TITLE: 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.
QUESTION [7 upvotes]: I came across a question which is as follows:
Find out the smallest number which leaves remainder of 1 when divided by 2, 3, 4, 5, 6 but divided by 7 completely.
What I did is given below step wise.
Step 1- Find out the LCM of 2, 3, 4, 5, 6 which is 60.
Step 2- Add 1 to 60 which is 61.
Step 3- Multiple 61 by 7 repeatedly till it fulfills the condition that remainder should be 1.
Step 4- I got the answer 146461 which seems to correct.
So now my question is:
1) Is this answer correct? If yes how to verify that this is smallest number which fulfills above condition?
2) I think this is not the best way to do this question. So Can anyone give a better way to solve this problem?
Thanks in advance

REPLY [2 votes]: 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.
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