Group Challenge: Q16 - Proposed Solution

16(a)
16(b) Method 1, using Algebraic Method


Find the smallest possible value of the whole number w if 432w is a multiple of 324.


Note that if 432w is a multiple of 324, we can express it as
432w = 324 x A where A is the factor, when multiply by 324 will give a number that is 432xw

In part (a), we know that 324 = 2 x 2 x 3 x 3 x 3 x 3
and 432 = 2 x 2 x 2 x 2 x 3 x 3 x 3

Then, we can rewrite
2 x 2 x 2 x 2 x 3 x 3 x 3 x w = 2 x 2 x 3 x 3 x 3 x 3 x A
Simplifying, we get
2 x 2 x w = 3 x A
Hence, 4 x w = 3 x A
the smallest number that we could have for w and A such that both sides are the same would be:
when w = 3 and A = 4.

Hence, w = 3

16(b) By listing and comparison

2 comments:

  1. so basically this i something like algebra right?so different from mac spicy's explanation...

    lovylim

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  2. Hi Lovy

    Yes, slightly different.
    I've introduced simple algebra in this.
    However, at the beginning, I made use of the product of factors to simplify the equation.

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