emmak wrote:
If a, b, and c are integers such that 0 < a < b < c < 10, is the product abc divisible by 3?
(1) If \(\frac{a}{1000}\) + \(\frac{b}{100}\) + \(\frac{c}{10}\) is expressed as a single fraction reduced to lowest terms, the denominator is 200.
(2) c – b < b – a
A++ to the question!
is the product abc divisible by 3? means is at least one a multiple of 3?
\(\frac{a}{1000}\) + \(\frac{b}{100}\) + \(\frac{c}{10}\) expressed as one fraction is
\(\frac{a+10b+100c}{1000}\) the factors of 200 are 2*5*2*5*2. To get the fraction to a 200 Den the sum must be a multiple of 5 and must NOT have a 2 or more 5s as factor, otherwise other semplification will be possbile.
1)the sum must be a multiple of 5, \(a+10b+100c\) if this is a multiple of 5 must end in 0 or 5.
(note that b will be the first digit of the tens and c will be the first digit of the hundreds and c is the unit). To end in 0 or 5 a must be 0 or 5. a cannot be 0 (0<a) so \(a=5\). Good
2) \(5+10b+100c\) must NOT have a 2 as factor or any more 5. divide by 5 \(1+2b+20c\) what remains after the first division MUST not be even or a multiple of 5.
This means that \(2b\neq{4}\) \(2b\neq{9}\) \(2b\neq{14}\) \(2b\neq{19}\)
and so on otherwise it will be divisibe: ie 2b=9 1+9+2C will be divisibe by 2 and 5.
Of all the values b cannot assume there is one that is interesting : \(2b\neq{14}\) \(b\neq{7}\) ( all other value of b are decimals of out of range 5-10)
So a=5 \(b\neq{7}\). With this info every combination abc will have a multiple of 3. The statement is SUFFICIENT.(2) c – b < b – a \(c+a<2b\) Not sufficient. b=3 c=5 a =1 YES b=4 c=5 a=1 NO.