The lowest number that is a multiple of both 60 and the integer n is 180. Find the smallest possible value of n.

Question

I have one solution but I think it's just a wild guessed one. Tell me if I am correct and also if not, then how should it be done?

What I have done is divided 180 by 60 to get 3. Then take lcm of 60 and multiples of 3 one by one until I get 180 as the lcm.

So I would first take the lcm of 60 and 3, then of 60 and 6, then of 60 and 9 after which I would get 180.

Is this the correct way to solve it?

P.S This is just a 1 mark question of one of the paper of O Levels in which calculators are not allowed and not more than 1 or 2 blank lines are given for working :/. Keep that in mind. :P

Since lcm$(60,n) = 180$, $n$ must be a divisor of $180$. But if $n$ is a divisor of $60$, then lcm$(60,n) = 60$ The smallest divisor of $180$ that is not a divisor of $60$ is $9$, and lcm$(60,9) = 180$. — Steven Alexis Gregory, Apr 02 '16 at 17:26

score 11 · Answer 1 · 2016-03-28T20:19:58.323

11

$$180=2^2.3^2.5$$ $$60=2^2.3.5$$ so $n$ must have the same primes as $180$ $$n=2^i3^j5^k$$ where $i,j\leq 2$ and $k\leq 1$

$60$ is not a multiple of $n$, otherwise their $lcm$ would have been $60$. Therefore, we must have $$n\equiv 0 \pmod {3^2}$$ It follows the smallest $n$: $$n=9$$

edited Mar 28 '16 at 20:19

answered Mar 28 '16 at 20:10

This is an O levels question :P We do not know about mod and identities. So.... – Mohammad Areeb Siddiqui Mar 28 '16 at 20:25
3

The mod statement is equivalent to "Therefore, $n$ must be a multiple of 9." – Teepeemm Mar 28 '16 at 22:03
1

You may want to expand the "Therefore". The leap is not immediately clear. – Thorbjørn Ravn Andersen Mar 29 '16 at 07:03
@ThorbjørnRavnAndersen can you please explain? :) – Mohammad Areeb Siddiqui Mar 29 '16 at 16:14
@MohammadAreebSiddiqui The "would have been 60. Therefore, we must have n==0 mod 3^2" snippet is not immediately self evident. – Thorbjørn Ravn Andersen Mar 29 '16 at 20:16

Steven Alexis Gregory · Answer 2 · 2016-03-28T20:48:59.170

8

$$\operatorname{lcm}(60,n) = 180$$

$$\operatorname{lcm}(2^2 \cdot 3^1 \cdot 5^1, 2^{n_2} \cdot 3^{n_3} \cdot 5^{n_5}) = 2^2 \cdot 3^2 \cdot 5^1$$

$$ 2^{\max(2,n_2)} \cdot 3^{\max(1,n_3)} \cdot 5^{\max(1,n_5)} = 2^2 \cdot 3^2 \cdot 5^1$$

\begin{align} \max(2,n_2) &= 2\\ \max(1,n_3) &= 2 \\ \max(1,n_5) &= 1 \end{align}

The smallest possible exponents are thus

\begin{align} n_2 &= 0\\ n_3 &= 2 \\ n_5 &= 0 \end{align}

So $ n = 2^0 \cdot 3^2 \cdot 5^0 = 9$.

edited Mar 28 '16 at 20:48

answered Mar 28 '16 at 20:43

Steven Alexis Gregory

27,619

i did not get it. how did the min(2, n2) became 2? – Mohammad Areeb Siddiqui Mar 28 '16 at 20:45
Ohhh. Got it now. One more thing. If max(2,n2) = 2 then why n2=0 why not less than 0? – Mohammad Areeb Siddiqui Mar 28 '16 at 20:50
@MohammadAreebSiddiqui - Every positive integer can be expresses as a product $\prod_{p \in P} p^{\pi_p}$ where $P$ is the set of prime numbers, $\pi_p \ge 0$, and all but finitely many of the $\pi_p$ are equal to $0$. – Steven Alexis Gregory Mar 28 '16 at 20:57
2

