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Math Olympiad PRMO USA Math Olympiad

Good numbers Problem | PRMO-2018 | Question 22

Try this beautiful good numbers problem from Number theory from PRMO 2018, Question 22.

Good numbers Problem – PRMO 2018, Question 22


A positive integer $k$ is said to be good if there exists a partition of ${1,2,3, \ldots, 20}$ into disjoint proper subsets such that the sum of the numbers in each subset of the partition is $k$. How many good numbers are there?

  • $4$
  • $6$
  • $8$
  • $10$
  • $2$

Key Concepts


Number theorm

good numbers

subset

Check the Answer


Answer:$6$

PRMO-2018, Problem 22

Pre College Mathematics

Try with Hints


What is good numbers ?

A good number is a number in which every digit is larger than the sum of digits of its right (all less significant bits than it). For example, 732 is a good number, $7>3+2$ and $3>2$ .

Given that $k$ is said to be good if there exists a partition of ${1,2,3, \ldots, 20}$ into disjoint proper subsets such that the sum of the numbers in each subset of the partition is $k$. Now at first we have to find out sum of these integers ${1,2,3, \ldots, 20}$. Later create some partitions such that two partitions be disjoint set and sum of the numbers of these partitions be good numbers

Can you now finish the problem ……….

Sum of numbers equals to $\frac{20 \times 21}{2}=210 \& 210=2 \times 3 \times 5 \times 7$

So $\mathrm{K}$ can be 21,30,35,47,70,105

Can you finish the problem……..

Case 1 :

$\mathrm{A}=\{1,2,3,4,5,16,17,18,19,20\}$, $\mathrm{B}=\{6,7,8,9,10,11,12,13,14,15\}$

Case 2 :

$A=\{20,19,18,13\}$, $B=\{17,16,15,12,10\}$, $C=\{1,2,3,4,5,6,7,8,9,11,14\}$

Case 3 :

$\mathrm{A}=\{20,10,12\}$, $\mathrm{B}=\{18,11,13\}$, $\mathrm{C}=\{16,15,9,2\}$, $\mathrm{D}=\{19,8,7,5,3\}$, $\mathrm{E}=\{1,4,6,14,17\}$

Case 4 :

$A=\{20,10\}, B=\{19,11\}$,$ C=\{18,12\}, D=\{17,13\}$,$ E=\{16,14\}$, $F=\{1,15,5\},$
$G=\{2,3,4,6,7,8\}$

Case 5 :

$A=\{20,15\}$, $B=\{19,16\}$, $C=\{18,17\}$, $D=\{14,13,8\}$, $E=\{12,11,10,2\},$
$F=\{1,3,4,5,6,7,9\}$

Case 6 :

$A=\{1,20\}$,$ B=\{2,19\}$, $C=\{3,18\} \ldots \ldots \ldots \ldots$, $J=\{10,11\}$

Therefore Good numbers equal to $6$

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Math Olympiad PRMO USA Math Olympiad

Polynomial Problem | PRMO-2018 | Question 30

Try this beautiful Polynomial Problem from Number theorm from PRMO 2018, Question 30.

Polynomial Problem – PRMO 2018, Question 30


Let $P(x)=a_{0}+a_{1} x+a_{2} x^{2}+\ldots .+a_{n} x^{n}$ be a polynomial in which $a_{i}$ is non-negative integer for each $\mathrm{i} \in{0,1,2,3, \ldots, \mathrm{n}} .$ If $\mathrm{P}(1)=4$ and $\mathrm{P}(5)=136,$ what is the value of $\mathrm{P}(3) ?$

  • $30$
  • $34$
  • $36$
  • $39$
  • $42$

Key Concepts


Number theorm

Polynomial

integer

Check the Answer


Answer:$34$

PRMO-2018, Problem 30

Pre College Mathematics

Try with Hints


Given that $P(x)=a_{0}+a_{1} x+a_{2} x^{2}+\ldots .+a_{n} x^{n}$ where $\mathrm{P}(1)=4$ and $\mathrm{P}(5)=136$. Now we have to find out $P(3)$.

