An easy problem.
a , b , c are three different, non-zero, digits, such that the sum of the 6 numbers occurring with them satisfies
a b c + a c b + b a c + b c a + c a b + c b a = 1 3 3 2
What are the values of a , b , c ?
This problem appeared at the ONMAPS 2010
This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try
refreshing the page, (b) enabling javascript if it is disabled on your browser and,
finally, (c)
loading the
non-javascript version of this page
. We're sorry about the hassle.
4 solutions
Rewriting is: ( 1 0 0 a + 1 0 b + c ) + ( 1 0 0 a + 1 0 c + b ) + ( 1 0 0 b + 1 0 a + c ) + ( 1 0 0 b + 1 0 c + a ) + ( 1 0 0 c + 1 0 a + b ) + ( 1 0 0 c + 1 0 b + a ) = 1 3 3 2 which is 2 2 2 a + 2 2 2 b + 2 2 2 c = 1 3 3 2 ⟹ 2 2 2 ( a + b + c ) = 1 3 3 2 ⟹ a + b + c = 6 As a , b , c are different and non-zero DIGITS, a + b + c = 1 + 2 + 3 = 6 so the answer is 1 , 2 , 3
so much complication, just do this: 1=a, 2=b, and 3=c abc=123 plus acb=132 plus bac=213 plus bca=231 plus cab= 312 plus cba=321. That is 1332
But your process doesn´t have any algebraic eplanation
If you see the unit digit of the six numbers and add them you will get 2(a+b+c) we know that the sum of these numbers ends with 2 so a+b+c should end with 1 or 6.Only 1,2,3 satisfy this.
1=a, 2=b, and 3=c abc=123 plus acb=132 plus bac=213 plus bca=231 plus cab= 312 plus cba=321. That is 1332
if we see the unit digits ,adding them we get 2a+2b+2c = unit digit 2. so as only1,2,3 satisfies. easy guess!