Last updated on May 26th, 2025
The smallest positive integer that divides the numbers with no numbers left behind is the LCM of 11 and 13. Did you know? We apply LCM unknowingly in everyday situations like setting alarms and to synchronize traffic lights and when making music. In this article, let’s now learn to find LCMs of 11 and 13.
We can find the LCM using listing multiples method, prime factorization method and the long division method. These methods are explained here, apply a method that fits your understanding well.
Step 1: List the multiples of each of the numbers;
11 = 11,22,33,…143
13 = 13,26,39,…143
Step 2: Find the smallest number in both the lists
LCM (11,13)=143
Step 1:Prime factorize the numbers
11 = 11×1
13 = 13×1
Step 2:find highest powers
Step 3:Multiply the highest powers of the numbers
LCM(11,13) = 143
Listed here are a few mistakes children may make when trying to find the LCM due to confusion or due to unclear understanding. Be mindful, understand, learn and avoid!
For any two integers a and b, the LCM is related to their product by the equation: LCM(a, b)×GCF(a, b)=a×b Use this formula to verify if the LCM of 11 and 13, when divided by 11, is equal to the product of 13 and the GCF of 11 and 13.
LCM(11,13)=143,GCF(11,13)=1
Verify:
LCM(11,13)/11=143/11=13
Now, check if:
13×GCF(11,13)=13×1=13
Since both sides are equal, the condition is verified.
This problem emphasizes verifying relationships involving LCM and GCF using fundamental number theory concepts.
If N is the least number divisible by both 11 and 13, and the sum of the digits of N equals 8, what is N?
The least number divisible by both 11 and 13 is their LCM:
LCM(11,13)=143
The sum of the digits of 143 is:
1+4+3=8
Thus, N=143
Here, the LCM of 11 and 13 is found to be 143, and then the sum of its digits is verified to meet the condition given in the problem.
A light flashes every 11 seconds and another light flashes every 13 seconds. Both lights flash at the same time. After how many seconds will they flash together again?
The time after which both lights will flash together again is the LCM of 11 and 13, which is 143 seconds.
This problem models a situation of repeated events. The lights will flash together at intervals corresponding to the LCM of their individual flashing times.
A number X is divisible by both 11 and 13. The number is less than 500. What is the largest possible value of X?
To find the largest number less than 500 divisible by both 11 and 13, we compute the LCM of 11 and 13, which is 143.
Now divide 500 by 143:
500143≈3.5
The largest integer is 3, so the largest number divisible by both is:
143×3=429
Since we are looking for a number divisible by both 11 and 13, we use the LCM. Then, we find the largest multiple of the LCM that is less than 500.
Hiralee Lalitkumar Makwana has almost two years of teaching experience. She is a number ninja as she loves numbers. Her interest in numbers can be seen in the way she cracks math puzzles and hidden patterns.
: She loves to read number jokes and games.