Math Problem Statement

A milkman has milk of three varieties. He has 403L, 465L & 651L of the three varieties of milk with him.

  1. What is the largest size of bottle in which he can bottle each of the three types of milk completely without mixing the milk?
  2. What is the minimum number of bottles required?
  3. How many different sizes of bottles (with the integral number of litres) can be used in order to bottle all three varieties of milk? (write comma-separated integers with spaces, eg: x, y, z) ( Sapient hiring drive 2018)

Solution

  1. Largest size of bottle:
    To find the largest size of bottle that can bottle each variety completely, we need the greatest common divisor (GCD) of the three amounts (403, 465, and 651 liters).

    The GCD of 403, 465, and 651 is 31 liters. Therefore, the largest size of a bottle that can hold each variety without leaving any remainder is 31 liters.

  2. Minimum number of bottles required:
    Each type of milk needs to be bottled in 31-liter bottles. To find the number of bottles required for each type:

    • For 403 liters: 40331=13\frac{403}{31} = 13 bottles
    • For 465 liters: 46531=15\frac{465}{31} = 15 bottles
    • For 651 liters: 65131=21\frac{651}{31} = 21 bottles

    Total bottles required = 13+15+21=4913 + 15 + 21 = 49 bottles.

  3. Different sizes of bottles:
    We are looking for all the divisors of the GCD (31 liters) since they would allow complete bottling of all three quantities.

    Divisors of 31 are: 1, 31

    So, the answer is: 1, 31

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Math Problem Analysis

Mathematical Concepts

Number Theory
Greatest Common Divisor (GCD)
Divisors

Formulas

GCD formula for three numbers
Division to calculate minimum number of bottles

Theorems

Euclidean algorithm for GCD

Suitable Grade Level

Grades 8-10