When 16 is subtracted from a three-digit number, and the resulting difference is divided by 2, the result is a three-digit number whose digits are those of the original number, but in reverse order. The sum of the three digits is 20. Find the original number.
\(\frac{100x+10y+z-16}{2}=100z+10y+x \qquad and \qquad x+y+z=20\\~\\ \)
Where x,y and z are single digit non-negative integers. And x and z cannot be 0
Looking at
\(\frac{100x+10y+z-16}{2}=50x+5y+\frac{z}{2}-8\)
Z must be 2,4,6 or 8
Z, 10+Z | 2,12 | 4,14 | 6,16 | 8,18 |
z/2, (10+Z)/2 | 1,6 | 2,7 | 3,8 | 4,9 |
z/2-8 (last digit) | 3,8 | 4,9 | 5,0 | 6,1 |
X | 3 or 8 | 4 or 9 | 5 or 0 | 6 or 1 |
x+y+z=20 therefor Y= | no soln | 7 (using the 9) | 9 (using the 5) | 6(using the 6) |
possible initial numbers | 974 | 596 | 668 |
So my possibilities so far are
974 and 596 abd 668
do they work though?
the digits add to 20, that is good.
Try 974
(974-16)/2 = 479 The digits are reversed so that is perfect
Try 596
(596-16)/2 = 290 that is no good
Try 668
(668-16)/2 = 326 that is no good either.
So the only original number that works is 974
Coding:
\frac{100x+10y+z-16}{2}=100z+10y+x \qquad and \qquad x+y+z=20\\~\\