2(999999999999)\)

can you finish the problem?

Therefore \((999….9)^3\)=\((99999999999) 7 (00000000000) 2(999999999999)\)…….

total numbers of Nines are (11+12) and (7+2)=9(another one) …..so total (11+12+1)=24 nines and the sum be \((24\times 9)\)=\(216\)

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Sum of two digit numbers | PRMO-2016 | Problem 7

Try this beautiful problem from Algebra based on Sum of two digit numbers from PRMO 2016.

Sum of two digit numbers | PRMO | Problem 7


Let s(n) and p(n) denote the sum of all digits of n and the products of all the digits of n(when written in decimal form),respectively.Find the sum of all two digits natural numbers n such that \(n=s(n)+p(n)\)

  • $560$
  • $531$
  • $654$

Key Concepts


Algebra

number system

addition

Check the Answer


Answer:$531$

PRMO-2016, Problem 7

Pre College Mathematics

Try with Hints


Let \(n\) is a number of two digits ,ten’s place \(x\) and unit place is \(y\).so \(n=10x +y\).given that \(s(n)\)= sum of all digits \(\Rightarrow s(n)=x+y\) and \(p(n)\)=product of all digits=\(xy\)

now the given condition is \(n=s(n)+p(n)\)

Can you now finish the problem ……….

From \(n=s(n)+p(n)\) condition we have,

\(n=s(n)+p(n)\) \(\Rightarrow 10x+y=x+y+xy \Rightarrow 9x=xy \Rightarrow y=9\) and the value of\(x\) be any digit….

Can you finish the problem……..

Therefore all two digits numbers are \(19,29,39,49,59,69,79,89,99\) and sum=\(19+29+39+49+59+69+79+89+99=531\)

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Problem on Positive Integers | PRMO-2019 | Problem 26

Try this beautiful problem from Algebra PRMO 2019 based on Positive Integers.

Positive Integers | PRMO | Problem 26


Positive integers x,y,z satisfy xy+z=160 compute smallest possible value of x+yz.

  • 24
  • 50
  • 29
  • 34

Key Concepts


Algebra

Integer

sum

Check the Answer


Answer:50

PRMO-2019, Problem 26

Higher Algebra by Hall and Knight

Try with Hints


x+yz=\(\frac{160-z}{y}\)+yz

=\(\frac{160}{y}+\frac{z(y^{2}-1)}{y}=\frac{160-z}{y}+\frac{zy^{2}}{y}=\frac{160-z}{y}+zy\)

for particular value of z, \(x+yz \geq 2\sqrt{z(160-z)}\)

or, least value=\(2\sqrt{z(160-z)}\) but an integer also

for least value z is also

case I z=1, \(x+yz=\frac{159}{y}+y\) or, min value at y=3 which is 56

case II z=2, \(x+yz=\frac{158}{y}+2y\) or, min value at y =2 which is 83 (not taken)

case III z=3, \(x+yz=\frac{157}{y}+3y\) or, min value at y=1 which is 160 (not taken)

case IV z=4, \(x+yz=\frac{156}{y}+4y\) or, min at y=6 which is 50 (taken)

case V z=5, \(x+yz=\frac{155}{y}+5y\) or, minimum value at y=5 which is 56 (not taken)

case VI z=6, \(x+yz=\frac{154}{y}+6y\) \( \geq 2\sqrt{924}\)>50

smallest possible value =50.

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Smallest positive value | Algebra | PRMO-2019 | Problem 13

Try this beautiful problem from Algebra PRMO 2019 based on smallest positive value

Smallest positive value| PRMO | Problem 13


Each of the numbers \(x_1, x_2,……….x_{101}\) is \(±1\). What is the smallest positive value of \(\displaystyle\sum_{1\leq i \leq j \leq {101}} x_i x_j\) ?

  • $24$
  • $10$
  • $34$

Key Concepts


Algebra

Integer

sum

Check the Answer


Answer:\(10\)

PRMO-2019, Problem 13

Pre College Mathematics

Try with Hints


\(S=\displaystyle\sum_{1\leq i \leq j \leq {101}} x_i x_j\) .

we have \((x_1+x_2+x_3+….+x_{101})^2={x_1}^2+{x_2}^2+…..+{x_{101}}^2+2S\)

\(\Rightarrow 2S\)=\((\displaystyle\sum_{i=1}^{101} {x_i})^2-\displaystyle\sum_{j=1}^{101} {x_j}^2\)

Can you now finish the problem ……….

Since we have \(x_i=\pm 1\) so \({x_i}^2=1\)

so \(2S=(\displaystyle\sum_{i=1}^{101} {x_i})^2-{101}\)

Since \(\displaystyle\sum_{i=1}^{101} {x_i}\) will be an integer

so \((\displaystyle\sum_{i=1}^{101} {x_i})^2\) will be a perfect square .

For smalll positive \(S\), \((\displaystyle\sum_{i=1}^{101} {x_i})^2\)must be smallest perfect square greater than \({101}\)

So \((\displaystyle\sum_{i=1}^{101} {x_i})^2={121}\)

\(\Rightarrow (\displaystyle\sum_{i=1}^{101} {x_i})\)=\({11}\) or \({-11}\)

Can you finish the problem……..

We can verify that the desired sum can be achieved by putting \(45\) \(x_i\)’s to be –1 and \(56\) \(x_i\)’s to be \(1\) So, \(2S = 121 – 101 = 20\)

\(\Rightarrow s=10\)

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Regular polygon | Combinatorics | PRMO-2019 | Problem 15

Try this beautiful problem from combinatorics PRMO 2019 based on Regular polygon

Regular polygon| PRMO | Problem 15


In how many ways can a pair of parallel diagonals of a regular polygon of \(10\) sides be selected

  • $24$
  • $45$
  • $34$

Key Concepts


Combinatorics

Regular polygon

geometry

Check the Answer


Answer:\(45\)

PRMO-2019, Problem 15

Pre College Mathematics

Try with Hints


regular polygon

The above diagram is a diagram of Regular Polygon .we have to draw the diagonals as shown in above.we joined the diagonals such that all the diagonals will be parallel

Can you now finish the problem ……….

