Bmo1 1997 - Q1

Can someone verify my solution please?

1) "N is a four-digit integer, not ending in zero, and R(N) is the four-digit integer obtained by reversing the digits of N; for example, R(3275) = 5723.

Determine all such integers N for which R(N) = 4N + 3."

Let the first digit of N be 'a', the second digit be 'b', the third digit be 'c', and the fourth digit be 'd'.

=> N = 1000a + 100b + 10c + d

=> R(N) = a + 10b + 100c + 1000d

The question states that R(N) = 4N + 3, so:

a + 10b + 100c + 1000d = 4(1000a + 100b + 10c + d) + 3
a + 10b + 100c + 1000d = 4000a + 400b + 40c + 4d + 3
3999a + 390b - 60c - 996d + 3 = 0
1333a + 130b - 20c - 332d + 1 = 0

If a >= 3, the above equation is not solvable for single digit positive integer values of b, c and d.

Hence a <= 2.

If a = 2, 1333a + 130b - 20c - 332d + 1 = 1 mod 2. 0 = 0 mod 2, so a =/= 2.

=> a = 1.

So, 1333 + 130b - 20c - 332d + 1 = 0

=> 130b - 20c - 332d = -1334

d = 2 mod 5, or else the equation would be unsolvable for integer values of 'b' and 'c'.

If d = 2, 130b - 20c - 664 = -1334

130b - 20c = -670, which again is impossible to solve for the values of 'b' and 'c' that are possible. Hence d =/= 2.

=> d = 7.

130b - 20c - 332d = -1334
130b - 20c - 2324 = -1334
130b - 20c = 990
13b - 2c = 99

If 'b' <= 7, the equation has no solutions for the values of 'c' that are possible. Hence b >= 8.

If b = 8, 13b - 2c = 0 mod 2. 99 = 1 mod 2, so b =/= 8.

=> b = 9.

117 - 2c = 99

=> c = 9

=> N = 1997

Thanks!

~~Simba

Reply 1

dvs

Can't see anything wrong with it. (And in olympiads it's usually a good sign when your answers are the same as the year.)

Reply 2

Simba

Thanks a lot :smile:

*Reps*!

Reply 3

yusufu

dvs

Can't see anything wrong with it. (And in olympiads it's usually a good sign when your answers are the same as the year.)

Reply 4

Simba

Lol, yes, I noticed that they seem to have an infatuation with including the year at least once in each olympiad. At least it shows they're trying with the questions :p: