A number when divided by 28, leaves a remainder of 7. How do you find the remainder when the same number is divided by 35?

Below are the first 11 (positive) numbers that leave a remainder of #7# when divided by #28#. Note that the description implies that the numbers are of the from #28n+7# for some (positive) integer #n#.

Look at the first three columns:

#{:(bb" n ",bb" 28n+7 ",bb" 35q+r ",bb" R(n/5)"), (" "0," "" "7,35xx0+color(red)(7)," "" "0), (" "1," "" "35,35xx1+color(red)(0)," "" "1), (" "2," "" "63,35xx1+color(red)(28)," "" "2), (" "3," "" "91,35xx2+color(red)(21)," "" "3), (" "4," "" "119,35xx3+color(red)(14)," "" "4), (" "5," "" "147,35xx5+color(red)(7)," "" "0), (" "6," "" "175,35xx5+color(red)(0)," "" "1), (" "7," "" "203,35xx6+color(red)(28)," "" "2), (" "8," "" "231,35xx7+color(red)(21)," "" "3), (" "9," "" "259,35xx8+color(red)(14)," "" "4), (" "10," "" "287,35xx9+color(red)(7)," "" "0) :}#

First of all, it is now clear that knowing only the remainder when a number is divided by #28# is not enough to uniquely determine the remainder when it is divided by #35#.

But there is a pattern. So we don't need complete information about the original number or about #n#. (If we know one of those, we could find the other.)

The pattern has a cycle of 5 values, so it makes sense to look at the fourth column, the remainder when #n# is divided by #5#.

Writing the result and proving it for all possible #n# is left to the reader.