Therefore if we put $x=1$ and $x=5$ then we will get two relations . Using these relations we can find out $a_0$ , $a_1$, $a_2$ .

Can you now finish the problem ……….

$a_{0}+a_{1}+a_{2}+\ldots \ldots+a_{n}=4$
$\Rightarrow a_{i} \leq 4$
$a_{0}+5 a_{1}+5^{2} a_{2}+\ldots+a 5^{n} a_{n}=136$
$\Rightarrow a_{0}=1+5 \lambda \Rightarrow a_{0}=1$

Can you finish the problem……..

Hence $5 a_{1}+5^{2} a_{2}+\ldots \ldots+5^{n} a_{n}=135$
$a_{1}+5 a_{2}+\ldots 5^{n-1} a_{n-1}=27$
$\Rightarrow a_{1}=5 \lambda+2 \Rightarrow a_{1}=2$
$\Rightarrow 5 a_{2}+\ldots .5^{n-1} a_{n-1}=25$
$a_{2}+5 a_{3}+\ldots .5^{n-2} a_{n-2}=5$
$\Rightarrow a_{2}=5 \lambda \Rightarrow a_{2}=0$
$a_{3}+5 a_{4}+\ldots \ldots \ldots+5^{n-3} a_{n-3}=1$
$a_{3}=1$
$\Rightarrow a_{4}+5 a_{5}+\ldots .+5^{n-4} a_{n-3}=0$
$a_{4}=a_{5}=\ldots . a_{n}=0$
Hence $P(n)=x^{3}+2 x+1$
$P(3)=34$

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Algebra Arithmetic Math Olympiad USA Math Olympiad

Numbers of positive integers | AIME I, 2012 | Question 1

Try this beautiful problem from the American Invitational Mathematics Examination, AIME 2012 based on Numbers of positive integers.

Numbers of positive integers – AIME 2012


Find the number of positive integers with three not necessarily distinct digits, \(abc\), with \(a \neq 0\) and \(c \neq 0\) such that both \(abc\) and \(cba\) are multiples of \(4\).

  • is 107
  • is 40
  • is 840
  • cannot be determined from the given information

Key Concepts


Integers

Number Theory

Algebra

Check the Answer


Answer: is 40.

AIME, 2012, Question 1.

Elementary Number Theory by David Burton .

Try with Hints


Here a number divisible by 4 if a units with tens place digit is divisible by 4

Then case 1 for 10b+a and for 10b+c gives 0(mod4) with a pair of a and c for every b

[ since abc and cba divisible by 4 only when the last two digits is divisible by 4 that is 10b+c and 10b+a is divisible by 4]

and case II 2(mod4) with a pair of a and c for every b

Then combining both cases we get for every b gives a pair of a s and a pair of c s

So for 10 b’s with 2 a’s and 2 c’s for every b gives \(10 \times 2 \times 2\)

Then number of ways \(10 \times 2 \times 2\) = 40 ways.

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I.S.I. and C.M.I. Entrance

Understanding Statistical Regularity Through Random Walks | Cheenta Probability Series

This is another blog of the Cheenta Probability Series. Let’s give a formal definition of statistical regularity to bring some seriousness into account.

**10 min read**

“The Law of Statistical Regularity formulated in the mathematical theory of probability lays down that a moderately large number of items chosen at random from a very large group are almost sure to have the characteristics of the large group.”  ~ W.I.King

Starting off with statistical regularity

So, let’s give a formal definition of this to bring some seriousness into account. If a sequence of independent experiments is held under the same specified conditions, the proportion of occurrences of a given event stabilize as the number of experiments becomes larger. This is ideally what is known to be statistical regularity.

It is an umbrella term that covers the law of large numbers, all central limit theorems and ergodic theorems.

But keeping in mind that we cover stuff for undergraduates mainly,we would not get into the aforementioned topics.

