If the HCF of 408 and 1032 is expressible in the form 1032 m – 408 × 5,

Solution:

We need to find $m$ if the H.C.F of 408 and 1032 is expressible in the form $1032 m-408 \times 5$.

Given integers are 408 and 1032 where $408<1032$.

By applying Euclid's division lemma, we get $1032=408 \times 2+216$.

Since the remainder $\neq 0$, so apply division lemma on divisor 408 and remainder 216

$408=216 \times 1+192$

Since the remainder $\neq 0$, so apply division lemma on divisor 216 and remainder 192

$216=192 \times 1+24$

Since the remainder $\neq 0$, so apply division lemma on divisor 192 and remainder 24

$192=24 \times 8+0$

We observe that remainder is 0. So the last divisor is the H.C.F of 408 and 1032.

Therefore,

$24=1032 m-408 \times 5$

$\Rightarrow \quad 1032 m=24+408 \times 5$

$\Rightarrow \quad 1032 m=24+2040$

$\Rightarrow \quad 1032 m=2064$

$\Rightarrow \quad m=\frac{2064}{1032}$

$\Rightarrow \quad m=2$.