New Year Puzzle

Number Theory Level 4

A four-digit number, $\overline{abcd}$ (in base 10) can be expressed as $\overline{abcd}=1000a+100b+10c+d$ . If we change this formula to $998a+102b+98c+a+1$ , for $1000<\overline{abcd}<9999$ , how many times does $\overline{abcd}=998a+102b+98c+d+1$ ?

The answer is 0.

1 solution

Fin Moorhouse
Jan 9, 2016

For all the numbers $\overline{abcd}$ , $998a+102b+98c$ is even. If the number is odd, then $d+1$ is even, so $998a+102b+98c+d+1$ is also even, so the number can't be odd. If it is even, then $d+1$ is odd, so $998a+102b+98c+d+1$ is also odd, so the number can't be even either. Therefore there are no solutions to this question.

Noooo look at your question. It says 998a + 102b + 98c + a + 1 not 998a + 102b + 98c + d + 1. Can you correct your question statement? Thanks.

Park Sejin - 4 years, 9 months ago

Oops, thanks for pointing that out. Corrected.

Fin Moorhouse - 4 years, 9 months ago

New Year Puzzle

The answer is 0.

1 solution

0 pending reports