Shared Birthdays

The woman was 60 when the grandson was born. For each subsequent year we have (Grandmother age, Grandson age) as $(61,1),(62,2),(63,3),(64,4),(65,5),(66,6)$ .

This is only possible if the age of grandmother fall between two prime numbers that are at lest six apart. Such numbers are:-31-37, 47-53, 53-59, 61-67, 73-79, 83-89. The consecutive numbers should have factors that are also consecutive.
$31=1*31....32=2*16...33=3*11.....34=2*17......XXXXXXX\\47=1*47.....48=2*24.....49=7*7.....XXXXXXX\\53=1*53....54=2.27....55-5*11.....XXXXXXX\\61=\color{#D61F06}{1} *61.....62=\color{#D61F06}{2} *31.....63=\color{#D61F06}{3} *21....\\64=\color{#D61F06}{4} *16..... 65=\color{#D61F06}{5} *13.....66=\color{#D61F06}{6} *11........................OK$

As Niranjan posits, the Grandmother's ages must be between two primes that are at least 6 apart. If $p$ is a prime then $p$ can be written as $6k \pm 1$ (Why? Hint: modular arithmetic).

The chain of six numbers: $6k + 1$ , $6k + 2$ , $6k + 3$ , $6k + 4$ , $6k + 5$ , $6k + 6$ is a convenient starting place because $6k + 1$ is always divisible by 1, $6k + 2$ is always divisible by 2, $6k + 3$ is always divisible by 3, and $6k + 6$ is always divisible by 6. In fact, the grandson's birthdays are likely to be 1, 2, 3, 4, 5, and 6. $6k + 4$ is divisible by 4 if $k$ is even and $6k + 5$ is divisible by 5 if $k$ is a multiple of 5. Hence the lowest value $k$ can be is 10 (the grandmother cannot be the same age as the grandson!). $k=10$ indeed works as a solution.

Thus the answer is $\boxed{66}$ .

In 6 consecutive years their ages will have to have a factor of five at one point. By starting with assumption of grandsons ages as 1 through 6, when boy is 5 grandmother's age must end in 5 or 0. But in previous year the boys age is 4 and no multiples of 4 have a 9 in the units place, so grandma's age ends in 5 when kid is 5. When kid is 6 grandma's age ends in 6 and is divisible by 6 (36, 66, 96, . . .) When kids is 4 grandma's age ends in 4 but from the above list 34 and 94 are not divisible by 4, but 64 is. The remaining numbers quickly check out. If original assumption of boys ages did not include 5, next possible age would have to include 10. But 9 and 10 only go into 99 and 100 for reasonable ages and 10 and 11 only go into consecutive ages at 120 and 121. Grandma's age will continue to be less reasonable as boys age increase.

To find the solution easily,we can assume the boy's age to be 1,2,....6.So the woman's age at the end of the 6th year is of the form 2k,3k,4k+2,5k+1 and 6x (where x is not divisible by 2)-- like 36,66,96...we find 66 satisfies the requirements.

You are assuming that the common birthday is not 29 February otherwise the 6 birthdays would be spread over 24 years since they celebrate once every four years.