Question

A trader has three different types of oils of volume 870 l, 812 l and 638 l. Find the least number of containers of equal size required to store all the oil without getting mixed.

Hint:

An easy way to find the quotients after dividing by the HCF is to look at the leftover prime factors in your factorization:
- For 870: \(3 \times 5 = 15\)
- For 812: \(2 \times 7 = 14\)
- For 638: \(11\)
This saves you from doing long division calculations!

Answer 1

Step 1: Find the HCF using the Euclidean algorithm instead of prime factorisation.
\[ 870 = 812 \times 1 + 58 \]
So $\text{HCF}(870, 812) = \text{HCF}(812, 58)$. Since $812 = 58 \times 14 + 0$, this HCF is $58$.
Step 2: Bring in the third number.
\[ 638 = 58 \times 11 + 0 \]
Since this divides exactly, the overall HCF of all three volumes is $58$ litres.
Step 3: Find the number of containers for each oil and add them up.
\[ \frac{870}{58} = 15, \quad \frac{812}{58} = 14, \quad \frac{638}{58} = 11 \]
\[ \text{Total containers} = 15 + 14 + 11 = 40 \]
\[ \boxed{40\ \text{containers}} \]