What Is the Other Number If HCF Is 33 and LCM Is 264?

[chwala]

Homework Statement

see attached.

Relevant Equations

lcm and hcf

This is the problem,

ok, one of the numbers is $66= 2×3×11$ we are told that the hcf = $33=3×11$ therefore, $11 and 3$ would constitute part of the other unknown number, also lcm {264] = {2×2×2×3×11}
the possible value for the other term would be ${3×11}$ times factors of $2^n$ where, 1≤n≤3
we do not have a solution in the given options.

 

[PeroK]

Homework Statement:: see attached.
Relevant Equations:: lcm and hcf

This is the problem,

View attachment 289435

ok, one of the numbers is $66= 2×3×11$ we are told that the hcf = $33=3×11$ therefore, $11 and 3$ would constitute part of the other unknown number, also lcm {264] = {2×2×2×3×11}
the possible value for the other term would be ${3×11}$ times factors of $2^n$ where, 1≤n≤3
we do not have a solution in the given options.

Where are you getting these crazy problems?

 

[chwala]

these are crazy guys...got this worksheet from Google...lol

 

[mathwonk]

isnt theee product of two numbers equal to the product of their LCM and HCF?

 

[chwala]

isnt theee product of two numbers equal to the product of their LCM and HCF?

This may be true but how does it work for this case?
the product of the lcm and hcf is equal to $8712$, given that the other number is $66$ then it follows that the other unknown number is $132$ but the lcm $(66,132)=132$ and not $264$ as required...and further the hcf of these two numbers is $66$ ...that's why we concluded that it was a crazy question...

 

[mathwonk]

hmmmmm. i think i see your question is not what the answer is, i.e. apparently 33 and 264, but why this answer 264 does not appear in the choices.?

 

[chwala]

hmmmmm. i think i see your question is not what the answer is, i.e. apparently 33 and 264, but why this answer 264 does not appear in the choices.?

Not only that...we are informed that ...first number divided by $2$ quotient is $33$... the first number can only be $66$. The question is/was wrongly framed.

 

[mathwonk]

ok; i assumed " completely divided by 2" could just mean after dividing out as many powers of 2 as possible, which might be none. but errors are not so hard to find in math. my own works are littered with them. in fact i once refereed a booklet issued by the local school district, or even the state, intended to help high school math teachers prepare for a certification exam, and it had some howlers. good work finding these.