77$\times$11 = 770 + 77 = 847, which does not have a third digit of 3

77$\times$22 = 847 + 847 = 1694, which does not have a third digit of 3

77$\times$33 = 1694 + 847 = 2541, which does not have a third digit of 3

77$\times$44 = 2541 + 847 = 3388, which does not have a third digit of 3

77$\times$55 = 3388 + 847 = 4235, which does have a third digit of 3

So the product was 4235.

Writing out the mutliplication like this,

All of the mutliplications will be $7\times k$, plus carried digits and the $0$ as shown below. The third digit in the product will come from the tens digit of the pink box below added to the units digts in the green box.

So the last digit of the number which is the tens digit of $7k$ plus twice the units digit of $7k$ is a $3$.

If $k=1$, then $7k=7$, and the tens digit plus twice the units digit is $0+14=14$ which has last digit 4.

If $k=2$, then $7k=14$, and the tens digit plus twice the units digit is $1+8=9$ which has last digit 9.

If $k=3$, then $7k=21$, and the tens digit plus twice the units digit is $2+2=4$ which has last digit 4.

If $k=4$, then $7k=28$, and the tens digit plus twice the units digit is $2+16=18$ which has last digit 8.

If $k=5$, then $7k=35$, and the tens digit plus twice the units digit is $3+10=13$ which has last digit 3.

So $k=5$, and the product is $4235$.

Two-digit numbers with identical digits are precisely the multiples of 11. So 77 has been multiplied by 11$n$ for some value of $n$.

77$\times$11$n$ = (77$\times$11)$n$ = 847$n$. So 847$n$ = _ _ 3 _ .

The 800 will not contribute to this digit, so we can consider the 47 times table.

47$\times$2 =

47$\times$3 = 1

47$\times$4 = 1

47$\times$5 = 2

So $n$ = 5, and the product is 847$\times$5 = 4235.

You can find more short problems, arranged by curriculum topic, in our short problems collection.