Find a three-digit number N = abc satisfying the
following conditions:
(i) the digits a, b, c are distinct;
(ii) N is equal to the sum of all two-digit numbers formed from the three
digits a, b, c.
hence
...(1)
from condition ii)
...(2)
now from (1) and (2)
...(*)
ok now take (mod 2)
that means a,b both even or both odd since their sum is even
now (*) mod 3
that means a+b+c is a multiple of 3 so N is a multiple of 3
now take (*) (mod 11)
if you take (mod 4)
you will have
a and b are distinct numbers so the difference between a and b is a multiple of 4,
mod 5
but 3 is primitive with 5 not common divisors between them
so
let's try some numbers let a be "1", b have two choices 5 or 9 false in the each probability
if a "2" , b = 6 , c=3 these fail in equation (mod 3)
if a "3" , b =7 , c =5 fail in (mod 11)
if a "4" , b =8 , c =7 fail in (mod 3)
if a "5" , b =9 , c =4 works in (mod 2,3,4,5,11) now justify the answer
is this true
594 = 33(5)+33(9)+33(4)
33*18 = 594 true
nice question
Starting from here
I find that if we put
we have
Therefore , is the multiple of
and is the multiple of .
I attempt to find out by putting and finally I obtain or but given that they are distinct so is the solution . Moreover , since , , .Therefore , the number is
Hello, Statsnoob2718!
I used a straight algebra approach.
It takes longer, but requires neither modulo arithmetic nor guessing.
Find a three-digit number satisfying the
following conditions:
(i) the digits are distinct,
(ii) is equal to the sum of all two-digit numbers formed from the digits .
The sum of the possible two-digits numbers is: .
This sum equals
We have: . .[1]
Since is a digit, then is an integer
. .
. . .[2]
Since is a digit, then is an integer
. .
Since is an integer, then is an integer.
. . Let
Substitute into [2]: .
Substitute into [1]: .
Hence, we have: .
Since are digits, must be 0.
. . Therefore: .