Richard Von Mises first mathematically demonstrated the idea of statistical regularity by pinpointing that no method for forming a subsequence of a random sequence (an infinite sequence of 0’s and 1’s) improves the odds for a specific event.

For instance, a sequence of fair coin tosses produces equal and independent 50/50 chances for heads and tails. A simple system of betting on heads every 3rd, 7th, or 21st toss, etc., does not change the odds of winning in the long run.

This is famously known as the “Impossibility of a gambling system“.

The Random Walk

This is in itself a topic in advanced probability theory and stochastic processes , but I will try to keep it simple here. Let’s consider a game  in which a player starts at the point \( x=0 \) and at each move, is required to take a step either forward (toward  \(+x\) ) or backward (toward \(−x\)). The choice is to be made randomly (maybe by tossing a coin). How shall we describe the resulting motion? In general, this problem is closely related to the coin tossing problem.

First let us look at  a few examples of a random walk. We may characterize the walker’s progress by the net distance \(D_N \) traveled in N steps. We illustrate the graph of the random walker in three instances below:

The progress made in a random walk

 What can we say about such a motion? We might first ask: “How far does he get on the average?” We must expect that his average progress will be zero, since he is equally likely to go either forward or backward. But we have the feeling that as \( N \) increases, he is more likely to have strayed farther from the starting point. We might, therefore, ask what is his average distance travelled in absolute value, that is, what is the average of \(|D|\). It is, however, more convenient to deal with another measure of “progress,” the square of the distance:\( D^2 \) is positive for either positive or negative motion, and is therefore a reasonable measure of such random wandering.

Now, till now we have not defined expected value of a quantity or a variable, which we sure will in the upcoming blog posts. For now , by “expected value” we mean the probable value (our best guess), which we can think of as the expected average behavior in many repeated sequences.

We represent such an expected value by \( ⟨D_N ^2⟩ \), and may refer to it also as the “mean square distance.” After one step, \( D^2 \)  is always \( +1 \), so we have certainly \( ⟨D_1 ^2⟩=1 \).

Now , we have an obvious recursion between \(D_N\) and \(D_{N-1} \). More specifically, \(D_N = D_{N-1} +1 \) or \(D_N=D_{N-1}-1 \).

Thus, squaring, \(D_N ^2 = D_{N-1} ^2 + 2 D_{N-1}+1 \) or, \(D_N^2 = D_{N-1} ^2 – 2 D_{N-1}+1 \)

In a number of independent sequences, we expect to obtain each value one-half of the time, so our average expectation is just the average of the two possible values. The expected value of \(D_N ^2 \) is  \(D_{N−1} ^2+1 \) . In general, we should expect for \(D_{N−1} ^2 \) its “expected value” \( ⟨D_{N−1} ^2 ⟩ \) (by definition!). So \( ⟨D_N ^2⟩=⟨D_{N−1} ^2 ⟩+1 \).

We have already seen that \(⟨D_1 ^2⟩=1\); it follows then that \( ⟨D_N ^2 ⟩=N \).

Damn, that was easy. Pretty simple right?

Now let’s draw an analogy of this game with a simple coin tossing experiment ( which many authors use as the prototype for demonstrating regularity !). Yeah, we were thinking slightly in an unorthodox manner ;).

To appropriately represent drifting away from the origin, in a random walk, we can use the Root Mean Square distance:

\( D_R = \sqrt{ ⟨D ^2⟩ } =\sqrt{N} \).

If we imagine the direction of each step to be in correspondence with the appearance of heads or tails in a coin toss, then \( D = N_H−N_T \) , the difference in the number of heads and tails. Since \( N_H+N_T=N \), the total number of steps (and tosses), we have \( D=2N_H−N \).

Now, it’s time to merge our intuition into reality. If the coin is honest or fair, what do you expect?

In \( N \) tosses, you should get \(\frac{N}{2} \) heads right?

So, lets observe the difference \( N_H – \frac{N}{2} =\frac{D}{2} \).

