# To find: The smallest the smallest positive integer such that if we divide it by two, three and four, the remainder is one but seven divides the number evenly.

Congruence
To find:
The smallest the smallest positive integer such that if we divide it by two, three and four, the remainder is one but seven divides the number evenly.

2021-04-15
Suppose the required smallest positive integer is x, then from the given information, three congruence equations are
$$\displaystyle{x}\equiv{1}\pm{o}{d}{2},{x}\equiv{1}\pm{o}{d}{3},{x}\equiv{1}\pm{o}{d}{4}$$.
The congruence $$\displaystyle{x}\equiv{1}\pm{o}{d}{2}$$ means if x divided by 2 the remainder is 1.
So the number x is one of the number from the following list:
1,3,5,7,9, 11, 13, 15, 17, 19, 21, 23, 25, 27, 29, 31, 33, 35, 37, 39,41, 43, 45, 47, 49,51...
Similarly
The congruence $$\displaystyle{x}\equiv{1}\pm{o}{d}{3}$$ means if x divided by 3 the remainder is 1.
So the number x is one of the number from the following list:
1,4,7, 10, 13, 16, 19, 22, 25, 28, 31, 34, 37, 40, 43, 46, 49...
The congruence $$\displaystyle{x}\equiv{1}\pm{o}{d}{4}$$ means if x divided by 4 the remainder is 1.
So the number x is one of the number from the following list:
1,5, 9, 13, 17,21, 25, 29, 33, 37,41, 45, 49, 53, 57,...
Let a and b be the two numbers, a divides b evenly means the remainder is zero
The smallest number that is found in above three list is 1, but 1 is not divisible by seven evenly since the remainder is not zero so take next number
The next smallest number found in above three lists is 49 and 49 is divisible by 7 evenly. So the next smallest number that solves the three congruences and evenly divided by 7 is 49
Final Statement:
The smallest positive integer with the given condition is 49.