Nsejs 2015
Number Theory
Level
3
How many four digit numbers divisible by twenty nine have the sum of their digits 29?
The answer is 5.
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1 solution
Let the digits beb,c,d,29−b−c−db,c,d,29−b−c−d As 0≤b,c,d≤9,0<b+c+d≤27<290≤b,c,d≤9,0<b+c+d≤27<29 Now, (29−b−c−d)+10b+100c+1000d=29+9b+99c+999d≡−20b+12c−16d (29−b−c−d)+10b+100c+1000d=29+9b+99c+999d≡−20b+12c−16d We need 29∣(5b−3c+4d)≡−24b−3c+33d⟺29∣(c+8b−11d)29∣(5b−3c+4d)≡−24b−3c+33d⟺29∣(c+8b−11d) For 29∣(8b−11d)⟹8b−11d=(29k−c)(11⋅3−8⋅4)29∣(8b−11d)⟹8b−11d=(29k−c)(11⋅3−8⋅4) ⟺8(b+116k−4c)=11(87k+d−3c)⟺b+116k=11m+4c⟺8(b+116k−4c)=11(87k+d−3c)⟺b+116k=11m+4c ⟺b≡5k+4c(mod11)⟺b≡5k+4c(mod11) and d≡3c+k(mod11)d≡3c+k(mod11) Test for c;0≤c≤9c;0≤c≤9 ensuring 0≤b,c,d≤90≤b,c,d≤9 and b+c+d≥20