The RMS deviation is given by \( (N_H- \frac{N}{2})_{\text{RMS}}=\frac{\sqrt{N}}{2} \).

Thus, an actual \( N_H \)  deviates from \( \frac{N}{2} \) by about \( \frac{\sqrt{N}}{2} \), or the fraction to deviate by \( \frac{1}{N} \frac{\sqrt{N}}{2} \).

So, the larger the \(N\) , the closer we expect the fraction \( \frac{N_H}{N} \) to \( \frac{1}{2} \).

Sounds familiar right? we circled back to statistical regularity again!

The fraction of tosses that gave heads in a particular sequence of tosses

 Unfortunately, for any given run or combination of runs there is no guarantee that the observed deviation will be even near the expected deviation. There is always the finite chance that a large fluctuation—a long string of heads or tails—will give an arbitrarily large deviation. All we can say is that if the deviation is near the expected \( \frac{1}{2 \sqrt{N}} \) , we have no reason to suspect the honesty of the coin. If it is much larger, we may be suspicious, but cannot prove, that the coin is loaded (or that the tosser is clever!).

If you still suspect the fairness of the coin, you should probably learn the Physics of Coin Tossing which Uttaran Chatterjee would address in the next blog.

Another Simple Game of Chance

This is mainly for our reader friends who like to visualize through coding.

Let’s consider a simple game of chance using a spinner. But this game is somewhat kind to the gambler.In our game,the payoff in each of several repeated plays is determined by spinning the spinner. We pay an entry fee for each play of the game and then receive the payoff indicated by the spinner. Let the payoff on the spinner be distributed uniformly around the circle; i.e. if the angle after the spin is \( \theta \), he receives \( \frac{\theta}{2 \pi} \) rupees. Thus, our payoff on one play is \(U\) rupees, where \(U\) is a random number taking values in \([0,1]\). Clearly, this is gambling for the poor :P.

Let us simulate the game to see what cumulative payoffs the gambler might receive, not counting the entry fees obviously, if he plays the game repeatedly.

Construct partial sums \( S_k = U_1+U_2+…+U_k, 1 \le k \le n \).

The successive partial sums form a random walk, with \(U_n\) being the \(n^{th} \) step and \(S_n\) being the position after \(n\) steps.

R Codes and plots

walk <- function(j)
{
  uniforms <- runif(10^j)
  
  firstsums <- cumsum(uniforms)
  
  sums <- c(0, firstsums)
  
  index <- order(sums)-1
  
  plot(index,sums,main="Random walk for for the partial sums",xlab="Number of trials",ylab="winnings",col=j)
  
}
par(mfrow=c(2,2))

walk(1)

walk(2)

walk(3)

walk(4)

We now present the 4 plots for 4 values of \(N\) , \(10,100,1000,10000 \).

Possible realizations of the first \(10^j \) steps

For small \(n\), that is for \(n=10\), we see irregularly spaced points increasing to the right, but as \(n\) increases, the spacing between the points becomes blurred and regularity emerges: the plots approach a straight line with slope equal to \( \frac{1}{2} \), the mean of a single step \(U_k\). If we look from a macroscopic point of view, ignoring the units on the axes, we see that the plots become independent of \(n\) as \(n\) increases. This is what regularity signifies.

Finally, here comes Number Theory!!

Trust me, I just can’t get enough of the Prime Number Theorem. Here is a short problem to think about for future researchers in this field.

Suppose a random walker starts at the point \( S_0 = 2 \), and walks according to the following rules:

1. If the walker is on the \( n^{th} \) prime number \( p_n \), he moves to either
\( p_n + 1 \) or \( p_{n+1} \) with equal probability.

2. If the walker is on a composite number \(x \), he moves to one of the prime factors of \(x \), each with probability \( \frac{1}{\omega(x) }\), where \(\omega(n) \) denotes the number of distinct prime factors of \(n\).

The random walk is given by the sequence of moves \(S_n\).

What can you say about the quantity \( \mathbb{P}(\sup_{n \ge 0} S_n = \infty) \) ?

