This kind of problem is pretty tricky if you've never seen anything
like it before.
What you're looking for is a number that leaves a remainder of 1 when
divided by 2, 3, 4, 5, or 6, but that is evenly divisible by 7. So as
you already noticed, you only need to look at multiples of 7. Let's
try 7. We get off to a good start. 7 leaves a remainder of 1 when
divided by 2, 3, or 6. Unfortunately it doesn't work for 4. So let's
see what we would have to do to get to a number which leaves a
remainder of 1 when you divide it by 4, without losing what we've
already got.
Well, if you take a number that leaves a remainder of 1 when you
divide by 2 and add 2 to that number, you get another number that
leaves a remainder of 1. So if you add any multiple of 2, you still
get a number that leaves a remainder of 1 when you divide by 2.
Simliarly, if you take a number that leaves a remainder of 1 when you
divide it by 3 and add any multiple of 3 to it, you get another number
that leaves a remainder of 1 when you divide it by 3.
Anything that leaves a remainder of 1 when divided by 2 or 3
automatically leaves a remainder of 1 when you divide by 6, so we get
that for free.
So if you take 7 and add anything which is a multiple of 2 and 3 and 7
to it, you get another multiple of 7 which leaves a remainder of 1
when you divide it by 2 or 3. So we don't lose anything if we add a
multiple of 42 (that's 2 x 3 x 7) to 7. So lets' see if we can
satisfy another of our conditions.
7 + 42 = 49. Hey, that's good! We've got a number that leaves a
remainder of 1 when you divide by 4. It still doesn't work for 5,
though. Can we use the same trick again?
If we add a multiple of 4 we still get a number that leaves a
remainder of 1 when you divide by 4. So if we add a multiple of
3 x 4 x 7, we'll get another number that satsifies all the conditions
so far. There's only one to go!
Okay, 49 + 84 = 133. No good.
133 + 84 = 217. No good.
217 + 84 = 301. Bingo!!
So 301 is a possible solution. Are there any others? Well, yes,
because you can add a multiple of 3 x 4 x 5 x 7 to get another one.
But usually what you're looking for is the smallest solution, so 301
is probably the answer you want.
thanx
heyaaaa