A number when divided by 2 3 4 5 6 leaves a remainder of 1 2 3 4 5, it is divisible by 7, then least number
Answers
Answer:
Step-by-step explanation:
301 is the lowest, with other numbers being in the form of 301 + a multiple of 420 and a general solution below:
Explanation:
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
2
×
2
×
3
×
5
=
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
5
×
60
. 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
⋮
⋮
⋮
⋮
+7
+420
+420
+60
⋮
⋮
⋮
⋮
⎞
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎠
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).