An Elementary Number Puzzle | AIME 1988 Problem 9
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- Опубліковано 16 жов 2024
- #Math #MathOlympiad #NumberTheory
In this video we solve a problem from the AIME 1988.
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I share Maths problems and Maths topics from well-known contests, exams and also from viewers around the world. Apart from sharing solutions to these problems, I also share my intuitions and first thoughts when I tried to solve these problems.
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I present the easiest explanation for those who may not feel at ease with congruences.
Step 1: Calculate the cube of every one digit number and conclude that 2^3=8 is the only cube with last digit 8. So x must have last digit 2.
Step 2: Calculate 12^3, 22^3,..., 92^3 and conclude that 42 and 92
are the only two digit numbers whose cubes end in 88.
Step 3:
(i) Calculate 042^3, 142^3, 242^3, ... , 942^3 and conclude 442 and 942 have cubes ending in 888.
(ii) Calculate 092^3, 192^3, 292^3, ... , 992^3, and conclude 192 and 692 have cubes ending in 888.
(iii) We know now that every x ending in 192, 442, 692, 942 has cube ending in 888 and there is no other such x. So, the smallest x is 192.
Yeah, back in the day my teacher was so tired he even fell asleep during the lesson. Of course no one understood congruences and I just shut the video immediately when I see them. I tried this route as well but it's difficult to calculate these cubes in head. So I took a cube of sum, ignored all exceeding values since we consider only 3 digits. a0 is obviously 2. So what's left is just 200a2+600a1²+100a1+20a1+8. 2nd digit is 2a1, so a1 can be 4 or 9. 3rd digit is 2a2+6a1²+a1 for a1=5. For a1=4 it's 2a2, so a2 is 4 or 9. For a1=9 it's 2a2+6, so a2 is 1 or 6. The smallest is 192. You can lose some values solving in memory, paper and pencil helps a lot.
Wth, thats too long bro
Use algebra (a+b)³=a³+3a²b+3ab²+b²...(*)
Last digit
Because 888 divide by 8 then the integer divide by 2 then b=2, assume a=xyz0
N=xyz0+2
Second digit
From (*), the second digit is only contributed by 3ab² that is equal 3*4*xyz0, thereforr 12*z= 8 mod 10 and get z= 4 or 9
First digit
First digit result is contributed by 3a²b+3ab² with b=2 and a=xyz0 , z= 4 or 9
This results for z=4
6+4+2y=8 mod 10
10+2y=8 mod 10
y=4 or 9
For z=9
6+10+2y=8 mod 10
y=1 or 6
We get
N=x442, x942, x192, x692 with x is non negatif integer
Smallest number 192
Heart filled with joy . Good explanation .God bless you .
I did this differently. Easy to show n ends in 2. Let n = 100a + 10b + 2. Cube this and we have
n^3 = 8 + 120b + 600b^2 + 1200a + higher powers of 10.
120b ends in 80 so b = 4 or 9.
Try each of these and we can find a in a similar way and we get for b=4, a=4 or 9. For b=9, a=1 or 6.
(These are just the single digit values of a).
n is the smallest of 442, 942,1l 192, 692. So n=192.
I started doing this in my head and got as far as 42 and 92. Then I watched the video and most of it was about finding the 100s digit. I was sure there was a quicker way! 👍
Hehe, I remember this problem from when I took the AIME in 10th grade (1988).
I answered 442 and got it wrong. Although 442 ^ 3 does end in 888, it is not the smallest such number.
@@amagilly Great
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Indian School Of Physics
idk about chemistry
Could the next 1997 AIME be problem 14
Can i suggest you to solve a geometry problem which questioned 2021 turkish math olympiad first step of team selection test
Which one
@@jinxedmoon8152 in booklet A, the 5th problem
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@@jinxedmoon8152 alright, Kayra_5000#1453
@@kemalkayraergin5655 can i have the booklet?
Thank you for the solution ☺️
888 is a good number ⭐
My solution : set x^3 = 888 (mod 1000) ------> use Hensel's lemma solve x^3 = 13 (mod 5^3) ------> use Chinese Remainder Theorem to x = 0 (mod 8) and x = 67 (mod 125) -----> we got x = 192 (mod 1000) that implies x = 1000n + 192 and the smallest positive integer is 192 at n=0 obviously.
Nice idea, but you miss that 442^3 = (250 + 192)^3 = 250^3 + 3*192*250^2 + 3*192^2*250 + 192^3 = (divisible by 1000) + (divisible by 1000 because 192 is even) + (divisible by 1000 because 192^2 is divisible by 4) + 192^3 ending with triple 8. By a similar argument, also 692^3 and 942^3 ends with 888, and this Chinese Remainder Theorem method misses these too.
Instead one can consider all x such that x^3 = 0 (mod 8), and all even x satisfy.
What is "AIME" ???
Why those expressions in the very beginning?
To prove that it suffices to look at the last three digits of our answer. Ie, to prove that if a solution exists, there also exists a solution between 0 and 1000.
I saw this problem in a random book, and I was bored, and I went further and found 1942^3 = 7323988888 with five 8's at the end
⭐️
After noticing that the last digit is 2, Then only those cubes ending 88 has the possibility of ending in 888. We use 2^3 = 8, 12^3 = 1728 , 22^3 = 10648, 8^3 = 512 and 18^3 = 5832 and know the ending for the rest: 32 = 50 - 18, 42 = 50 - 8, 52 = 50 + 2, 62 = 50 + 12, 72 = 50 + 22, 82 = 100 - 18, 92 = 100 - 8. We see only 42 and 92 has 88 as last two digits. eg 92^3 = (100-8)^3 where 8^3 = 512 implies 92^3 ends in 88. 62^3 =(50 + 12)^3 has last 2 digits as same as 12^3. 18^3 = 5832 ends in 32 implies 32^3 ends 68. . From here onwards I prefer n = 42 (mod 50) as in the video.
There is no 2-digit number, when cubed produces ...888 in the end.It must be a 3-digit number. Obviously the last digit is 2.In middle the required digit must be such when multiplied by 2 , must produce 8 in units place. It is either 4 or 9.
Blud spent 13 mins on an elementary number puzzle 💀
192 so easy
the beginning of the video was unnecessary because all aime solutions are integers from 0 to 999 inclusive
Let that integer be x mod (10) give us x ≡ 2 mod (10). Checking x mod (100), x is either 92 or 42 mod (100). If x mod (100) is 92, x is 192 mod (1000) and of x mod(100) is 42, x is 442 mod(100). Therefore the smallest integer is 192
There are two more solutions (neither of which is the smallest): 942 and 692.
Wow 😮😮, 😮
Cant you solve this without using mods at all though??
Where the hell do you get x is congruent to 2 mod 10 from when youbjave bo idea what x is??
@@leif1075 because x³ is 8 mod 10, x can only be 2 mod 10. No x satisfies the 1st equation without satisfying the 2nd.