Math Olympiad Question | Nice Algebraic Equation | You should know this logic!!

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[adammizaushev]

My solution was the same, but in slightly different words
We need to count pairs of factors making 144 in product
To do that, we first split 144 into prime factors:
144 = 2^4 * 3^2
And we just count the possibilities to take all the prime factor to the left or to the right side:
Since there are four 2s, we can take zero of them to the left (and all the four to the right), one to the left (and three to the right), and so on, up to all the four to the left (and zero to the right). So if a prime factor appears in the power of N, there are N+1 possibilities to split it to two groups.
2^4 => 4+1=5 possibilities
3^2 => 2+1=3 possibilities
Now, the counts of possibilities for each factor are just multiplied:
5 * 3 = 15
But I also need to take into account the negative case, which makes the same count of 15 possibilities.
Thence, the answer is 15+15=30

[pxdyfot]

×y = 144 = 2^4 × 3^2

5×3 = 15
But we have to consider -
So, answer 5×3×2 = 30

[cyrangan]

This is WAY simpler than this explanation. Cross multiply, and xy=144. Now simply write out the factors of 144. The factors themselves are the x and y pairs. The factors are 1, 2, 3, 4, 6, 8, 9, 12, 16, 18, 24, 36, 48, 72, and 144. Now just pair them up from the outside in, then reverse them, then “negativize” them, and remember to include 12, 12 and -12, -12 only once. Answer is 30 pairs.

[laogui]

much simpler and quicker than all that. With pairs (a, b), a<=b, the lesser divisor will be <=12, 1, 2, 3, 4, 6, 8, 9 and 12: the first seven count twice, (a, b) and (b, a) plus one for (12, 12) - 15; if -ve is allowed double it - 30

[abhinavchoudhary]

Thank you mam...your each and every videos teach me a new thing.

[amarendrabhowmick]

You are doing yeoman's job.and you are definitely good in maths .appreciate you.

[mohammadsalauddin]

Thanks I learned some new things from today

[cigydoz]

When I went to Olympiads, it was required to prove any formula outside the basic ones. Please give the proof about the number of factors in an exponential multiplication.

[Set_Get]

learnt something i didnt know before. thank you.

[sequid-zlyq]

very nice videos, I love watching to learn new things

[bobdolphin]

Since one pair of factors for 144 is 12 x 12, the number of positive ordered pairs must be odd, so half the number of integer pairs must be odd and the number of integer pairs must be an even number. Eliminate 15 as not even. Eliminate 12 as half of 12 is still even. Eliminate 10, as you can easily think of more than five pairs of factors. Therefore 30 must be the correct answer. Much more challenging without multiple choice answers.

[admin]

Find the prime factors of 144 directly and increase the powers by 1, then multiply these values ​​and then multiply the value you found by 2 because we need to take negative numbers into account as well, it's that simple. Solution in 5 seconds

[coloreilmare]

XY= 1.2.2.2.2.3.3 so we have 6 numbers in multiplication. X is allowed to take 1 number in 6 numbers so we have 6 different value for x. And also y can take 1 number in other 5 numbers so we have 5 situation. multiple them and we have 6.5 is "30" ( I don't know whether my brain is thinking in the right way)

[gregoiredetours]

30 solutions (1*144, 2*72, 3*48, 4*36, 6*24, 8*18, 9*16, 12*12)

[johnhebert]

Neat trick. I don't think they knew this back in my day.

[alpborakirte]

xy = 144 = 2^4 x 3^2 (4+1)(2+1)=15 there are also negative pairs so 2x15=30

[dneary]

It's 30. xy = 144 = 2^4 3^2 - so there are (4+1)(2+1) = 15 positive factors, and the same number of negative factors - and ordered pairs (x, y) can have x be any of these factors, with y = 144/x

[nelsonamador]

n!/r!(n-r) n = 6 and r = 2, n is number of possible primes and r number of groups in this case X and Y, so they are two. 6!/2!(6-2)! = 15

[atillakaragazi]

The answer should be 28. (a, b) and (b, a) are different ordered pairs unless a=b. In this case, you are double counting (12, 12) and (-12, -12).

[italixgaming]

You overcomplicated this poor little thing.
Our problem can be written as xy=144.
We notice that x and y must have the same sign, and that for each ordered pair (x, y) of positive integers solution of the problem we have one pair of negative integers that is solution: (-y, -x). And vice versa. So we can just suppose for now that x and y are positive, and we know that we'll find exactly half of the solutions of the global problem.
Now let's write 144 as product of prime factors: 144=2^4.3².
So any divisor of 144 must be written as 2^p.3^q where p can be equal to 0, 1, 2, 3 or 4 and p can be equal to 0, 1 or 2.
And we'll be able to write :
So for each divisor x of 144 there is exactly one divisor y of 144 such as xy=144.
Since we have 5 choices for p and 3 choices for q - whatever the first choice is - we have a total of 15 choices for x and then 15 ordered pairs (x, y).
And since there are as many ordered pairs of negative integers, our answer is 30.