How many pairs of consecutive integers? How many pairs of consecutive integers?
How many pairs of consecutive integers between and inclusive 1000 to 2000 is no carry required when the two integers added ?
for example: 1001+1002=1003 is a valid counting but 1078+1079 is not.
 A: Let the smaller of the two consecutive integers have the digits $1abc$ (the one's digit is $c$, the $10$'s digit is $b$, the $100$'s digit is $a$). There are several possibilities.
1) $c<9$, so the next integer is $1ab(c+1)$ (the one's digit is $c+1$, the $10$'s digit is $b$, the $100$'s digit is $a$). To add those two without any carries in the addition process we must have
$$0\le a<5,\ 0\le b<5,\ 0\le c<5$$
I'm sure you can count all those possibilities. I get $5^3=125$.
2) $c=9$ and $b<9$. Then the next integer is $1a(b+1)0$ (the one's digit is $0$, the $10$'s digit is $b+1$, the $100$'s digit is $a$). Here to get no carry we must have
$$0\le a<5,\ 0\le b<5,\ c=9$$
I'm also sure you can count all those possibilities. I get $5^2=25$.
3) $c=9,\ b=9$, and $a<9$. Then the next integer is $1(a+1)00$ (the one's digit is $0$, the $10$'s digit is $0$, the $100$'s digit is $a+1$). Here to get no carry we must have
$$0\le a<5,\ b=9,\ c=9$$
I'm also sure you can count all those possibilities. I get $5^1=5$.
4) $c=9,\ b=9,\ a=9$. Then the next integer is $2000$. Here to get no carry we must have
$$a=9,\ b=9,\ c=9$$
I'm also sure you can count all those possibilities. I get $5^0=1$.
Then add up the counts for those four cases. I get $125+25+5+1=156$.
This problem is small enough to be checkable. I made an Excel spreadsheet, checking all the numbers from $1000$ through $1999$ for the smaller of the two numbers, and I got the same answer of $156$. I also get that answer when I use the single-line Python code
print(sum(all(int(c) + int(d) < 10 for c, d in zip(str(n), str(n+1)))
          for n in range(1000, 2000)))

