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
For product a*b*c to be divisible by 3, at least one of the three integers should be 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.
\(\frac{a}{1000}\) + \(\frac{b}{100}\) + \(\frac{c}{10}=\frac{a+10b+100c}{1000}=\frac{x}{200}\)
100c+10b+a=c00+b0+a=cba
So cba and 1000 have only factor 5 in common.
This tells us that 100c+10b+a or 3-digit number cba is a multiple of 5 but not of 25 or 2.a) 100c+10b+a will be multiple of 5,
only when a is 5 as 0<a<10.
b) Numbers ending in 25, 75 are multiples of 25, so
b cannot be 7, that is number cba cannot be c75.
So b can be 6, and c any of the remaining digits 7,8 or 9.
As b is a multiple of 3, product a*b*c will be divisible by 3. b can also be 8, then c is 9.
As c is a multiple of 3, product a*b*c will be divisible by 3.Product a*b*c is always divisible by 3.
Sufficient
(2) c – b < b – a
If c is 8 and b is 7
a) a is 4……answer is No
b) a is 3…..answer is yes
Insufficient
A