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The LCM of165, 176, 385 and 495 is k. When is divided by the HCF of the numbers, the quotientis p. What is the value of p?

A 2520

B 5040

C 6720

D 3360

i have familiar with basic concepts of lcm and hcf but these numbers seem terrifying to me.

user64742
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Manish Kumar Balayan
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  • Hint; work with the prime factorings of each number. – lulu Jan 28 '20 at 15:49
  • Do you know how to find lcm of those numbers? – Praphulla Koushik Jan 28 '20 at 15:49
  • i could break them into factors and then find their lcm but i would take time and in exam time and pressure it isnt easy. there is a formula used in my exam syllabus that product of two numbers=product of their lcm and hcf. i think we have to use this somehow . – Manish Kumar Balayan Jan 28 '20 at 15:53
  • There are similar related formulas for more than two terms... see here. I don't see it immediately being useful however for your problem. You would have $165\cdot 176\cdot 385\cdot 495 = \gcd(176\cdot 385\cdot 495, 165\cdot 385\cdot 495, 165\cdot 176\cdot 495, 165\cdot 176\cdot 385)\cdot \text{lcm}(165, 176, 385, 495)$ or similar. – JMoravitz Jan 28 '20 at 16:08
  • Personally, I find it much easier to recognize these as all being multiples of $11$, explicitly with the second as being $2^4\times 11$ and the others being odd multiples of $11$, so it is quick to see that the $\gcd$ is $11$ itself. So, dividing each number by $11$ the question is effectively asking for $\text{lcm}(15,16,35,45)$ which should be relatively painless. – JMoravitz Jan 28 '20 at 16:11
  • Hint: the (obvious) power of $2$ in the LCM/GCD excludes all but one choice - see my answer. – Bill Dubuque Jan 28 '20 at 17:26

2 Answers2

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Easy mental way: clearly the power of $\,2\,$ in ${\small \rm\color{#0a0}{LCM}/\color{#c00}{GCD}} = \color{#0a0}4\!-\!\color{#c00}0,\,$ but $\,2^{\large 4}\!\nmid 2520,\,$ $2^{\large 5}\!\mid 3360\mid 6720 $

Bill Dubuque
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  • That's a good standard test taking strategy, to eliminate possibilities, but if there was a "none of the above" option, it doesn't help you decide between them. Depending on the available options, finishing the problem may be a better approach than eliminating the bad answers. – JMoravitz Jan 28 '20 at 17:29
  • @JMoravitz But there is no "none of the above" here (and we could continue this way if we needed to). It is highly likely that the problem was designed to be quickly solvable as above. – Bill Dubuque Jan 28 '20 at 17:30
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Spot by inspection or divisibility tests that each number is a multiple of $11$ and the second number is $16\times 11 = 2^4\times 11$ while the other numbers are all odd multiples of $11$.

It follows that the $\gcd$ of the numbers is then $11$.

Recognize that $\text{lcm}(a,b,c,d)/\gcd(a,b,c,d) = \text{lcm}(\frac{a}{\gcd(a,b,c,d)},\frac{b}{\gcd(a,b,c,d)},\dots,\frac{d}{\gcd(a,b,c,d)})$ so the problem simplifies as $\text{lcm}(15,16,35,45) = \text{lcm}(3\cdot 5, 2^4,5\cdot 7, 3^2\cdot 5)$

We can then see that the result will be $2^4\times 3^2\times 5\times 7 = 5040$

JMoravitz
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