Let S(n) be the sum of the digits of a natural number n, e.g. S(168) = 15.
Call a number n "special" if it is evenly divisible by S(n), e.g. 36 is special because S(36) = 9 and 9 divides 36.
Prove that there cannot be a sequence of 22 consecutive numbers that are all special.
Bonus: Prove that there cannot be a sequence of 21 consecutive special numbers.
(In reply to
Solution, not bonus by Jer)
Suppose there are 21 consecutive special numbers. Then, they must be of the form x90, x91, ..., x99, y00, y01, ..., y09, y10. Since sod(y01)=sod(y10)=sod(y)+1, then sod(y)+1y01 and sod(y)+1y10. Since y10y01=9, then sod(y)+19. Therefore, sod(y)+1=1, 3, or 9. Then, sod(y)=0, 2, or 8. If sod(y)=0, then y=0. That is impossible because x=y1 and x is nonnegative. Suppose sod(y)=2. Then, sod(y02)=sod(y)+2=2+2=4. Then, 4y02. That is impossible because no multiple of 4 ends in 02. Suppose sod(y)=8. Then, sod(y02)=sod(y)+2=8+2=10. Then, 10y02. That is impossible because no multiple of 10 ends in 2. Therefore, there are no 21 consecutive special numbers.

Posted by Math Man
on 20210616 20:22:54 