B1 Give the smallest positive integer that is divisible by 26, ends in 26, and for which the sum of the digits equals 26.
B1 Because the number ends in 26, it has the form a...b26, so a...b00 is also divisible by 26 and it must be true that a...b is divisible by 13, since 100 is already divisible by 2. Since the sum of the digits must equal 26, the sum of the digits of a...b must equal 18. This means that a...b is also divisible by 9. Therefore, a...b must be divisible by $9 * 13=117$. We need to find the smallest multiple of 117 where the sum of the digits equals 18. 117, 134, 351, 468 bingo!, so 46826.