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Find the smallest positive integer x so that the four numbers represented by x/3, (x+2)/4, (x+3)/7, and (x+5)/11 are all integers.

The answer is: 270

An explanation of how to solve this problem would be most appreciated.

Thank you

x has to be even so that x+2 is a multiple of 4

x+5 is a multiple of 11 so x can be 6,28,50,72,94,116,138,etc

x+3 is a multiple of 7 so x can be 4,18,32,46,60,74,88,102,116

so 116 is the first number that occurs in each series.

However, 116 is not a multiple of 3. If you keep going, the next number is 270 which is the number you want.

Algebra

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