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 2a. Writing these digits in reverse is equivalent to multiplying the second number, 2a, by 10 so that the number you write down is actually equal to 20a + a ie 21a 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 10a + b. Multiplying this by 3 gives 30a + 3b. If you now write these in reverse order then you are effectively multiplying the second number, 30a + 3b, by 100 so what we now have is
100(30a + 3b) + 10a + b
3000a + 300b + 10a + b
= 3010a + 301b
and this is always divisible by 7
3. Likewise, a three digit number with digits a, b and c is actually equivalent to 100a + 10b + c. Multiplying this by 4 gives 400a + 40b + 4c. If you now write these in reverse order then you are effectively multiplying the second number, 400a + 40b + 4c, by 100 so what we now have is
100(400a + 40b + 4c) + 100a + 10b + c
= 40000a + 4000b + 400c + 100a + 10b + 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
1000a + 100b + 10c + d
then, multiplying by five, we have
5000a + 500b + 50c + 5d
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(5000a + 500b + 50c + 5d) + 1000a + 100b + 10c + d
= 50000000a + 5000000b + 500000c + 50000d + 1000a + 100b + 10c + 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...;-)
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