3. Sofie and her grandmother are both born on January 1. The grandmother's age is an integer multiple of her granddaughter Sofie's age for six consecutive years. This does not hold for the seventh year. A few years later, the grandmother's age is once again an integer multiple of Sofie's age.

How old can the grandmother be?
A) 63
B) 66
C) 70
D) 90
E) 91
A3. C) 70 In six consecutive years, the age of grandma is an integer multiple of the age of her granddaughter Sofie. In the first year of these six consecutive years, we call the age of grandma $g$ and that of Sofie $s$. Then $g$ is divisible by $s$, and also $g-s$ is then divisible by $s$. In the following year, the age of grandma is $g+1$ and that of Sofie $s+1$. Then $g+1$ is divisible by $s+1$, but $(g+1)-(s+1)=g-s$ is also divisible by $s+1$. The year after that, the age of grandma is $g+2$ and that of Sofie $s+2$. Then $g+2$ is divisible by $s+2$ and also $(g+2)-(s+2)=g-s$ is divisible by $s+2$.

If we continue for three more years, we find that $g-s$ is divisible by the consecutive numbers $s, s+1, s+2, s+3, s+4$ and $s+5$. If the numbers 5, 6, and 7 are included, then $g-s$ would be at least $5 \cdot 6 \cdot 7=210$ and grandma would thus be more than 210 years old. That is impossible. If $s$ is at least 6 (so that 5 is no longer included) and at most 9, then you have the numbers 9, 10, and 11, and grandma would be more than 990 years old. If $s$ is at least 10 and at most 13, then you have 13, 14, and 15, and grandma would have to be even older. And so on. The only realistic option is therefore to take $s=1$, so that the numbers are from 1 to 6. (Note that $s=0$ is not possible, because then both Sofie and grandma would be 0 years old.) Now $g-s$ must be at least a multiple of $5 \cdot 6=30$. That is not divisible by 4, so $g-s$ must even be a multiple of 60. Indeed, that is divisible by the numbers from 1 to 6. (This is actually the least common multiple of these numbers.) We conclude that $g-s=60$, because if $g-s=120$ or a larger multiple of 60, then grandma would be inhumanly old.

Since $g-s=60$ and $s=1$, grandma is 61 years old and Sofie is 1 in the first year. Indeed, 61 is a multiple of 1. Further, 62 is a multiple of 2, 63 is a multiple of 3, 64 is a multiple of 4, 65 is a multiple of 5, and 66 is a multiple of 6. In the seventh year, we get 67, which is not a multiple of 7. In the two years after that, we get 68 and 69, which are not multiples of 8 and 9, respectively. But when grandma is 70, Sofie is 10, and their ages are again multiples of each other.