Give it a try.

Food For Thought

For our gambling game described above,if the expected payoff is \( \frac{1}{2} \) rupees each play of the game,the gamme is fair if the fee to play is \( \frac{1}{2} \) rupees.

Make a minor modification of this simulation by repeating the experiment after subtracting the mean \( \frac{1}{2} \) from each step of the random walk.

Now, plot the centered random walk (i.e. centered partial sums \(S_k – \frac{k}{2} \) for the same values of \(n\) as before.

Do you observe the same plots?

References

1.Experiencing Statistical Regularity – Ward Whitt

2.The Problem of The Random Walk- K.Pearson

3.An Introduction To Probability Theory and its applications – W.Feller

Stay tuned and keep following this series for getting more interesting perspectives on standard probability results and terms. I bet you too will see probability theory in a different light!

Keep learning, keep thinking!

Cheers..

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Algebra Arithmetic Calculus Math Olympiad USA Math Olympiad

Arithmetic Sequence Problem | AIME I, 2012 | Question 2

Try this beautiful problem from the American Invitational Mathematics Examination, AIME 2012 based on Arithmetic Sequence.

Arithmetic Sequence Problem – AIME 2012


The terms of an arithmetic sequence add to \(715\). The first term of the sequence is increased by \(1\), the second term is increased by \(3\), the third term is increased by \(5\), and in general, the \(k\)th term is increased by the \(k\)th odd positive integer. The terms of the new sequence add to \(836\). Find the sum of the first, last, and middle terms of the original sequence.

  • is 107
  • is 195
  • is 840
  • cannot be determined from the given information

Key Concepts


Series

Number Theory

Algebra

Check the Answer


Answer: is 195.

AIME, 2012, Question 2.

Elementary Number Theory by David Burton .

Try with Hints


First hint

After the adding of the odd numbers, the total of the sequence increases by \(836 – 715 = 121 = 11^2\).

Second Hint

Since the sum of the first \(n\) positive odd numbers is \(n^2\), there must be \(11\) terms in the sequence, so the mean of the sequence is \(\frac{715}{11} = 65\).

Final Step

Since the first, last, and middle terms are centered around the mean, then \(65 \times 3 = 195\)

Hence option B correct.

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Algebra Arithmetic Math Olympiad USA Math Olympiad

Arithmetic Mean | AIME I, 2015 | Question 12

Try this beautiful problem from the American Invitational Mathematics Examination, AIME, 2015 based on Arithmetic Mean.

Arithmetic Mean of Number Theory – AIME 2015


Consider all 1000-element subsets of the set {1, 2, 3, … , 2015}. From each such subset choose the least element. The arithmetic mean of all of these least elements is \(\frac{p}{q}\), where \(p\) and \(q\) are relatively prime positive integers. Find \(p + q\).

  • is 107
  • is 431
  • is 840
  • cannot be determined from the given information

Key Concepts


Inequalities

Algebra

Number Theory

Check the Answer


Answer: is 431.

AIME, 2015, Question 12

Elementary Number Theory by David Burton

Try with Hints


Each 1000-element subset \({ a_1, a_2,a_3,…,a_{1000}}\) of \({1,2,3,…,2015}\) with \(a_1<a_2<a_3<…<a_{1000}\) contributes \(a_1\) to sum of least element of each subset and set \({a_1+1,a_2+1,a_3+1,…,a_{1000}+1}\). \(a_1\) ways to choose a positive integer \(k\) such that \(k<a_1+1<a_2+1,a_3+1<…<a_{1000}+1\) (\(k\) can be anything from \(1\) to \(a_1\) inclusive

Thus, the number of ways to choose the set \({k,a_1+1,a_2+1,a_3+1,…,a_{1000}+1}\) is equal to the sum. But choosing a set \({k,a_1+1,a_2+1,a_3+1,…,a_{1000}+1}\) is same as choosing a 1001-element subset from \({1,2,3,…,2016}\)!

average =\(\frac{2016}{1001}\)=\(\frac{288}{143}\). Then \(p+q=288+143={431}\)

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Algebra Arithmetic Functional Equations Math Olympiad Math Olympiad Videos USA Math Olympiad

Logarithms and Equations | AIME I, 2000 | Question 9

Try this beautiful problem from the American Invitational Mathematics Examination, AIME I, 2000 based on Logarithms and Equations.

