Below, the standard addition algorithm is being used to add three two-digit numbers.

4z+27+x5=y14 42 + 27 +45 = 114 x = 4 , y = 1 , z = 2

4 = z + 7 + 5 mod 10 = z + 2 mod 10, so z must = 2.

14 = 42 + 27 + x5 mod 100 = (x + 7)*10 + 4 mod 100, so x + 7 = 1 mod 10 which means x = 4.

Finally 42 + 27 + 45 = 114, so y = 1.

Now we have x*y*z = 4*1*2 = 8.

–4z

–27

+x5

———

y14

z = 2 because 7+5+z will give 4 only if z = 2. So we have

–42

–27

+x5

———

y14

Similarly only if x = 4 (carrying 1), we can get 1 in tens place in the sum

–4z

–27

+45

———

y14

So finally y = 1

So (x)(y)(z) = 8

Standard Addition Algorithm