Neka je skup
A = { a , a + 1 , a + 2 , a + 3 , … , a + m − 1 } , A=\{a,a+1,a+2,a+3,\ldots,a+m-1\}, A = { a , a + 1 , a + 2 , a + 3 , … , a + m − 1 } , B = { b , b + 1 , b + 2 , b + 3 , … , b + m − 1 , b + m , … , b + 2 m − 1 } . B=\{b,b+1,b+2,b+3,\ldots,b+m-1,b+m,\ldots,b+2m-1\}. B = { b , b + 1 , b + 2 , b + 3 , … , b + m − 1 , b + m , … , b + 2 m − 1 } . Vrijedi
a + ( a + 1 ) + ( a + 2 ) + ( a + 3 ) + ⋯ + ( a + m − 1 ) = 2 m , b + ( b + 1 ) + ( b + 2 ) + ( b + 3 ) + ⋯ + ( b + 2 m − 1 ) = m , ∣ b + 2 m − 1 − ( a + m − 1 ) ∣ = 99. \begin{aligned} a+(a+1)+(a+2)+(a+3)+\cdots+(a+m-1)&=2m,\\ b+(b+1)+(b+2)+(b+3)+\cdots+(b+2m-1)&=m,\\ \left|b+2m-1-(a+m-1)\right|&=99. \end{aligned} a + ( a + 1 ) + ( a + 2 ) + ( a + 3 ) + ⋯ + ( a + m − 1 ) b + ( b + 1 ) + ( b + 2 ) + ( b + 3 ) + ⋯ + ( b + 2 m − 1 ) ∣ b + 2 m − 1 − ( a + m − 1 ) ∣ = 2 m , = m , = 99. Odnosno, rješavamo sustav jednadžbi
m a + ( 1 + 2 + 3 + ⋯ + m − 1 ) = 2 m , 2 m b + ( 1 + 2 + 3 + ⋯ + m − 1 + ⋯ + 2 m − 1 ) = m , ∣ b − a + m ∣ = 99. \begin{aligned} ma+(1+2+3+\cdots+m-1)&=2m,\\ 2mb+(1+2+3+\cdots+m-1+\cdots+2m-1)&=m,\\ |b-a+m|&=99. \end{aligned} ma + ( 1 + 2 + 3 + ⋯ + m − 1 ) 2 mb + ( 1 + 2 + 3 + ⋯ + m − 1 + ⋯ + 2 m − 1 ) ∣ b − a + m ∣ = 2 m , = m , = 99. Iz prve jednadžbe proizlazi
m a + ( 1 + m − 1 ) ( m − 1 ) 2 = 2 m , ma+\frac{(1+m-1)(m-1)}{2}=2m, ma + 2 ( 1 + m − 1 ) ( m − 1 ) = 2 m , pa
2 m a + m ( m − 1 ) = 4 m , a = 5 − m 2 . 2ma+m(m-1)=4m, \qquad a=\frac{5-m}{2}. 2 ma + m ( m − 1 ) = 4 m , a = 2 5 − m . Iz druge jednadžbe sustava proizlazi
2 m b + ( 1 + 2 m − 1 ) ( 2 m − 1 ) 2 = m , 2mb+\frac{(1+2m-1)(2m-1)}{2}=m, 2 mb + 2 ( 1 + 2 m − 1 ) ( 2 m − 1 ) = m , pa
4 m b + 2 m ( 2 m − 1 ) = 2 m , 4mb+2m(2m-1)=2m, 4 mb + 2 m ( 2 m − 1 ) = 2 m , odnosno 4 m b + 4 m 2 = 4 m 4mb+4m^2=4m 4 mb + 4 m 2 = 4 m b = 1 − m b=1-m b = 1 − m
Tada iz treće jednadžbe slijedi
∣ b − a + m ∣ = 99 , \left|b-a+m\right|=99, ∣ b − a + m ∣ = 99 , odnosno
∣ 1 − m − 5 − m 2 + m ∣ = 99 , ∣ m − 3 ∣ = 198. \left|1-m-\frac{5-m}{2}+m\right|=99, \qquad |m-3|=198. 1 − m − 2 5 − m + m = 99 , ∣ m − 3∣ = 198. Odatle je m = 201 m=201 m = 201 m = − 195 m=-195 m = − 195 m > 0 m>0 m > 0 m = 201 m=201 m = 201 m m m a a a b b b
a = 5 − 201 2 = − 98 , b = − 200. a=\frac{5-201}{2}=-98,\qquad b=-200. a = 2 5 − 201 = − 98 , b = − 200. Zato je
A = { − 98 , − 97 , − 96 , − 95 , … , 102 } , A=\{-98,-97,-96,-95,\ldots,102\}, A = { − 98 , − 97 , − 96 , − 95 , … , 102 } , B = { − 200 , − 199 , − 198 , − 197 , … , 201 } . B=\{-200,-199,-198,-197,\ldots,201\}. B = { − 200 , − 199 , − 198 , − 197 , … , 201 } . 