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In how many ways can we pick three different numbers out of the group \(1,2,3,\ldots,100\)such that the largest number is larger than the product of the two smaller ones? (The order in which we pick the numbers does not matter.)

 Jan 20, 2021
 #1
avatar+266 
0

Not sure but this is what I have...

1≤a

b

a2<100a2<100 so a<10a<10.

ab<100ab<100 so ab≤99ab≤99 so b≤99ab≤99a so b≤⌊99a⌋b≤⌊99a⌋.

b+1≤c≤100b+1≤c≤100.

 

∑9a=1(∑⌊99a⌋b=a+1(∑100c=ab+11))=∑a=19(∑b=a+1⌊99a⌋(∑c=ab+11001))=.

∑9a=1(∑⌊99a⌋b=a+1[100−ab])∑a=19(∑b=a+1⌊99a⌋[100−ab])

∑99b=2(100−b)+∑49b=3(100−2b)+∑33b=4(100−3b)+∑24b=5(100−4b)+∑19b=6(100−5b)+∑16b=7(100−6b)+∑14b=8(100−7b)+∑12b=9(100−8b)+∑11b=10(100−9b)=∑b=299(100−b)+∑b=349(100−2b)+∑b=433(100−3b)+∑b=524(100−4b)+∑b=619(100−5b)+∑b=716(100−6b)+∑b=814(100−7b)+∑b=912(100−8b)+∑b=1011(100−9b)=

∑98k=1k+∑94k=2;+2k+∑88k=1;+3k+∑80k=4;+4k+∑70k=5;+5k+∑58k=4;+6k+∑44k=2;+7k+∑28k=4;+8k+∑10k=1;+9k=∑k=198k+∑k=2;+294k+∑k=1;+388k+∑k=4;+480k+∑k=5;+570k+∑k=4;+658k+∑k=2;+744k+∑k=4;+828k+∑k=1;+910k=

∑98k=1k+∑47k=12k+∑29k=0;(3k+1)+∑20k=14k+∑14k=15k+∑9k=06k+4+∑7k=07k+2+∑308k+4+∑1k=09k+1=∑k=198k+∑k=1472k+∑k=0;29(3k+1)+∑k=1204k+∑k=1145k+∑k=096k+4+∑k=077k+2+∑038k+4+∑k=019k+1=

98∗992+2∗47∗482+329∗302+30+4∗20∗212+514∗152+69∗102+4∗10+7∗7∗82+2∗8+83∗42+4∗4+9∗1∗22+2∗1=98∗992+2∗47∗482+329∗302+30+4∗20∗212+514∗152+69∗102+4∗10+7∗7∗82+2∗8+83∗42+4∗4+9∗1∗22+2∗1=

49∗99+47∗48+3∗29∗15+30+2∗20∗21+5∗7∗15+3∗9∗5+40+7∗7∗4+16+4∗3∗4+16+9+2=49∗99+47∗48+3∗29∗15+30+2∗20∗21+5∗7∗15+3∗9∗5+40+7∗7∗4+16+4∗3∗4+16+9+2=

49∗99+47∗48+29∗45+30+20∗42+35∗15+27∗5+40+49∗4+32+48+11=10269

(Okay, I don't know how to simplify)

 

cheekycheekycheeky

 Jan 20, 2021
 #2
avatar+266 
+1

Oh, sorry, seems like the LateX didn't work, so it might have some missing parts.

DewdropDancer  Jan 20, 2021

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