When a number is divided by #2#, #3#, #4#, #5# or #6#, we always get a remainder of #1#. But on dividing a number by #7#, we find the number is divisible? What is the smallest such number? What are other such numbers? and how to solve such problems?

1 Answer

301 is the lowest, with other numbers being in the form of 301 + a multiple of 420 and a general solution below:


I read this problem and haven't been able to let it go - thanks for asking it (although I may not sleep tonight...)

We're looking for a number that is divisible by 2, 3, 4, 5, and 6 (meaning that N is a multiple of those numbers) and that number +1 being divisible by 7.

We can make progress on this by seeing that the Lowest Common Multiple of these numbers is #2xx2xx3xx5=60#. Now the key is to find a multiple of 60 such that when we add 1 to it, the number becomes divisible by 7.

After some trial and error, I found that the lowest number that works is 300, which is #5xx60#. When we take #300+1=301, 301-:7=43#.

With some more trial and error, I found that the next multiples of 60 increase by 7, so 5, 12, 19,... and that the resulting multiple of 7 increases by a factor of 60, so 43, 103, 163, etc.

To summarize so far, we have:

#(("factor with 60", "multiple", "+1", "/7"),(5,300, 301, 43),(12, 720, 721, 103),(19,1140,1141, 163),(vdots, vdots, vdots, vdots),("+7","+420","+420","+60"),(vdots, vdots, vdots, vdots))#

So we have the smallest number (301) and other such numbers (721. 1141, etc). And we have a pattern we can follow for other questions of this sort:

  • find the LCM of the numbers where there is a remainder

  • find the lowest occurrence where, by adding the remainder to the LCM, it's evenly divisible by the other number (my method being trial and error)

  • find the pattern of factors (which appears to be additive, using the factor of the "other side" - the "factor along with 60" increasing by 7, while the "/7" increases by 60).