@MohammadAreebSiddiqui - Because $2^{-1}$ is not an integer. – Steven Alexis Gregory Mar 28 '16 at 23:20
Yeah I figured that out while sitting in the examination hall today. :P – Mohammad Areeb Siddiqui Mar 29 '16 at 16:15
Very clear answer by the way. Thanks alot. :) – Mohammad Areeb Siddiqui Mar 29 '16 at 16:16
Also, see Bitfool's answer. He suggests that the number could be negative does that mean your assumptions are not right? – Mohammad Areeb Siddiqui Mar 29 '16 at 16:19
@MohammadAreebSiddiqui - No, for two reasons. (1.) The default domain for problems like this is the non negative integers, and (2.) "smallest" in this case refers to the smallest divisor of 180. See THIS – Steven Alexis Gregory Mar 29 '16 at 16:59

score 5 · Answer 3 · answered Mar 29 '16 at 01:51

$n$ is the smallest number that divides 180 but not 60. (If it divided both, 180 couldn't be the least common multiple because $60 < 180$.)

The primes in $60$ are $(2, 2, 3, 5)$ and the ones in $180$ are $(2, 2, 3, 3, 5)$.

If $n$ divides $180$, its primes are contained in $(2, 2, 3, 3, 5)$.

The smallest list contained in $(2, 2, 3, 3, 5)$ but not in $(2, 2, 3, 5)$ is $(3, 3)$, and $3 \cdot 3 = 9$.

score 3 · Answer 4 · answered Mar 28 '16 at 19:55

Given two positive integers $A,B$, you have $AB=gcd(A,B) lcm(A,B)$.

Here, $A=n, B=60$ and $lcm(A,B)=180$. Hence, calling $g$ the gcd of $n$ and $60$: $60n=180g$, or $n=3g$.

Thus $n$ is a multiple of $3$. But then, the gcd is also a multiple of $3$, since $3$ divides $60$. Hence $n$ is a multiple of $9$. The smallest multiple of $9$ is $9$ itself, and you can check it works.

score 3 · Answer 5 · answered Mar 28 '16 at 23:12

Totally without the use of any mathematics:

x = 1, 2, 3, 4, 5, 6 are already divisors of 60, so the lowest common multiple of 60 and x is 60.

180 is not a multiple of x = 7 or 8, therefore it is not the lowest common multiple of 60 and x; it is not a multiple of x at all.

Let x = 9: Yes, the lowest common multiple of 60 and 9 is 180. Because the multiples of 60 are 60, 120, 180, 240, ... and the first of those that is also a multiple of 9 is 180.

(In the general case, factoring the numbers that correspond to 60 and 180 into prime factors will give you the solution quickly).

What makes you think that arithmetic is not mathematics? – Steven Alexis Gregory Apr 02 '16 at 17:04 — Steven Alexis Gregory, Apr 02 '16 at 17:04

score 3 · Answer 6 · answered Mar 29 '16 at 14:38

3

Even though there are lots of correct answers (even one accepted) I will add another. The OP asks

Is this the correct way to solve it?

and adds the important information

P.S This is just a 1 mark question of one of the paper of O Levels in which calculators are not allowed and not more than 1 or 2 blank lines are given for working :/. Keep that in mind. :P

I think that in this case his method is just the right one to use. It requires minimum use of fancy stuff and takes minimal time when there's time pressure. It may well be what the examiners had in mind when they asked the question.

answered Mar 29 '16 at 14:38

Ethan Bolker

103,433

So my approach is mathematically right, right? – Mohammad Areeb Siddiqui Mar 29 '16 at 16:20
Yes___________________ (the long dash is there because a comment must have at least 15 (nonblank) characters). – Ethan Bolker Mar 29 '16 at 16:22

score 2 · Answer 7 · answered Mar 28 '16 at 20:07

2

You can do a prime factorization of each number:

$$60 = 2^2 \cdot 3^1 \cdot 5^1$$

$$180 = 2^2 \cdot 3^2 \cdot 5^1$$

The number $180$ is the least common multiple of $60$ and this other number. As the LCM, it takes on the largest exponent for each prime of any of the numbers.

