 Question

Hi Shifus, kindly enlighten me for the following question please. Thank you 🙏

The HCF & LCM of 264 and P are 12 and 5544 respectively. Find the smallest possible value of P.

Source: Assessment

264 = 2^3 x 3 x 11
12 = 2^2 x 3
5544 = 2^3 x 3^2 x 7 x 11
P = 2^2 x 3^2 x 7 = 252

Ans : 252.

Hi alfretztay, thank u for replying.

I understand how to derive the index notations for the first 3 rows. What I can’t understand is how did u deduce that

P= 2^2 x 3^2 x 7

Sorry to trouble u to further explain. Thank u so much for your help and time. 🙏

From “HCF & LCM of 264 and P are 12 and 5544 respectively”,

264 = 2^3 x 3 x 11
12 = 2^2 x 3 (P includes 2^2 x 3)
5544 = 2^3 x 3^2 x 7 x 11 (P includes 3^2 x 7 after considering factors of 264)
P = 2^2 x 3^2 x 7 = 252

Ans : 252.

Hi Alfretztay,

ok thank u. I think I got it from your explaination.

Thank u very much.

Hi altp10,

you are most welcome.

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Factorise the numbers:

2   |   264  ,  P

2   |   132   ,  P/2

3    |   66    ,   P/4

____________

22    ,   P/ 12

Note: the left side needed to be 2, 2, 3 because the HCF is 12 = 2x2x3. It cannot be 2, 2, 2,… even though 132 can be divided by 2.

2 x 2 x 3 x 22 x (P/12)  = 5544 (LCM)

22P = 5544

P = 252