Find a natural number by the Euclid's division

Find the natural number that simultaneously satisfy the conditions:

a) divided by 4 the remainig is 3

b) divided by 10, the remaning is 1;

c) divided by 12, the remaining is 3;

d) the sum of quotients of the three divisions from points a), b) and c) is higher by 16 than one third of the number.

 Solution:

In this issue we use the Euclidean division.

This theorem says the dividend (D) is equal to the product of the divisor (Q) and the quotient (C) added the remaining (R)

D = Q  ∙ C + R

If we denote the number  with the letter D and apply the above theorem we can write conditions:

a) D = 4 ∙ C1 + 3

b) D = 10 ∙ C2 + 1 ∙

c) D = 12 ∙ C3 + 3

For the last condition we write: C1 + C2 + C3 = D / 3 + 16

      a)       D = 4 ∙ C1 + 3

               D - 3 = 4 ∙ C1

               C1 = (D-3) / 4

       b)     D = 10 ∙ C2 + 1 ∙

                D - 1 = 10 ∙ C2

                C2 = (D-1) / 10

        c)     D = 12 ∙ C3 + 3

                D - 3 = 12 ∙ C3

                C3 = (D-3) / 12

Replacing expressions obtained for C1, C2 and C3 to the condition d) we obtain:

                    (D-3) / 4 + (D - 1) / 10 + (D - 3) / 12= D / 3 + 16

The common denominator of these fractions:

4 = 2∙2

10 = 2 ∙ 5

12 = 2∙2 ∙ 3

 3 = 3∙1

The common denominator = 2∙2 ∙ 3 ​​∙ 5 = 60

Multiply each fraction reach 60: first by 15, the second by 6, third by 5, fourth by 20 and the free term (whole number) by 60:

15 (D-3) + 6 (D-1) + 5 (D-3) =20 D  + 16 ∙ 60

15 D - 45 +6 D -6 +5 D – 15 = 20 D + 960

26 D - 20 D = 960 + 45 + 6 + 15

  6 D = 1026

      D = 171

Solution: The number required is 171