Polygon 2
Fig.2
Polygon 1
Fig. 1

If we joined the diagonals (shown in Fig. 1), i.e \((P_3 \to P_10)\),\((P_4\to P_9)\),\((P_5 \to P_8)\) then then we have 3 diagonals.so we have 5\(4 \choose 2\) ways=\(15\) ways.

If we joined the diagonals (shown in Fig.2), i.e \((P_1 \to P_3)\),\((P_10\to P_4)\),\((P_9\to P_5)\),\((P_8\to P_6)\)then we have \(4\)diagonals.so we have 5\(3 \choose 2\) ways=\(30\) ways.

Therefore total numbers of ways that can a pair of parallel diagonals of a regular polygon of \(10\) sides be selected is \(15+30=45\)

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Problem on Real numbers | Algebra | PRMO-2017 | Problem 18

Try this beautiful problem from Algebra PRMO 2017 based on real numbers.

Problem on Real numbers | PRMO | Problem 18


If the real numbers \(x\), \(y\), \(z\) are such that \(x^2 + 4y^2 + 16z^2 = 48\) and \(xy + 4yz + 2zx = 24\). what is the
value of \(x^2 + y^2 + z^2\) ?

  • $24$
  • $21$
  • $34$

Key Concepts


Algebra

Equation

Check the Answer


Answer:\(21\)

PRMO-2017, Problem 18

Pre College Mathematics

Try with Hints


The given equation are

\(x^2 + 4y^2 + 16z^2 = 48\)
\(\Rightarrow (x)2 + (2y)2 + (4z)2 = 48\)
\(2xy + 8yz + 4zx = 48\)
adding tis equations we have to solve the problem….

Can you now finish the problem ……….

Now we can say that
\((x)^2 + (2y)^2 + (4z)^2 – (2xy) – (8yz) – (4zx) = 0\)
\(\Rightarrow [(x – 2y)2 + (2y – 4z)2 + (x – 4y)2)] = 0\)
\(x = 2y = 4z \)

\(\Rightarrow \frac{x}{4}=\frac{y}{2}=z\)

Can you finish the problem……..

Therefore we may say that,

\((x, y, z) = (4m, 2m, m)\)
\(x^2 + 4y^2 + 16z^2 = 48\)

\(16m^2 + 16m^2 + 16m^2 = 48\)
so \(m^2 = 1\)
\(x^2 + y^2 + z^2 = 21m^2 = 21\)

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Roots of Equations | PRMO-2016 | Problem 8

Try this beautiful problem from Algebra based on roots of equations.

Roots of Equations | PRMO | Problem 8


Suppose that \(a\) and \(b\) are real numbers such that \(ab \neq 1\) and the equations \(120 a^2 -120a+1=0\) and \(b^2-120b+120=0\) hold. Find the value of \(\frac{1+b+ab}{a}\)

  • $200$
  • $240$
  • $300$

Key Concepts


Algebra

quadratic equation

Roots

Check the Answer


Answer:$240$

PRMO-2016, Problem 8

Pre College Mathematics

Try with Hints


The given equations are \(120 a^2 -120a+1=0\) and \(b^2-120b+120=0\).we have to find out the values of \(a\) and \(b\)….

Let \(x,y\) be the roots of the equation \(120 a^2 -120a+1=0\)then \(\frac{1}{x},\frac{1}{y}\) be the roots of the equations of \(b^2-120b+120=0\).can you find out the value of \(a\) & \(b\)

Can you now finish the problem ……….

From two equations after sim[lificatiopn we get…\(a=x\) and \(b=\frac{1}{y}\) (as \(ab \neq 1)\)

Can you finish the problem……..

\(\frac{1+b+ab}{a}\)=\(\frac { 1+\frac{1}{y} +\frac{x}{y}}{x}\)=\(\frac{(x+y)+1}{xy}=240\)

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Can we prove that the length of any side of a triangle is not more than half of its perimeter?

Can we Prove that ……..


The length of any side of a triangle is not more than half of its perimeter

Key Concepts


Triangle Inequality

Perimeter

Geometry

Check the Answer


Answer: Yes we can definitely prove that by Triangle Inequality

Mathematical Circles – Chapter 6 – Inequalities Problem 3

Mathematical Circles by Dmitri Fomin , Sergey Genkin , Llia Itenberg

Try with Hints


We can start this sum by using this picture below

The length of the three sides of this triangle are a,b and c. So if we apply triangle inequality which implies that the length of one side of a triangle is less than the sum of the lengths of the two sides of that triangle. In reference to the theorem

b + c > a

Proof based on triangle

So can you try to do the rest of the sum ????????

According to the question we have to find the perimeter at first

Perimeter is the sum of the length of all sides of the triangle = a + b + c

And the length of each side is a or b or c.

We have to prove : a + b + c > length of any one side

This can be one of the most important hint for this problem. Try to do the rest of the sum …………………………..

Here is the rest of the sum :

As stated above if we use triangle inequality :

b + c > a

Lets add a to both the sides

a + b + c > a + a

a + b + c > 2 a

The left hand side of the above inequality is the perimeter of this triangle.

perimeter > 2 a

So , \(\frac {perimeter}{2} > a \)

\(\frac {perimeter}{2} \) = semi perimeter

Hence this is proved that the length of one side of a triangle is less than half of its perimeter.

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