Always divisible by 7?

Question

A three-digit number is converted to a six-digit number by writing the digits again next to it.
For example, the number 235 is converted to 235235.
Why is it that the new number can always be divided by 7?

Answer 1

By writing a three-digit number next to itself, you are multiplying that three-digit number by 1001. Since 1001=7*143, the new number created is divisible by 7 (as well as 143!).

Answer 2

A three-digit number with digits "abc" could be written as 100*a + 10*b + c. The six digit number then would be written as 100000*a + 10000*b + 1000*c + 100*a + 10*b + c. After factoring we obtain (100000 + 100)*a + (10000 + 10)*b + (1000 + 1)*c = 100100*a + 10010*b + 1001*c. Now factor again to obtain 1001*(100*a + 10*b + c). Of these two factors 1001 = 7* 143, so the original six-digit number will be divisible by 7 because one of its factors is divisible by 7.

Answer 3

By keeping start 3 digits 'as end 3 digits' we get "six digit numbers" that are definitely divisible by 1001.

Examples are 127127= 127*1001

437437= 437*1001 (or any other similar number!)

1001 being 7*11*13 all 'said six digit numbers' having same-digits in 'start and end groups' are divisible by 7,11 and 13. Further said 3-digits number itself is another factor, which can be factored again if it is not a prime number!

It is a classic example being followed in Vedic mathematics for mental factoring!

Similarly 8-digit numbers having same " start and end digit equal-groups" are divisible by 10001, which has two prime number factors 73 and 137. Naturally all said 8-digit numbers are divisible by 73 and 137.

It appears that a study of numbers like 11, 0101, 001001 or a higher order has been carefully done by Vedic mathematicians (ancient Indian) and we can still read two sutras relating 1001 factors 7 and 13 and another sutra relating 100000001 which is 05882353*17<----as 1/17 in "Vedic Mathematics"


Regards!

Answer 4

235235 = 235(1001) = 235( 11•13•7)

Answer 5

The number abc is in fact

a*100 + b*10 +c

therefore

abcabc is in fact

a*100000 + b*10000 + c*1000 + a*100 + b*10 + c =
= a*(100100) + b*(10010) + c*(1001) =
= 1001*(100a+10b+c) = 7*11*13*(100a+10b+c)

your number can be divided by 7, 11 and 13 all the time

(and inf fact abcabc is abc multiplied by 1001)

Answer 6

Mathematically it's easy to answer:
if your number is xyz (like 235), then xyzxyz (like 235235) can be expressed as:
x(1+1,000)+y(10+10,000)+z(100+100,000),
or x(1001)+y(10010)+z(100100)
or 1001*(x+10y+100z)
because 1001 is divisible by 7, then the whole number will be divisible by 7.

Answer 7

235235 = 235(1001) = 235( 11•13•7)

Answer 8

The original number is 100h +10t + u
The new number is 100000h +10000t + 1000u +100h+10t+u.
This equals 100100h + 10010t +1001u
= 7(14300h+1430t+143u)
Thus the number is always divisible by 7

Answer 9

♠ xyzxyz = 100000x +10000y +1000z +100x +10y +z =
=100100x +10010y +1001z =14300*7*x +1430*7*y +143*7z=
= 7*(14300*x +1430*y +143*z);

Answer 10

cus maths works