Find Positive Integers with GCD 10 and LCM 100

To find all positive integers $a$ and $b$ with greatest common divisor $10$ and least common multiple $100$, factor out their shared divisor. The remaining factors must be relatively prime, and their product is fixed by the LCM. This reduces the problem to checking a short list of factor pairs.

Factor out the greatest common divisor

Since $\gcd(a,b)=10$, both numbers are multiples of $10$. Write

$$a=10x,\qquad b=10y$$

for positive integers $x$ and $y$. Dividing both numbers by their greatest common divisor leaves numbers with no common factor greater than $1$, so

$$\gcd(x,y)=1.$$

The LCM can then be expressed in terms of $x$ and $y$. For any positive integers $x,y$,

$$\operatorname{lcm}(10x,10y)=10\operatorname{lcm}(x,y).$$

Because $x$ and $y$ are coprime, $\operatorname{lcm}(x,y)=xy$. Therefore,

$$\operatorname{lcm}(a,b)=10xy.$$

The required LCM is $100$, so $10xy=100$, or $xy=10$.

Find the coprime factor pairs

The positive factor pairs whose product is $10$ are $(1,10)$ and $(2,5)$, along with their reversals. Each pair is coprime:

$$\gcd(1,10)=1,\qquad \gcd(2,5)=1.$$

Thus the possibilities for $(x,y)$ are

$$(1,10),\ (10,1),\ (2,5),\ (5,2).$$

Multiplying each coordinate by $10$ gives the complete list:

$$\boxed{(a,b)=(10,100),\ (100,10),\ (20,50),\ (50,20).}$$

These are ordered pairs, so reversing the entries counts as a separate solution. If the question treats the pair as unordered, there are just two possibilities: $\{10,100\}$ and $\{20,50\}$.

Check the candidates

For $(10,100)$, the smaller number divides the larger, so the greatest common divisor is $10$ and the LCM is $100$. For $(20,50)$, the Euclidean algorithm gives

$$50=2\cdot20+10,\qquad 20=2\cdot10,$$

so $\gcd(20,50)=10$. Their LCM is

$$\operatorname{lcm}(20,50)=\frac{20\cdot50}{10}=100.$$

Swapping the entries does not change either the GCD or the LCM, so the reversed pairs also satisfy both conditions.

Why the list is complete

The calculation $xy=10$ restricts the possibilities to positive factor pairs of $10$. The additional condition $\gcd(x,y)=1$ rules out any factor pair with a common divisor greater than $1$. Here both factor pairs are coprime, so there are no other cases to consider.

More generally, if positive integers have GCD $g$ and LCM $L$, write them as $gx$ and $gy$, where $\gcd(x,y)=1$. Their LCM is then $gxy$, which means $xy=L/g$. This gives a useful general test: $g$ must divide $L$, and each solution corresponds to a coprime factor pair of $L/g$.

Common mistakes

• Using all factor pairs without checking coprimality. After factoring out the GCD, the two remaining factors must be relatively prime. For this example, the pairs $(1,10)$ and $(2,5)$ pass that test.
• Counting only one ordering. If the problem asks for all pairs $(a,b)$, include both orders unless it explicitly treats the pair as unordered.
• Using the product formula without its condition. The identity $\operatorname{lcm}(x,y)=xy$ applies only when $x$ and $y$ are coprime. In general, $\gcd(x,y)\operatorname{lcm}(x,y)=xy$.
• Stopping at the factor pairs. The pairs found are for $x,y$, not $a,b$; multiply each entry by $10$ to recover the requested integers.