When it is divided by 2, 3, 4, 5, or 6 there is a remainder of 1, but when it is divided by 7 there is no remainder?

Question

I need to know the lowest number that this happens too.

Answer 1

find the LCM of 2, 3, 4, 5, and 6

2=2

3=3

4=2*2

5=5

6=2*3

LCM=2*3*2*5=60

add 1 to 60 to get 61 but 61 is not divisible by 7

add 1 to 120 (the next multiple of 60) to get 121 but 121 is not divisible by 7

add 1 to 180 (the next multiple of 60) but 181 is not divisible by 7

add 1 to 240 but 241 is not divisible by 7

add 1 to 300 to get 301 and 301 is divisible by 7; 301/7=43 and when 301 is divided by either 2, 3, 4, 5, or 6 there is a remainder of 1 because 300 is a multiple of 2, 3, 4, 5, and 6

take the LCM of 60 and multiply it by 7 to get 420

420 is divisible by 7 (as well as by 2, 3, 4, 5, and 6)

if you add 420 to 301 you get a multiple of 7 because both numbers are divisible by 7

721 is divisible by 7 but 420 +300=720 which is divisible by 2, 3, 4, 5, and 6

therefore, if you add 1 to 720 you get 721 which when divided by 2, 3, 4, 5, and 6 leaves a remainder of 1

therefore 301 and 721 satisfy the conditions

add another 420 to 721 to get 1141 which is divisible by 7 but leaves a remainder of 1 when divided by the other numbers

when you add a multiple of 2, 3, 4, 5, 6, and 7 to 301 you always get a multiple of 7 but which is 1 more than a multiple of 2, 3, 4, 5, and 6

therefore you can keep adding 420 to get the next number in the list of numbers satisfying the conditions

301, 721, 1141, 1561,...

Answer 2

SUBMITTED TWICE CONSECUTIVELY BY ALLY.  I SUGGEST THAT THE OTHER ONE BE ANSWERED BY INTERESTED TUTORS.

Answer 3

the desired number hassato be a multiple of 7 and must be odd (because div. by 2 gives remainder of 1). There are more restrictions, but let's start with just these.  Candidates are 7, 21, 35, 49, 63, 77, 91, etc.

 

Divisibility by 5 with remainder 1 means candidates (odd) are 11, 21, 31, 41, 51, 61, 71, 81, 91, etc.

 

Candidates in both list are: 21, 91, ...

 

This is a START on how to approach the problem.  (You may have to extend the two lists above.)

 

Be sure to test the candidate that you eventually select against all of the criteria (rem. 1 if div. by 2, 3, 4, 5 or 6)