Hcf of 184230 and 276 by …

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Sia ? 6 years, 6 months ago
Given numbers are 184, 230, and 276.
Applying Euclid's division lemma to 184 and 230, we get
230 = 184 {tex}\times{/tex} 1 + 46
184 = 46 {tex}\times{/tex} 4 + 0
The remainder at this stage is zero.
So, the divisor at this or the remainder at the previous stage i.e., 46 is the HCF of 184 and 230.
Also,
276 = 46 {tex}\times{/tex} 6 + 0
∴ HCF of 276 and 46 is 46
HCF (184, 230, 276) = 46
Hence, the required HCF of 184, 230 and 276 is 46.
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