how many 4-digit numbers are there such that the thousands digit is equal to the sum of the other three digits?
firstfollowingfirstfollowingfirstfollowingfirstfollowingfirstfollowingfirstfollowingfirstfollowingfirstfollowingfirstfollowingfirstfollowingdigitdigitsdigitdigitsdigitdigitsdigitdigitsdigitdigitsdigitdigitsdigitdigitsdigitdigitsdigitdigitsdigitdigitsabcdabcdabcdabcdabcdabcdabcdabcdabcdabcdb+c+d<10b+c+d<10b+c+d<10b+c+d<10b+c+d<10b+c+d<10b+c+d<10b+c+d<10b+c+d<10b+c+d<100000110022003300440055006600770088009900100121013201430154016501760187019801200231024202530254027502860297023003410352036303540385039603400451046204730454049504500561057205830554056006710682069306700781079207800891089009101021103210431054106510761087109810201131114211531164117511861197113012411252126312741285129612401351136213731384139513501461147214831494146015711582159315701681169216801791179018202031204220532064207520862097203021412152216321742185219621402251226222732284229522502361237223832394236024712482249324702581259225802691269027303041305230633074308530963040315131623173318431953150326132723283329432603371338233933370348134923480359135903640405140624073408440954050416141724183419441604271428242934270438143924380449144904550506150725083509450605171518251935170528152925280539153905460607160826093607061816192618062916290637070817092708071917190728080918090819090sum=55sum=45sum=36sum=28sum=21sum=15sum=10sum=6sum=3sum=1
sum = 55 + 45 + 36 + 28 + 21 + 15 + 10 + 6 + 3 + 1 = 220
sum - 1 (0000) = 219
how many 4-digit numbers are there such that the thousands digit is equal to the sum of the other three digits?
4-digit-number:~abcd=a103+b102+c10+da=b+c+d⇒(b+c+d)103+bcd10<10000⇒(b+c+d)103<10000−bcd10⇒(b+c+d)<10000−bcd10103⇒(b+c+d)<10−bcd101000⏟0…0.999⇒(b+c+d)<10
b+c+d<10(b=0)cd000001002003004005006007008009010011012013014015016017018020021022023024025026027030031032033034035036040041042043044045050051052053054060061062063070071072080081090sumb=0=1+2+3+4+⋯+10
bsum01+2+3+4+5+6+7+8+9+10=5511+2+3+4+5+6+7+8+9=4521+2+3+4+5+6+7+8=3631+2+3+4+5+6+7=2841+2+3+4+5+6=2151+2+3+4+5=1561+2+3+4=1071+2+3=681+2=391=1sum=220
There are 220 - (abcd = 0000) = 219 4-digit numbers
Hallo Melody, i think add for the first digit 5 the following digits 221 with 3 numbers of ways and you get
216+3 also 219 4-digit numbers
Thanks Heureka, I did miss that one.
I don't actually understand what you have done :/
firstfollowingfirstfollowingfirstfollowingfirstfollowingfirstfollowingfirstfollowingfirstfollowingfirstfollowingfirstfollowingfirstfollowingdigitdigitsdigitdigitsdigitdigitsdigitdigitsdigitdigitsdigitdigitsdigitdigitsdigitdigitsdigitdigitsdigitdigitsabcdabcdabcdabcdabcdabcdabcdabcdabcdabcdb+c+d<10b+c+d<10b+c+d<10b+c+d<10b+c+d<10b+c+d<10b+c+d<10b+c+d<10b+c+d<10b+c+d<100000110022003300440055006600770088009900100121013201430154016501760187019801200231024202530254027502860297023003410352036303540385039603400451046204730454049504500561057205830554056006710682069306700781079207800891089009101021103210431054106510761087109810201131114211531164117511861197113012411252126312741285129612401351136213731384139513501461147214831494146015711582159315701681169216801791179018202031204220532064207520862097203021412152216321742185219621402251226222732284229522502361237223832394236024712482249324702581259225802691269027303041305230633074308530963040315131623173318431953150326132723283329432603371338233933370348134923480359135903640405140624073408440954050416141724183419441604271428242934270438143924380449144904550506150725083509450605171518251935170528152925280539153905460607160826093607061816192618062916290637070817092708071917190728080918090819090sum=55sum=45sum=36sum=28sum=21sum=15sum=10sum=6sum=3sum=1
sum = 55 + 45 + 36 + 28 + 21 + 15 + 10 + 6 + 3 + 1 = 220
sum - 1 (0000) = 219