The statement “If we repeat a three-digit number twice, to form a six-digit number, the result will be divisible by 7, 11, and 13” is a mathematical property in number theory and mathematics. By repeating a three-digit number twice, we get a six-digit number having the last three digits the same as the first three digits. The number formed is divisible by the prime numbers 7, 11 and 13.

How to determine whether a number is divisible by 7, 11, and 13?

Since 7, 11, and 13 are all prime numbers, their product, 7 * 11 * 13 = 1001, is also a multiple of each of these prime numbers. Therefore, if a number is divisible by 1001, it is also divisible by 7, 11, and 13 individually, because 1001 is their least common multiple (LCM).

Is It True that 6 Digit number formed by repeating a 3-digit number twice is always divisible by 7, 11 and 13

Let us consider a three digit number 123

• Repeat it twice: When we repeat 123 twice, we get the six-digit number 123123.
• Check divisibility by 1001 (or 7, 11, and 13): The six digit number 123123 can be written as 123*1001. This implies 1001 is a divisor of 123123. Hence 123123 is divisible by 7, 11 and 13.

Let us consider another three digit number 752

• Repeat it twice: When we repeat 752twice, we get the six-digit number 752752.
• Check divisibility by 1001 (or 7, 11, and 13): The six digit number 123123 can be written as 752*1001. This implies 1001 is a divisor of 752752. Hence 752752is divisible by 7, 11 and 13.

Mathematical Proof to show 6 Digit number formed by repeating a 3-digit number twice is always divisible by 7, 11 and 13:

Consider a Three-digit Number:

Start by considering a 3-digit number, denoted as ‘n,’ which possesses of three digits x, y and z.

Let n is represented by:

n=xyz, here x,y and z are digits of number n.

Repeat the number twice:

Repeat the number n twice, that will result in a 6 digit number, i.e., n will become

n=xyzxyz, the last three digits will be same as first three digits.

Check divisbilty by 7, 11, and 13

Mathematically n can be expressed as

n= xyzxyz = 100000x + 10000y + 1000z + 100x + 10*y + z

Notice that we can factor out 1001 from this expression:

n=100000*x + 10000*y + 1000*z + 100*x + 10*y + z

n=100*x(1000 + 1) + 10*y(1000 + 1) + z(1000 + 1)

n= 1001 * (100*x + 10*y + z)

Key Insight:

At this point we observe that we can factor out of 1001 from n . Which 1001 is a divisor of n i.e., n is divisble by 1001.

Divisibility by 7, 11 and 13:

Since n is divisble by 1001, it means n is divisble by 7, 11 and 13 as 1001 is the least common multiple (LCM) of 7, 11 and 13.

Hence we can say that if we repeat a three-digit number twice, to form a six-digit number, the result will be divisible by 7, 11 and 13.