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.

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- April 26th 2010, 09:25 AMStatsnoob2718Find a three-digit number N = abc
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. - April 26th 2010, 12:33 PMAmer

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

(Itwasntme) nice question - April 26th 2010, 10:34 PMsimplependulum
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 - April 27th 2010, 08:56 AMSoroban
Hello, Statsnoob2718!

I used a straight algebra approach.

It takes longer, but requires neither modulo arithmetic nor guessing.

Quote:

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: .