Find all possible digits x, y, z such that the number 13xy45z is divisible
by 792.
(In reply to
Puzzle Answer by K Sengupta)
Let N=13xy45z
Now, 792=8*9*11. Since gcd(8,9,11)=1, it follows that N must be separately divisible by 8,9, and 11.
Now, N is divisible by 8
=> 45z is divisible by 8, and so z=6
So, N= 13xy456
N is divisible by 9 , if (1+3+x+y+4+5+6) = 19+x+y is divisible is divisible by 9
Therefore, x+y= 8, or 17 ..... (i)
Also, N is divisible by 11 if:
(6+4+x+1) - (5+y+3) is divisible by 11
or, 3+x-y is divisible by 11
or, x-y= -3, 8 ....... (ii)
From (i) and (ii), we must have:
(x+y, x-y)= (8,8), (17,-3)
The second case gives: y=10, a contradiction.
The first case gives: (x,y) = (8,0)
Consequently, (x,y,z) = (8,0,6) is the only possible solution to the given puzzle.
Edited on April 26, 2022, 1:12 am
Explanation to Puzzle Answer
