Three friends Riley, Sammy and Todd have a certain number of oranges in their possession.

It is known that:

(1) The number of oranges possessed by each of the three friends are two digit positive integers.

(2) The number of oranges possessed by Sammy is obtained by adding the two digits of the number of oranges possessed by Riley and summing the result with the original number of oranges in Riley's possession.

(3) The number of oranges possessed by Todd is obtained by reversing the digits corresponding to the number of oranges in Sammy's possession.

(4) The total number of oranges possessed by the three friends is 272.

Determine analytically the number of oranges possessed by each of the three friends.

Let the number of oranges possessed respectively by Riley and Sammy be (10x+y) and (10a+b).
Then the total number of oranges in Todd's possession is (10b+a)

By the problem:
(i) 10x+ y + 11(a+b) = 272; (ii) 10a+ b = 11x + 2y

Reducing both sides of (i) in modulo 11, we have:
y-x = - 3(Mod 11).
Since x and y must be less than 10, this is possible if y- x = -3, 8.

If y -x = 8, then:
Either (y,x) = (9,1); giving a+b=23 in terms of (i). This is not feasible as max(a+b)< 9+9 = 18.
Or, (y,x) = (8,0); giving a+b = 24 in terms of (i), which is not feasible for similar reason.

If y-x = -3; then x = y+3, so that:
a+ b+ y = 22 and 10a + b -13y = 33 in terms of (i) and (ii).
Accordingly, 9a - 14y = 11.
Since, both a and y must be less than 10, this is feasible only if (a,y) = (9,5)
Hence, x =5+3 = 8, so that b = 22 - 8 - 5 = 9.

Consequently, the respective number of oranges in the possession of Riley, Sammy and Todd are 85, 98 and 89.

******************************************

Also refer to the respective solutions posted by Federico Kereki and Ady TZIDON in the Comments section.

Comments: (
You must be logged in to post comments.)