Find the least number which when divided by 2, 3, 4, 5, 6 leaves a remainder of 1 but it is divided by 7 completely.

Solution 1:

You definitely need $n\equiv 1\pmod {60}$ and $n\equiv 0\pmod 7$. So the trick is to apply the Chinese remainder theorem. Solve $60x+7y=1$ with $(x,y)=(2,-17)$.

Then $n\equiv 1\cdot 7\cdot (-17)+0\cdot 60\cdot 2\pmod{420}$, or $n\equiv -119\pmod{420}$. The smallest such positive number is $420-119=301$.

Solution 2:

Consider $60k+1,$ $$60(7n)+1=420n+1$$ $$60(7n+1)+1=420n+61$$ $$60(7n+2)+1=420n+121$$ $$60(7n+3)+1=420n+181$$ $$60(7n+4)+1=420n+241$$ $$60(7n+5)+1=420n+301$$ $$60(7n+6)+1=420n+361.$$ Now by divisibility test for 7, among $1, 61, 121, 181, 241, 301, 361$, only $301$ is divisible by $7.$

Therefore any number of the form $420n+301$ satisfies the given requirements.

Solution 3:

Here's another way to tackle it. You already figured out that you're looking for $60k+1$. When you multiply $60$ by $k$ you want a predecessor to a multiplication of $7$.

$60 \equiv 4 \pmod 7$ and $5 \times 4 = 20$ which is a predecessor to a multiplication of $7$.

So $k=5$.

Solution 4:

Since no-one has mentioned it in their answers so far, your step 3 is wrong, or at least ill-advised, in that you are coming across a lot of answers that do not meet your carefully-set-up satisfaction of the first five conditions. For example, $61\times 7 = 427$ does not meet the desired remainders for $4$ or $5$.

The problem is that you have abandoned the safety of $n\equiv 1 \bmod 60$. The way to retain this in a simple search is to add 60 repeatedly looking for divisibility by 7.

We can do a bit better than that, though. We can translate into smaller numbers to make life easier for ourselves. $60 \equiv 4 \bmod 7$ (and of course $61 \equiv 5 \bmod 7$) so we can ask: how many 4s do we need to add to 5 before the answer is divisible by 7?

Again we can slog through the possibilities but the multiples of 7 are easier to cope with. We add two 4s (13 - and maybe see that we've reached $6 \bmod 7$) and add another two 4s to reach $21 \equiv 7 \equiv 0 \bmod 7$ - adding four 4s altogether. So back on our Actual Problem, we need to add four 60s - $4\times 60=240$ to our original $61$ so $ 240+61=301$ for the smallest positive solution.

Note that $60$ is your interval between satisfying those first 5 conditions, but we could also have started our search at $1$ rathe than $61$, with (of course) the same eventual result.