Question

Maths puzzle - i completely failed on this one but could have kicked myself when i looked up the answer:
Find a five digit number which if you place 1 in front of it gives a number which is one third of the number obtained when you add 1 to the end of this number.

Answer 1

If I were to say that the last digit of that 5 digit number is 7, would that be correct?

Answer 2

hmm - i dont remember the number - just the method

Answer 3

"the number obtained when you add 1 to the end of this number." does this number refer to the first number of the one to which we have added 1 in front of?

Answer 4

yes - when i say add i dont me add arithmetically - just make a six digit number

Answer 5

King GEorge - yes

Answer 6

I've almost got it. Give me about 2 more minutes.

Answer 7

and by adding 1 in front that also makes a six digit number - this number is 1/3 of the second 6 digit number

Answer 8

The solution is 42857

Answer 9

yep - good job

Answer 10

It's rather easy to find E, since \(3E \equiv 1 \mod 10\). Hence, \(E=7\)

Answer 11

right - good thinking

Answer 12

With a little bit of work, you then get that \[3D \equiv5 \mod 10\]\[3C\equiv 4 \mod 10\]\[3B\equiv 6 \mod 10\]\[3A \equiv 2 \mod 10\]Hence, my solution follows.

Answer 13

the method they used was
using the equation
3(100000 + x) = 10x + 1
solve for x

Answer 14

That's a much simpler way to do it. My way is just more easily generalized to harder problems.

Answer 15

right - i cant believe i didnt think of the simple way

Answer 16

plus, the above way, you have to find 299999/7. Without a calculator this could be hard. My way required no numbers above 24