Relative ages

Here's my solution which doesn't involve guesswork.

Let the grandmother be $x$ years old, and let the kid be $k$ years old when this first happens. Then, we are given that

$\frac{ x+k} { k} , \frac{ x + k + 1 } { k + 1}, \frac{ x + k + 2} { k + 2}, \frac{ x + k + 3} { k + 3}, \frac{ x + k + 4 } { k + 4}, \frac{ x + k + 5} { k + 5},$

are all integers. Subtracting 1 from each of them, we get that

$\frac{ x } { k } , \frac{ x } { k+1 } , \frac{ x } { k+2 } , \frac{ x } { k+3 } , \frac{ x } { k+4 } , \frac{ x } { k+5}$

are all integers. Hence, (k, k+1, k+2, k+3, k+4, k+5 ) must all divide $x$ . This implies that the LCM of these 6 consecutive integers divide $x$ .

Let's try $k = 1$ . Then $LCM ( 1, 2, 3, 4, 5, 6) = 60$ . We get that $x = 60$ is a solution, and in fact so is $x = 60 n$ . This is why I added in the restriction that $x < 100$ .

Let's try $k = 2$ . Then $LCM (2, 3, 4, 5, 6, 7) = 420$ , which would be too old. Similarly, for any larger $k$ , $x$ will be too large. Hence, the age is 60.

So what we are looking for is at least $5$ consecutive composite numbers. Use this as a reference to simplify our calculation.

First six numbers that satisfy it are $(23, \ldots , 28)$ . We check that $23 = 1 \times 23, 24 = (1+1) \times 12, 25 = 5 \times 5$ , which is doesn't satisfy the condition.

Second six numbers are $(24, \ldots , 29 )$ , but factorization them also shows that they don't satisfy the condition for $25 = 5 \times 5$ . And furthermore, it's extremely too young to be a grandmother at this age.

Next is $(31, \ldots, 36 )$ but $4 \mid 37$ isn't true.

$32, \ldots 37$ but $4 \mid 37$ isn't true.

Again:

$(47, \ldots ,52)$ : but $49 = 7^2$

$(48, \ldots ,53)$ : $48 = 6 \times 8, 49 = (6+1) \times 7$ but $(6+2) \mid 50$ is not true

$(61, \ldots ,66)$ Checks out with $61 = 1 \times 61, 62 = (1+1) \times 31, 63 = (1+2) \times 21,$

$64 = (1+3) \times 16, 65 = (1+4) \times 13, 66 = (1+5) \times 11$

Hence, answer is $61 - 1 = \boxed{60}$

We can even check that up to age 119 (which is pretty old if she's still alive), only one solution exist.