The HCF of two numbers is 12 and their LCM is 672. If one of the numbers is 96. Find the other number.
Answers
Answer:
HCF(A,B) = 12 = 2^2 * 3
LCM(A,B) = 288 = 2^5 * 3^2
Step-by-step explanation:
For each prime factor (in this case 2 and 3) we can only choose the corresponding power of that factor in the HCF or the one in the LCM. One of the numbers must contain exactly 2^5, because else the LCM would contain a lower power of 2.
And the other number must have exactly 2^2, otherwise the HCF would not contain 2^2. So, we have only 2 choices for each prime factor. One of the numbers has the same power for that prime factor as the HCF, and the other number has the same power for that prime factor as the LCM. (So powers 3 or 4 are not possible!)
For 3, one of the numbers has the power 1, and the other has power 2.
There are always 2 choices for each prime factor (if the powers are different for HCF and LCM). If there is only one prime factor with a different power in HCF and LCM, then we have only one pair.
If there are n prime factors with different powers for HCF and LCM, then there are 2^(n-1) pairs.
In this case 2 and 3 have both different powers in HCF and LCM, so the number of different pairs is : 2^(2–1) = 2.
Those pairs are: (2^2 * 3, 2^5 * 3^2) and (2^5 * 3, 2^2 * 3^2). Or (12, 288) and (96, 36).
N.B. The other pairs that Jacob mentions, (24, 144) and (48, 72) are not correct because they have HCF 24.
Other example : how many pairs of numbers are there for which HCF = 320 and LCM = 63 million ?
Answer: HCF = 320 = 2^6 * 5.
LCM = 63 million = 2^6 * 3^2 * 5^6 * 7
There are 3 prime factors for which the power differs in HCF and LCM: 3, 5 and 7.
So the number of different pairs with HCF = 320 and LCM = 63 million equals 2^(3–1) = 4. Those pairs are: (Both numbers contain 2^6).
(2^6 * 3^2 * 5^6 * 7, 2^6 * 5)
(2^6 * 3^2 * 5^6, 2^6 * 5 * 7)
(2^6 * 5^6 * 7, 2^6 * 3^2 * 5)
(2^6 * 5^6, 2^6 * 3^2 * 5 * 7).
In numbers: (63 million, 320), (9 million, 2240), (7 million, 2880), (1 million, 20160)
HOPE THIS WILL HELP YOU ☺️
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