The sum of 128 and the square of its tens digit is 24 times its tens digit.
Is that really what you meant? With no mention of the units digit?
128+tē = 24t
tē-24t+128 = 0
(t-8)(t-16) = 0
t-8 = 0; t-16 = 0
t = 8; t = 16
So the tens digit is 8. it could be any number with the tens digit 8.
It could be 80, 87, 286, 6784, or any number that has 8 for its next to
the last digit. Are you sure you copied the problem right?
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Maybe you meant this:
The sum of 128 and the square of its tens digit is 24 times its UNITS digit.
Then 128+tē = 24u
tē = 24u-128
The smallest t could be is 0 and the largest it could be is 9
So:
0 ≦ t ≦ 9
so
0 ≦ tē ≦ 81
0 ≦ 24u-128 ≦ 81
128 ≦ 24u ≦ 209
5.3 < u < 8.7
So u can only be 6, 7 or 8
If u = 6, then tē = 24u-128 = 24(6)-128=144-128 = 16, and t=4.
So one solution is 46.
If u = 7, then tē = 24u-128 = 24(7)-128=168-128 = 40, which has
no integer square root. So we discard u=7.
If u = 8, then tē = 24u-128 = 24(8)-128=192-128 = 64, and t=8.
So the only other solution is 88.
If the above was what you mean, the two solutions are 46 and 88.
Edwin