Find a number with $3$ digits, knowing that the sum of its digits is $9$, their product is $24$ and also the number read from right to left is $\frac{27}{38}$ of the original.
Problem
Source: Cono Sur Olympiad, 1995, Problem#1
Tags: number theory
TRAN THAI HUNG
28.05.2006 19:48
It's easy.If i don't have any mistake then the solution is:
Call that number is abc We have:
$\ abc=\frac{27}{38}\ cba$ so $\ cba\equiv0\mod 38$
ANd $\ a+b+c=9$ so $\ cba\equiv0\mod9$
So cba have the form:9.38.k(0<k<3)
Try with k=1 and 2 we have k=1.
So abc=243
MAth is my life
PPonte
28.05.2006 23:42
What is mod 38?
rafaelguedez
03.06.2006 03:42
PPonte you say "a" is congruent to "b" mod 38 when 38 | a - b, congruence is usefull, you use it in most of number theory problems, read about it
1234567890987654321
14.06.2010 15:35
Let the 3-digit number be abc. We have a+b+c=9 and abc=24. Since 100c+10b+a=(27/38)(100a+10b+c),38 divides 100a+10b+c. Thus,by listing multiples of 38 we easily see that 342 satisfies all the conditions. Hence,the number is 342.
trimo
22.07.2017 13:25
abc = 24. So, one of them is a multiple of 3. Of course putting 6 won't satisfy. So, one of them is 3. Now the sum of the other two is 6 and their product is 8. So the other two numbers are 4 & 2. Also right to left is a multiple of 27. So, right to left will be 243. Hence the original number is 342.
MathIQ.
18.08.2024 21:29
$ABC=342$