# Identical digit multiplication

77 is multiplied by another two-digit number with identical digits. What is the product?

77 is multiplied by another two-digit number with identical digits.

The third digit of the product, counting from left to right, is a 3.

What is the product?

*This problem is adapted from the World Mathematics Championships*

**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,

Image

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.

Image

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 =

__9__4

47$\times$3 = 1

__4__1

47$\times$4 = 1

__8__8

47$\times$5 = 2

__3__5

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