Smallest number for given remainders

Let x = the number
 

When divided by 2 the remainder is 1
$\dfrac{x}{2} = A + \dfrac{1}{2}$   →   equation (1)
 

When divided by 3 the remainder is 2
$\dfrac{x}{3} = B + \dfrac{2}{3}$   →   equation (2)
 

When divided by 4 the remainder is 3
$\dfrac{x}{4} = C + \dfrac{3}{4}$   →   equation (3)
 

When divided by 5 the remainder is 4
$\dfrac{x}{5} = D + \dfrac{4}{5}$   →   equation (4)
 

When divided by 6 the remainder is 5
$\dfrac{x}{6} = E + \dfrac{5}{6}$   →   equation (5)
 

• From the above equations, x, A, B, C, D, and E must be whole numbers.
• From equation (1), x must be odd.
• From equation (4), x must be divisible by 5 + the remainder 4.
  • If it ends with 0: 0 + 4 = 4 (even).
  • If it ends with 5: 5 + 4 = 9 (odd)

Thus, x must end with 9.
 

Try x = 9

$\dfrac{9}{3} = B + \dfrac{2}{3}$

$B = \dfrac{7}{3}$   →   (not a whole number - not okay)

 

Try x = 19

$\dfrac{19}{3} = B + \dfrac{2}{3}$

$B = \dfrac{17}{3}$   →   (not a whole number - not okay)

 

Try x = 29

$\dfrac{29}{3} = B + \dfrac{2}{3}$

$B = 9$   →   (whole number - okay)
 

$\dfrac{29}{4} = C + \dfrac{3}{4}$

$C = \dfrac{13}{2}$   →   (not a whole number - not okay)

 

Try x = 39

$\dfrac{39}{3} = B + \dfrac{2}{3}$

$B = \dfrac{37}{3}$   →   (not a whole number - not okay)

 

Try x = 49

$\dfrac{49}{3} = B + \dfrac{2}{3}$

$B = \dfrac{47}{3}$   →   (not a whole number - not okay)

 

Try x = 59

$\dfrac{59}{3} = B + \dfrac{2}{3}$

$B = 19$   →   (whole number - okay)
 

$\dfrac{59}{4} = C + \dfrac{3}{4}$

$C = 14$   →   (whole number - okay)
 

$\dfrac{59}{6} = E + \dfrac{5}{6}$

$C = 9$   →   (whole number - okay)

 

Thus, x = 59           answer