Logarithms and Equations – AIME I 2000


\(log_{10}(2000xy)-log_{10}xlog_{10}y=4\) and \(log_{10}(2yz)-(log_{10}y)(log_{10}z)=1\) and \(log_{10}(zx)-(log_{10}z)(log_{10}x)=0\) has two solutions \((x_{1},y_{1},z_{1}) and (x_{2},y_{2},z_{2})\) find \(y_{1}+y_{2}\).

  • is 905
  • is 25
  • is 840
  • cannot be determined from the given information

Key Concepts


Logarithms

Theory of Equations

Number Theory

Check the Answer


Answer: is 25.

AIME I, 2000, Question 9

Polynomials by Barbeau

Try with Hints


First hint

Rearranging equations we get \(-logxlogy+logx+logy-1=3-log2000\) and \(-logylogz+logy+logz-1=-log2\) and \(-logxlogz+logx+logz-1=-1\)

Second Hint

taking p, q, r as logx, logy and logz, \((p-1)(q-1)=log2\) and \((q-1)(r-1)=log2\) and \( (p-1)(r-1)=1\) which is first system of equations and multiplying the first three equations of the first system gives \((p-1)^{2}(q-1)^{2}(r-1)^{2}=(log 2)^{2}\) gives \((p-1)(q-1)(r-1)=+-(log2)\) which is second equation

Final Step

from both equations (q-1)=+-(log2) gives (logy)+-(log2)=1 gives \(y_{1}=20\),\(y_{2}=5\) then \(y_{1}+y_{2}=25\).

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I.S.I. and C.M.I. Entrance IIT JAM Statistics ISI M.Stat PSB ISI MSAT Number Theory Probability

ISI MStat PSA 2019 Problem 18 | Probability and Digits

This problem is a very easy and cute problem of probability from ISI MStat PSA 2019 Problem 18.

Probability and Digits – ISI MStat Year 2019 PSA Problem 18


Draw one observation \(N\) at random from the set \(\{1,2, \ldots, 100\}\). What is the probability that the last digit of \(N^{2}\) is \(1\)?

  • \(\frac{1}{20}\)
  • \(\frac{1}{50}\)
  • \(\frac{1}{10}\)
  • \(\frac{1}{5}\)

Prerequisites


Last Digit of Natural Numbers

Basic Probability Theory

Combinatorics

Check the Answer


Answer: is \(\frac{1}{5}\)

ISI MStat 2019 PSA Problem Number 18

A First Course in Probability by Sheldon Ross

Try with Hints


Try to formulate the sample space. Observe that the sample space is not dependent on the number itself rather only on the last digits of the number \(N\).

Also, observe that the number of integers in \(\{1,2, \ldots, 100\}\) is uniformly distributed over the last digits. So the sample space can be taken as \(\{0,1,2, \ldots, 9\}\). So, the number of elements in the sample space is \(10\).

See the Food for Thought!

This step is easy.

Find out the cases for which \(N^2\) gives 1 as the last digit. Use the reduced last digit sample space.

  • 1 x 1
  • 3 x 7 (Since \(N^2\) and they must have the same last digit)
  • 7 x 3 (Since \(N^2\) and they must have the same last digit)
  • 9 x 9

So, there are 2 possible cases out of 10.

Therefore the probability = \( \frac{2}{10} = \frac{1}{5}\).

