Solution

(A) Yes, for example: \(72=12+24+36\). Let us check that the sum of any two of these numbers is divisible by the third number:

\[\begin{aligned} & \circ \quad(12+24): 36=36: 36=1 \\ & \circ \quad(12+36): 24=48: 24=2 \\ & \circ \quad(24+36): 12=60: 12=5 \end{aligned}\]

**(B)** We show that this is not possible. By contradiction, assume that \(a+b+n=73\) and \(a+b\) is divisible by \(n\). If \(a+b\) is divisible by some number \(n\), then the sum of all three numbers \(a+b+n\) is also divisible by \(n\). Therefore, \(73=a+b+n\), must be divisible by \(a\), \(b\), and \(n\). But the number \(73\) is prime, divisible only by \(1\) and \(73\). Hence three distinct numbers can only take two values \(1\) or \(73\), which is impossible.

Atrisinājums

(A) Jā, var šādi \(72=12+24+36\). Pārbaudām, vai katru divu šo skait|u summa dalās ar trešo skaitli:

\[\begin{aligned} & \circ \quad(12+24): 36=36: 36=1 \\ & \circ \quad(12+36): 24=48: 24=2 \\ & \circ \quad(24+36): 12=60: 12=5 \end{aligned}\]

**(B)** Pamatosim, ka prasītais nav iespējams. Ja divu skaitļu \(a\) un \(b\) summa dalās ar kādu skaitli \(n\), tad arī visu trīs šo skaitlu summa \(a+b+n\) dalās ar skaitli \(n\). Tātad visu trīs skaitļu summai jeb \(73\) jādalās ar jebkuru no trīs saskaitāmajiem. Bet skaitlis \(73\) ir pirmskaitlis, kas dalās tikai ar \(1\) un \(73\). Tātad šie trīs dažādie skaiți var pieņemt tikai vērtības \(1\) vai \(73\), kas nav iespējams.