Math, asked by akshitsharma9873, 11 months ago

How many pairs of number exist such that hcf is 4 and lcm is 64

Answers

Answered by VedaantArya
0

Answer:

Two ordered pairs, or a single unordered pair, of the elements: 4, 64.

Step-by-step explanation:

LCM = 64 = 2 * 2 * 2 * 2 * 2 * 2

HCF = 4 = 2 * 2

Let a, b be the required numbers.

a = 2 * 2 * f * g * ...

b = 2 * 2 * h * i * ...

a * b = LCM * HCF = 2 * 2 * 2 * 2 * f * g * h * i * ... = 64 * 4

Cancelling, we get:

f * g * h * i * ... = 2 * 2 * 2 * 2.

Since f, g, h, i are prime factors, the equation implies any other factors of a and b should be 2, and there must be a total of 4 more such factors.

But, if a and b both have another 2 (that is, f = 2 and h = 2), then another 2 shall be included in the HCF, and the HCF would change to 8, which isn't acceptable.

So, all the 4 extra 2's must belong to a single number out of a and b.

So, a = 2 * 2 * 2 * 2 * 2 * 2 = 64, and b = 2 * 2 = 4.

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