Trying out numbers

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.

Multiplying 77 by a number written $kk$

Writing out the mutliplication like this,

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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.

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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 $4235$.

Using the 11 times table

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 = 94

47$\times$3 = 141

47$\times$4 = 188

47$\times$5 = 235

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