Question

If LCM of two 6-digit integers has 10 digits, then their GCD has at most how many digits?

Answer 1

\[\text{lcm(a,b)} \times \gcd(a,b) = a\times b\]

Answer 2

The product of any two \(6\)-digit numbers can have\(11, 12, 13\) digits.

Since the \(\text{lcm}\) has \(10\) digits, \(\gcd\) can have \(1, 2,3,4\) digits.

Answer 3

how did you arrived at this conclusion that lcm has 10 digits so GCD at most has 4 digits

Answer 4

Is it possible for the product of a 10 digit number and a 5 digit number to have 13 digits ?

Answer 5

no

Answer 6

but the answer 4 is not correct

Answer 7

...as per to my laptop.

Answer 8

should be \(2\) then

Answer 9

\(\large \color{black}{\begin{align} \text{lcm(a,b)} \times \gcd(a,b) = a\times b\hspace{.33em}\\~\\
\text{10 digit} \times \gcd(a,b) = \text{12 digit}\hspace{.33em}\\~\\
\end{align}}\)

Answer 10

the product of two \(6\) digit natural numbes has \(12 ~~digits\)

Answer 11

http://www.wolframalpha.com/input/?i=999999*999999+%3D

Answer 12

Ok. I see on website where you re coming from .. so max can have 12 digit on multiplying two 6 digits numbers.... but how did you arrived that if it were not for this website wolf...

Answer 13

u calculate the product largest 6 digit numbers by hand also

Answer 14

hahha..... thought there was an easier way

Answer 15

but this doenst prove anything

Answer 16

both ans. 2, 4, are wrong

Answer 17

how did u know that

Answer 18

only 1 option remains now

Answer 19

Ans is 3 : In this case we know

\[ $a<10^6$ and $b<10^6$, so\\ $ab<10^{12}$.\]
We also know that
\[$\mathop{\text{lcm}}[a,b]\ge 10^9$\]

since the smallest 10-digit number is $10^9$
Therefore
$$\gcd(a,b) < \frac{10^{12}}{10^9} = 10^3,$$ so $\gcd(a,b)$ has at most $\boxed{3}$ digits.

Answer 20

\[so $\gcd(a,b)$ \\has at most\\ $\boxed{3}$ digits.\]

Answer 21

Ans is 3 : In this case we know

\(a<10^6\) and \(b<10^6\), so \(ab<10^{12}\)

We also know that
\(lcm[a,b]≥10^9\)


since the smallest 10-digit number is \(10^9\)
Therefore
\(\gcd(a,b)<\dfrac{10^{12}}{10^9}=10^3\)


so \(\gcd(a,b)\) has at most\(\boxed{3}\) digits.

Answer 22

looks good now