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, 2010

Number theory Difficulty 6.0 AIME, harder Prove it Ukraine

There are 16 consecutive positive integers written on the board. Andrew calculates their product and Olesya – their sum. Can it happen that in both numbers there coincide

a) three last digits,
b) four last digits?

Solution

Answer: a) yes; b) no.

It's obvious that a number received by Andrew is divisible by $16$ and by $125$ , because from $16$ consecutive numbers more than four are divisible by $2$ and at least $3$ are divisible by $5$ . It also implies that three last digits in Andrew's number are $0$ .

a) Let the numbers $a, a+1, a+2, \dots, a+15$ be written on the board. Then Olesya obtained the number $8(2a+15)$ . Putting $a=55$ we have that three last digits of this number are $0$ , so the answer to the case a) is 'yes'.

b) Since the number obtained by Andrew is divisible by $16$ , then it's true for the number obtained by four last digits of this number. If we assume that the answer to b) is 'yes' then the number obtained by four last digits of the number $8(2a+15)$ is divisible by $16$ . It follows that $8(2a+15) \div 16$ . This contradiction gives the answer 'no'.

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