The exponents on the $2$ and $5$ are the same in the LCM as for $60$, so the other number need not have any factors of $2$ or $5$.

But we do need two factors of $3$.

Hence ... ?

answered Mar 28 '16 at 20:07

John

26,582

why as the lcm it takes on the largest exponent for each prime of any of the numbers? – Mohammad Areeb Siddiqui Mar 28 '16 at 20:23
1

@MohammadAreebSiddiqui, if the $lcm (60,n)=180$ then $n$ does not divide $60$ but $n$ divides $180$. The only way that happens is if only $n$ is a multiple of $9$. You can verify that $9$ does not divide $60$ and divides $180$. Furthermore, that $lcm(9,60)=180$. Please see Numthcurious answer. – Mar 28 '16 at 20:29
In order for a number $M$ to be a multiple of another number $N$, the multiple $M$ has to have at least the prime factors of $N$. It can also have other factors. The smallest positive, nonzero multiple of a number $N$ is itself. It contains exactly the same prime factors as $N$. Now add another number into the mix. In order for the multiple to be the least common multiple, we need to add just enough of the missing prime factors so that they're both covered. In this case, the other number has to be $9$ because it contributes the two factors of $3$. – John Mar 28 '16 at 20:34
@John by two factors of 3 you mean the "two" threes, right? :P – Mohammad Areeb Siddiqui Mar 28 '16 at 20:46
Basically. I phrased it as "two factors of $3$" to mean $3 \cdot 3$. Factor indicates multiplication. "Two $3$'s" normally means multiplying $3$ by $2$, or adding $3$ to itself twice, which is something totally different than what I meant. – John Mar 28 '16 at 21:40

score 1 · Answer 8 · answered Mar 28 '16 at 20:37

1

$$180=2^2.3^2.5$$ $$60=2^2.3.5$$ so $n$ must have a similar prime factorization as $180$ $$n=2^i3^j5^k$$ where $i,j\leq 2$ and $k\leq 1$

$60$ is not a multiple of $n$, otherwise their $lcm$ would have been $60$. Therefore, $n$ must be a multiple of $9$. That is only to ensure that $n$ does not divide $60$. It follows the smallest $n$: $$n=9$$

answered Mar 28 '16 at 20:37

1

Credit to Numthcurious. – Mar 28 '16 at 20:38
how did u came to the conclusion that 9 is the multiple that should be the smallest if 60 should not be the lcm? – Mohammad Areeb Siddiqui Mar 28 '16 at 20:46
Remember you are looking for the smallest number $n$ that divides $180$ but that does not divide $60$. 9 is that number. – Mar 28 '16 at 21:12
So did u guessed it? – Mohammad Areeb Siddiqui Mar 28 '16 at 21:31
1

Of course not. You can check that 1,2,3,4,5,6 are common divisors of 60 and 180 but not 9. – Mar 29 '16 at 01:04

score 1 · Answer 9 · edited Mar 29 '16 at 03:23

It seems to me (that is, I offer no mathematical arguments to support my proposal) that $-180$ is the smallest possible integer for which $180$ is a multiple.

The question as posed does not seem to me to strictly insist or restrict that the answer be a positive integer; in fact, it seems to specifically allow for such possibility by avoiding usage of nearly all mathematical jargon except for "integer" (which may be negative or positive) and "multiple" (again, may be negative or positive).

Also $-180$ is clearly smaller than $9$, unless "smallest possible value of $n$" does not mean what it would appear to mean to most mortals familiar with a number line.

score 1 · Accepted Answer · answered May 24 '18 at 08:00

No need to big calculations. Simple justification is required. The number n should be factor of 180 but should not be factor of 60. The factors of 60 are 1,2,3,4,5,6,10,15,20,30 & 60. The factors of 180 are 1,2,3,4,5,6,9,10,12,15,18,20,30,36,45,60,90 & 180. It is clear that the lowest possible number which is a factor of 180 but not that of 60 is only 9 Hence answer is 9

The lowest number that is a multiple of both 60 and the integer n is 180. Find the smallest possible value of n.

10 Answers10