  • Observe that there is a little bit of handwaving in the First Step. Please make it more precise using the ideas of Probability that it is okay to use the sample space as the reduced version rather than \(\{1,2, \ldots, 100\}\).
  • Generalize the problem for \(\{1,2, \ldots, n\}\).
  • Generalize the problems for \(N^k\) for selecting an observation from \(\{1,2, \ldots, n\}\).
  • Generalize the problems for \(N^k\) for selecting an observation from \(\{1,2, \ldots, n\}\) for each of the digits from \(\{0,1,2, \ldots, 9\}\).
Outstanding Statistics Program with Applications

Outstanding Statistics Program with Applications

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Math Olympiad PRMO USA Math Olympiad

Ordered Pairs | PRMO-2019 | Problem 18

Try this beautiful problem from PRMO, 2019, Problem 18 based on Ordered Pairs.

Orderd Pairs | PRMO | Problem-18


How many ordered pairs \((a, b)\) of positive integers with \(a < b\) and \(100 \leq a\), \(b \leq 1000\) satisfy \(gcd (a, b) : lcm (a, b) = 1 : 495\) ?

  • $20$
  • $91$
  • $13$
  • \(23\)

Key Concepts


Number theory

Orderd Pair

LCM

Check the Answer


Answer:\(20\)

PRMO-2019, Problem 18

Pre College Mathematics

Try with Hints


At first we assume that \( a = xp\)
\(b = xq\)
where \(p\) & \(q\) are co-prime

Therefore ,

\(\frac{gcd(a,b)}{LCM(a ,b)} =\frac{495}{1}\)

\(\Rightarrow pq=495\)
Can you now finish the problem ……….

Therefore we can say that

\(pq = 5 \times 9 \times 11\)
\(p < q\)

when \( 5 < 99\) (for \(x = 20, a = 100, b = 1980 > 100\)),No solution
when \(9 < 55\) \((x = 12\) to \(x = 18)\),7 solution
when,\(11 < 45\) \((x = 10\) to \(x = 22)\),13 solution
Can you finish the problem……..

Therefore Total solutions = \(13 + 7=20\)

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AMC-8 Math Olympiad USA Math Olympiad

Largest Possible Value | PRMO-2019 | Problem 17

Try this beautiful problem from PRMO, 2019 based on Largest Possible Value.

Largest Possible Value | PRMO | Problem-17


Let a, b, c be distinct positive integers such that \(b + c – a\),\( c + a – b\) and \(a + b – c\) are all perfect squares.
What is the largest possible value of \(a + b + c\) smaller than \(100\)?

  • $20$
  • $91$
  • $13$

Key Concepts


Number theory

Perfect square

Integer

Check the Answer


Answer:\(91\)

PRMO-2019, Problem 17

Pre College Mathematics

Try with Hints


Let \(b + c – a = x^2\) … (i)
\(c + a – b = y^2\) … (ii)
\(a + b – c = z^2\) … (iii)

Now since \(a\),\( b\), \(c\) are distinct positive integers,
Therefore, \(x\), \(y\), \(z\) will also be positive integers,
add (i), (ii) and (iii)
\(a + b + c = x^2 + y^2 + z^2\)
Now, we need to find largest value of \(a + b + c or x^2 + y^2 + z^2\) less than \(100\)
Now, to get a, b, c all integers \(x\),\( y\), \(z\) all must be of same parity, i.e. either all three are even or all three
are odd.

Can you now finish the problem ……….

Let us maximize\(x^2 + y^2 + z^2\), for both cases.
If \(x\), \(y\), \(z \)are all even.
Therefore,

\(b + c – a = 8^2 = 64\)
\(c + a – b = 42 = 16\)
\(a + b – c = 22 = 4\)
Which on solving, give\( a = 10\),\( b = 34\), \(c = 40\) and \(a + b + c = 84\)
If x, y, z are all odd
\(\Rightarrow b + c – a = 92 = 81\)
\(c + a – b = 32 = 9\)
\(a + b – c = 12 = 1\)
Which on solving, give \(a = 5\) ,\(b = 41\), \(c = 45\) and\( a + b + c = 91\)

Can you finish the problem……..

Therefore Maximum value of \(a + b + c < 100 = 91\)

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