LCM implies multiplication?

Number Theory Level 3

$\LARGE \color{#3D99F6}{0.\bar{3}}, \color{#D61F06}{0.\bar{5}}$

What is the lowest common multiple of the two numbers given above?

Note that $0. \overline{ABC} = 0.ABCABCABC\ldots$ .

0.\bar{15}

1.\bar{6}

0.1\bar{5}

It does not exist

1.\bar{5}

4 solutions

Abhishek Sharma
May 5, 2015

Let $x=0.\bar{3}$ . $x=0.333\dots$ $10x=3.333\dots$ Subtracting $1$ from $2$ we get, $9x=3$ $x=\frac{1}{3}$ $0.\bar{3}=\frac{1}{3}$

Similarly, we get $0.\bar{5}=\frac{5}{9}$ .

Multiples of $0.\bar{3}$ are $\large \frac{1}{3}$ , $\large \frac{2}{3}$ , $\large1$ , $\large\frac{4}{3}$ , $\large\boxed{\frac{5}{3}}$ , $\dots$

Multiples of $0.\bar{5}$ are $\large\frac{5}{9}$ , $\large\frac{10}{9}$ , $\large\boxed{\frac{5}{3}}$ , $\dots$

Instead of doing this we could have simply used the formula - LCM OF $\frac{a}{b}$ & $\frac{c}{d}$ = $\frac{lcm of (a,c)}{hcf of (d,b)}$ . But by writing multiples we get an idea how the formula has been derived

Least common multiple ( LCM ) of $0.\bar{3}$ and $0.\bar{5}$ is $\large\frac{5}{3}$ or $\large1.\bar{6}$ .

Instead of writing multiples the other way to do it fast is directly calculating its lcm.

LCM OF $\frac{a}{b}$ & $\frac{c}{d}$ = $\frac{lcm of (a,c)}{hcf of (d,b)}$

A Former Brilliant Member - 6 years, 1 month ago

Yes Kalash , that really helps. :)

Nihar Mahajan - 6 years, 1 month ago

Yeah!! Much faster,And reliable.

A Former Brilliant Member - 6 years, 1 month ago

Much Faster :3 XD

Mehul Arora - 6 years, 1 month ago

Cheers! xD

Nihar Mahajan - 6 years, 1 month ago

@Nihar Mahajan – Cheers, Mate!

Mehul Arora - 6 years, 1 month ago

HCF, now that's a notation I've never seen before.

Vilim Lendvaj - 3 years, 2 months ago

Highest Common Factor... You're welcome.

Siddharth Rai - 3 years, 2 months ago

Great solution!!!

Souvik Aoc - 6 years, 1 month ago

Good solution.

Rama Devi - 6 years ago

I also did the second one. Is there any existence of other solution other than these

Vijayant Patel - 4 years, 3 months ago

Seriously wondering here . . .since when was the definition of LCM extended to include those of decimal numbers &/or fractions?

Peter Chan - 3 years, 9 months ago

Did the same way using formula .... The formula is very helpful

Ashutosh Kapre - 3 years, 7 months ago

“LCM is smallest positive integer” according to the best answer to the previous question in this thread (LCM of 2 & -4). What changed?

Phil Baines - 3 years, 3 months ago

good question, the people as well as the definition seems to change every single time.

For my understanding the final number should be divisble by both numbers. and to achieve that, it must have all the factors in it that both of the numbers have in it.

in the case here, it's the factors (1/3) and 5.

from my point there's no restrictions to wether it's integer, if it's positive or negative.

the 2 start numbers could even be e and i, for my taste.

but it seems some people here, and the brilliant team is included here, sometimes limit it to positive integers when they feel like it.' and then they ignore that , using fractions again.

so I have no idea whats going on in their minds!

Denis Schüle - 3 years ago

The LCM of two rationals makes absolutely no sense at all. I guess the next step is to find the LCM of sqrt(2) and π ?

JR Dierickx - 3 years, 2 months ago

I overlooked the bar above the 5 in 0.5. So I read it as finding the least common multiple of 1/3 and 1/2, which is 1. It puzzled me that 1 was not one of the answer choices.

Kermit Rose - 2 years, 5 months ago

Good Solution

Sundaranand Mahadevan - 5 months, 3 weeks ago

Gia Hoàng Phạm
Oct 6, 2018

$a=0.\underline{3}=0.\underline{1} \times 3=\frac{1}{9} \times 3=\frac{3}{9}$

$b=0.\underline{5}=0.\underline{1} \times 5=\frac{1}{9} \times 5=\frac{5}{9}$

$LCM(\frac{3}{9},\frac{5}{9})=\frac{LCM(3,5)}{9}=\frac{15}{9}=1+\frac{6}{9}=\boxed{\large{1.\underline{6}}}$

What does that 0.3_ denotes

Revanth Kanamarlapudi - 2 years, 4 months ago

0.3_ denotes $0.333333333\dots$

Gia Hoàng Phạm - 2 years, 4 months ago

Hossain Nahdi
Mar 31, 2021

First we observe that, $0.\overline{3} = 0.\overline{1} \times 3$ , and $0.\overline{5} = 0.\overline{1} \times 5$ . Thus, $gcd(0.\overline{3}, 0.\overline{5}) = 0.\overline{1}. \newline\newline$ We know that, $ab = gcd(a,b)\times lcm(a,b)$ . Thus, $\newline lcm(a,b) = \frac {ab} {gcd(a,b)} \newline or, lcm(a,b) =$ $\frac {3 \times 0.\overline{1} \times 5 \times 0.\overline{1}}{0.\overline{1}} \newline or,$ $lcm(a,b) = 15 \times 0.\overline{1} = 1.\overline{6} \newline \newline$ Thus, our answer is: $1.\overline{6}$ .

Michael Fitzgerald
Feb 20, 2018

# https://brilliant.org/practice/greatest-common-divisor-lowest-common-multiple-3/?p=4
from fractions import Fraction
irrational_1 = 1./3
irrational_2 = 5./9

max_denominator = 100
def gcd(*numbers):
    """Return the greatest common divisor of the given integers"""
    from fractions import gcd
    return reduce(gcd, numbers)
def lcm(*numbers):
    """Return lowest common multiple."""    
    def lcm(a, b):
        return (a * b) // gcd(a, b)
    return reduce(lcm, numbers, 1)

fract = lambda x: Fraction(x).limit_denominator(max_denominator)
x1 = int(fract(irrational_1).numerator)
x2 = int(fract(irrational_2).numerator)
y1 = int(fract(irrational_1).denominator)
y2 = int(fract(irrational_2).denominator)

lcm_denom = lcm(y1,y2)
lcm_num = lcm(lcm_denom*x1/y1,lcm_denom*x2/y2)
a = float(lcm_num)/lcm_denom

if a - lcm_num//lcm_denom > 0:          
    x1 = int(fract(a - lcm_num//lcm_denom).numerator)
    y1 = int(fract(a - lcm_num//lcm_denom).denominator)
    print '%0.12f or %s %d/%d' %(a, lcm_num//lcm_denom, x1, y1)
else:
    print int(a)

1	`1.666666666667 or 1 2/3`

LCM implies multiplication?

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