**21. Sevens**

1. Take a single digit number, say 3, then double it to get 6 and write the result as 63. Do you always get a number divisible by 7?

2. Take a two digit number, say 12, and multiply it by 3 to get 36 and write the result as 3612. Is this result always divisible by 7?

3. What happens when you follow the same method and multiply a three digit number by four? Try a four digit number multiplied by five. Can you explain these results?

Scroll down the page for the answer.

**1.** If the single digit number is *a* then multiplying it by 2 gives 2*a*. Writing these digits in reverse is equivalent to multiplying the second number, 2*a*, by 10 so that the number you write down is actually equal to 20*a* + *a* ie 21*a* and this is always divisible by 7 whatever the value of *a*.

**2.** In a similar way, a two digit number with digits *a* and *b* is actually equivalent to 10*a* + *b*. Multiplying this by 3 gives 30*a* + 3*b*. If you now write these in reverse order then you are effectively multiplying the second number, 30*a* + 3*b*, by 100 so what we now have is

100(30*a* + 3*b*) + 10*a* + *b*

which is

3000*a* + 300*b* + 10*a* + *b*

= 3010*a* + 301*b*

= 301(10*a* + *b*)

and this is always divisible by 7

**3.** Likewise, a three digit number with digits *a*, *b* and *c* is actually equivalent to 100*a* + 10*b* + *c*. Multiplying this by 4 gives 400*a* + 40*b* + 4*c*. If you now write these in reverse order then you are effectively multiplying the second number, 400*a* + 40*b* + 4*c*, by 100 so what we now have is

100(400*a* + 40*b* + 4*c*) + 100*a* + 10*b* + *c*

= 40000*a* + 4000*b* + 400*c* + 100*a* + 10*b* + *c*

= 40100*a* + 4010*b* + 401*c*

= 401(100*a* + 10*b* + *c*)

which, in general, is *not* divisible by 7...well, we never said it was going to be easy...

However, looking at a four digit number with digits *a, b, c* and *d* such that the number is

1000*a* + 100*b* + 10*c* + *d*

then, multiplying by five, we have

5000*a* + 500*b* + 50*c* + 5*d*

Reversing these numbers and writing them as before is equivalent to multiplying the second one by 10000 so that the result is equal to

10000(5000*a* + 500*b* + 50*c* + 5*d*) + 1000*a* + 100*b* + 10*c* + *d*

= 50000000*a* + 5000000*b* + 500000*c* + 50000*d* + 1000*a* + 100*b* + 10*c* + *d*

= 50001000*a* + 5000100*b* + 500010*c* + 50001*d*

= 50001(1000*a* + 100*b* + 10*c* + *d*)

and since 50001 is divisible by 7 then any such number will be

Of course, we haven't attempted to answer the general case where an *n* digit number is multiplied by *n* + 1 because to do so would spoil all your fun...;-)

Visit the puzzle archive for more puzzles and their solutions.