19. The OP stated they have registered the phone in their name (I didnt know you could do that when its a contract phone taken out in someone elses name?) so I doubt the original owner could get the phone barred once the details have been transferred over?

correct!!!!

]]>Okay and see you later.

]]>Anyway, I'll go now. Bye Bye !

]]>Thank you again ! I had the answer in my book, but I didn't want to see it,(and still haven't) just wanted it to be checked by someone who had experience ! Thanks !

]]>We can then conclude that there exists 11 ways possible to have 1000 with a sequence of numbers which contains at maximum 11 times the number 88 and at minimum 1 time the number 88.

Correct!

In total, you will have 14 ways of having one thousand.

Correct!

I can find no hole in your reasoning and you have the correct answer.

The simple technique involves the number of solutions to the Diophantine equation.

Given that 0 ≤ a,b,c ∈ N

Your answer is proved using generating functions.

]]>This is somewhat different than your other questions. Verification of your reasoning can be supplied by a simple technique. It will take time though.

]]>Several digits "8" are written and some "+" signs are inserted to get the sum 1000. Figure out how it is done (For example, if we try 88+88+8+88+8, we fail because we get only 280 insted of 1000.)

My answer(It may be long) :

1-First, the number 1000 is composed of 125 digits "8", because :

2-This limits us to 3 principal number in our sequence of numbers, which will help us build 1000.

8: He is composed of only one digit 8.

88: He is composed of 11 digit 8.

888: He is composed of 111 digit 8.

3-Now, there exists different ways to build the number 1000(Depending of the composition of your sequence of numbers.)

4-If your sequence of numbers is composed of the number 888(Which can only go one time in your sequence.), this results in this :

a) 125-111=14 numbers "8" left.

You could now continue your sequence by 88 because:

14-11=3 numbers "8" left. Then, finish off your sequence by 3 digits "8", because :

3-3=0 digits "8" left.

b) Or, you could have had 888 in your sequence, which would give :

125-111=14 chiffres '8' left.

And then, complete the sequence by 14 digits 8.

14-14=0 digits "8" left.

5)If your sequence contains one or many times the number 88, it should give :

a) Maximum of 88 composed in the number 1000: 11 times

11*88=968

digits "8" in total

125-121=4 digits '8'* left

To obtain 1000, we must add 4 times the digit 8

4-4=0 digits "8" left.

We can then conclude that there exists 11 ways possible to have 1000 with a sequence of numbers which contains at maximum 11 times the number 88 and at minimum 1 time the number 88.

Each time you will remove a "88" in your sequence, it will be needed to be replaced by 11 times the digit 8 so we can keep the answer equalling 1000.

Ex: 88+88+88+88+88+88+88+88+88+88+88+8+8+8+8=1000

Now, we will remove a "88"

88+88+88+88+88+88+88+88+88+88+**8+8+8+8+8+8+8+8+8+8+8**+8+8+8=8=1000

6- Finally, the last way of obtaining 1000 consists to write 125 times the number 8.

8+8+8+8...125 times

In total, you will have 14 ways of having one thousand.

Do you think I've figured out how it's done ? Or not ? No answers, just suggestions about what I wrote. Thank you!

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