Olympiad Question | Learn tips to solve Olympiad Math Question fast | Math Olympiad Preparation

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Olympiad Question | Learn tips to solve Olympiad Math Question fast | Math Olympiad Preparation

Olympiad Mathematical Question! | Learn Tips how to solve Olympiad Question without hassle and anxiety!

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[bernadetteuhlig]

I love your videos. First time i can watch videos without having the feeling of wasting my time afterwards.

I solved it like my 6th graders would do it, so I can use this in my next lesson:
840=2*2*2*3*5*7.
We call it Primfaktorzelegung in Germany. What is it called in english?
And than I sort it like
840=(2*3)*(2*5)*(2*7).
Now you see that a=2*3 and b=2*5 and c=2*7.

[Skank_and_Gutterboy]

Nice problem, fun to work! I took a different approach. I used cross-multiplication to put a and b in terms of c: a=(3/7)c and b=(5/7)c. Then I substituted these expressions into the equation abc=840: (3/7)c*(5/7)c*c = (15/49)*c^3 = 840.
c^3 = 2744 and c=14.

Finding c above answers the problem but I solved for a and b, then checked against the original equations just to verify. Now that we know c=14, a = (3/7)(14) = 6, and b = (5/7)(14) = 10. a/3 = 2, b/5 = 2, and c/7 = 2, that checks, they have to be equal to satisfy the 1st problem equation. abc = (6)(10)(14) = 840, that satisfies the 2nd equation.

[iZAPMath]

I love how you explained and the use of the colors! PreMath, you are the best!

[akuntumbal]

I like your method sir, it's easy to understand

[brendanharkin]

Elegant method. I went a different route; a=3c/7; b=5c/7 so abc = 15c^3/47= 840; c^3 = 840x49/15=2744 so c =14.

[anatoliy]

Thanks for a nice task with a simple its solution and a detailed explanation<)) You are an excellent teacher!

[shango]

That was just fabulous. Totally enjoyed your approach, very logical.

[shubhamojha]

First time pre math has taken so easy question. Very simple . Though you have brought easy question I say thankyou for your efforts

[govindashit]

So simple.
But I love your explanation. Thanks dear.

[juandelacruz]

I got the same answers but your method is much much better. My idea was to multiply each given by their reciprocal since multiplying a number by its reciprocal equals 1 (except for the number 0 of course) e.g. (a/3) (5/b) = 1....I also did it to the others then I equate 1 = 1. I had 3 new equations. There were lots of substitutions I made. I even got to extracting the cube root of c cube which is the cube root of 2744..then I got the rest of the answers from there. It was cumbersome.

[covenslayer]

I noticed that the denominators were all prime, so the only way they could be equivalent fractions is if the numerators were the same multiple of each respective denominator. The multiple couldn't be negative or 0 because 840 isn't negative or 0. So I started with multiple factor of 1 ==> 3 * 5 * 7 = 105 and multiple factor of 2 means 105 * 8 = 840 so 7 * 2 = 14 = c

[pk]

The value of c could also be negative 14 because any two of the numbers a, b, and c could be negative . The four possibilities for the triplet (a, b, c) are (6, 10, 14) ; (6, -10, -14) ; (-6, 10, -14) ; and (-6, -10, 14) . Another way to solve the problem is to solve for a in terms of c and b in terms of c ; then, substitute these values into abc = 840 ===> (15/49)c^3 = 840 ===> c^3=(14)^3 ===> c =14 .

[rahulpaul]

from relations we have
a=3c/7 and
And also abc =840
Or, 15c^3/7^2= 840
Or, c^3=7^3 × 2^3
So, c=14

[williamwingo]

Here's a different approach. No peeking, no calculator, all arithmetic done by hand:
a/3 = b/5 = c/7; abc = 840.
First, find b and c in terms of a, using the "product of means and extremes" rule in each case:
a/3 = b/5; 3b = 5a; b = 5a/3.
c/7 = a/3; 3c = 7a; c = 7a/3.
Now, abc = 840; substituting the values of b and c determined above:
(a)(5a/3)(7a/3) = 840; multiplying out the constants:
(35a^3)/9 = 840; solving for a^3:
a^3 = (840)(9)/35; collecting terms:
a^3 = (840/35)(9); carrying out the division 840/35:
a^3 = (24)(9); remembering that 24 = 8 x 3;
a^3 = (8)(3)(9) = (8)(27); solving for a:
a = cubrt[(8)(27)]; separating the two terms under the radical:
a = [cubrt(8)][cubrt(27)]; evaluating the cube roots mentally (honest):
a = (2)(3) = 6.
Returning to b and c:
b = 5a/3 = (5)(6)/3 = 30/3 = 10;
c = 7a/3 = (7)(6)/3 = (7)(2) = 14.
so a = 6; b = 10; and c = 14.
Check: abc should = 840. (6)(10)(14) = (6)(140) = 840.
Cheers. 🤠

[HappyFamilyOnline]

Really great👍
Thank you so much for your hard work😊😊

[narend]

what prompted you to assign to k ? Any theorems or postulates > Or ia it standard practice ? Please explain. Thanks

[Copernicusfreud]

I took the ratio a/3 = c/7 and solved for a and the ratio b/5 = c/7 and solved for b, and then plugged a and b into the abc = 840. I ended up with c^3 = 2, 744, and then c = 14. I did have to use the calculator to get the cube root of 2, 744.

[alster]

c=14

I paused the video and solved the rest on my own.

Fast forwarding to solving k, we have k=2

Since 105k³=840
k³=8
k= cbrt(8)
k= 2
Since c=7k
c=7(2)
c=14

[illyriumus]

Thank you! Just wondering...what grade level was this Olympiad question for?

[ashwanisharma]

You can easily do this qns by equating a/3=c/7, i.e a =3c/7 and similarly b=5c/7 now abc=840
Put a, b in terms of c, you get c³=8*7³, i.e. c= 14
I AM STUDYING IN CLASS 10TH CBSE (maths lover) SCORED 50/50 IN MATHS ..
OTHERWISE I LIKE YOUR MATHOD BUT I THINK IT IS EASY