a number leaves a remainder of 1 in each case when divided by 2, 3, 4, 5, 6. Find the number if it is divisible by 7.

1 thought on “a number leaves a remainder of 1 in each case when divided by 2, 3, 4, 5, 6. Find the number if it is divisible by 7.”

1. Answer:

     301

   Step-by-step explanation:

   The desired number will be a multiple of 7 that is 1 more than some multiple of the least common multiple of 2, 3, 4, 5, and 6. The value can be found by trial and error, or by using inverses modulo 7.

   LCM

   The least common multiple of 2, 3, 4, 5, 6 will be the product of their prime factors, each to the highest power:

     2 = 2; 3 = 3; 4 = 2²; 5 = 5; 6 = 2·3

   Then the LCM is 2²·3·5 = 60.

   Divisible by 7

   For some multiple n of 60, we want …

     (60n +1) mod 7 = 0

   This is equivalent to …

     4n mod 7 = 6

   The solution to this can be found by trial and error, or by making use of the mod-7 inverse of 4 (which is also found by trial and error).

   Here are a few trials:

     4·1 mod 7 = 4

     4·2 mod 7 = 1 . . . . . . this tells us the mod 7 inverse of 4 is 2

     4·3 mod 7 = 5

     4·4 mod 7 = 2

     4·5 mod 7 = 6 . . . . . . the value of n we’re looking for is 5

   That is …

      (60·5 +1) mod 7 = 301 mod 7 = 0

   The (smallest positive) number with the division properties we’re looking for is 301.

   __

   Additional comment

   Using 2 as the inverse of 4 mod 7, we can solve for n:

     (2)(4n) mod 7 = (2)(6) mod 7 . . . . multiply the equation by the inverse of 4

     n = 12 mod 7 = 5 . . . same as the value we found from “trial and error”

   Note that 301 is the smallest positive number with the required division properties. Any multiple of 60·7 = 420 added to 301 will also have the required division properties. For example, 721, 1141, 1561, ….

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