If a,b,c,d,e and f are distinct positive integers and

It gets pretty easy when both the statements are used.

If we take the numbers as 1, 2 ,3 ,4 ,5 ,6. ( Dont know if 0 can be a possibility )

10ac+bc= 100d+10e+f
or , c ( 10a + b) = 100d + 10e + f { putting the values of b and f )
or , c ( 10a + 4) = 100d + 10e + 2 {LHS will have 2 in the units place.
In order to match this the value of c
has to be 3 , because 3 * 4 = 12 }
or, 3 (10a + 4) = 100d + 10e + 2
or, 30a + 12 = 100d + 10e + 2
or, 30a + 10 = 100d + 10e
or, 3a + 1 = 10d + e {Now put a = 5 , d = 1 , and e = 6 }

This can be solved. So a = 5

However, my doubt is can't we come to the same conclusion without knowing the values of b and f ?

If a,b,c,d,e and f are distinct positive integers and 10ac+bc= 100d+10e+f, what is the value of a?

(1) a+b+c+d+e+f < 22
(2) b=4 and f=2

So I got soooo bored of writing applications that I had to find something to do with my brain. That is how I am back here. btw.. hope everyone's prep is going on well.

ok here is my attempt.

I satrted with statement B (looks like it has more info than A)
so if (10a+4)c = 100d+10e+2
=> c has to be equal to 3 (you multiply a number ending in 4 with a number ending in 2 the multiple has to be 3.
this quick step tells me b = 4 ; c = 3 & f =2.

but a, d,e could be any numbers

no statement A: this condition gives many possibilities.

combining
a+d+e < 13 (22-4-3-2).....(i)
also 30a+12 = 100d+10e+2
=> 3a +1 = 10d +e.....(ii)
case 1: d= 1
then a+e< 12
only possible value of a that will give each number as a distinct integer is 4

case 2: d=2
then a+e<11
no possible values

case3: d=3
a+e<10
no possible values.

Thus C.

Anyone else tried this? Any other ideas?

If the first statement is a+b+c+d+e+f <=22 then A will be the answer.

But with the given conditions as it is the answer would be E.

Hey kevin where r u?

Excellent approach, but were you too quick to dismiss A?

if a+b+c+...+f< 22, then (a,b,c,d,e,f) must be a permutation of the first six positive integers.

Intuitively, I know that (A) is sufficient, just can't seem to come up with the right combination in under two minutes. Gosh, the GMAT seems to be getting harder all the time

Keep in mind that this is NOT a GMAT question, so don't estimate the difficulty of the exam by the questions I post. GMATPrep is by far the best indicator. These questions are just for fun and meant to help us to look for creative solutions.

If a+b+c+d+e+f<22 and each is a distinct positive integer, it follows that (a,b,c,d,e,f) is a permutation of (1,2,3,4,5,6).

We are told that 10ac+bc=100d+10e+f

Let's write it as:

.ab
x.c
___
def

Focussing on the units digits, it's clear that none of them can be 1 or 5.
Could b and c be 2 and 3 or vice versa? If that case, f would be 6, the two digit number de would simply be ac, which is not possible since the six numbers have to be distinct. This makes it clear that bc>9, since we NEED a carry-over. Only 3 and 4 will do! I have to start a class! Can somebody continue?

Hey kevin, i think the above equation will not have a solution if all of them are distinct and the solution set is 1,2,3,4,5,6

so if b and c are 3 and 4 (or vice versa) f would be 2

Let's see if (b,c)=(3,4)

.a3
X 4
----
de2

The two digit number de is 4*a+1- a would have to be 5 or 6, but in each case d would be 2- not permissible

How about (b,c)=(4,3)?

.a4
X.3
----

de2

This time, the two digit number de is 3*a+1- a must be 5, so de is 16

54*3=162