If the HCF of 408 and 1032 is expressible in the form 1032 m−408×5, find m.
If the HCF of 408 and 1032 is expressible in the form 1032 m−408×5, find m.
We need to find m if the H.C.F of 408 and 1032 is expressible in the form 1032 m — 408 x 5
Given integers are 408 and 1032 where 408 < 1032
By applying Euclid’s division lemma, we get 1032 = 408x 2 + 216.
Since the remainder ≠ 0, so apply division lemma on divisor 408 and remainder 216
408 = 216 x 1 + 192.
Since the remainder ≠ 0, so apply division lemma on divisor 216 and remainder 192
216 = 192 x 1 + 24.
Since the remainder ≠ 0, so apply division lemma on divisor 192 and remainder 24
192 = 24 x 8 + 0.
We observe that remainder is 0. So the last divisor is the H.C.F of 408 and 1032.
Therefore,
24 = 1032m — 408 x 5
1032m = 24 + 408 x 5
1032m = 24 + 2040
1032m = 2064
m= 20641032
m = 2
Therefore, m = 2.
We need to find m if the H.C.F of 408 and 1032 is expressible in the form 1032 m — 408 x 5
Given integers are 408 and 1032 where 408 < 1032
By applying Euclid’s division lemma, we get 1032 = 408x 2 + 216.
Since the remainder ≠ 0, so apply division lemma on divisor 408 and remainder 216
408 = 216 x 1 + 192.
Since the remainder ≠ 0, so apply division lemma on divisor 216 and remainder 192
216 = 192 x 1 + 24.
Since the remainder ≠ 0, so apply division lemma on divisor 192 and remainder 24
192 = 24 x 8 + 0.
We observe that remainder is 0. So the last divisor is the H.C.F of 408 and 1032.
Therefore,
24 = 1032m — 408 x 5
1032m = 24 + 408 x 5
1032m = 24 + 2040
1032m = 2064
m= 20641032
m = 2
Therefore, m = 2.
