Alan Rench sent in this problem from a high school algebra book from 1915:

There is a number consisting of three digits, those in the tens' and units' places being the same.
The digit in the hundreds' place is 4 times that in the units' place.
If the order of the digits is reversed, the number is decreased by 594.

What is the number?



Solution to the Problem:


The original number is 822.

Let x = units digit.
Then x = tens digit.
Let y = hundreds digit.

The original number is 100y + 10x + x.
The number with digits reversed is equal to 100x + 10x + y.

So we can write two equations:
y = 4x
(100x + 10x + y) = (100y + 10x + x) - 594

Simplifying the second equation and then dividing by 99:
99x - 99y = - 594
x - y = -6

Substitute y = 4x to get:
-3x = -6
So x = 2 and y = 8.

The two numbers are 822 and 228 and the difference is 594.


Correctly solved by:

1. Ritwik Chaudhuri Santiniketan, West Bengal, India
2. Veena Mg Bangalore, Karnataka, India
3. Colin (Yowie) Bowey Beechworth, Victoria, Australia
4. Davit Banana Istanbul, Turkey
5. Jonathan Punke Bremen High School,
Midlothian, Illinois
6. Ivy Joseph Pune, Maharashtra, India
7. Dr. Hari Kishan D.N. College,
Meerut, Uttar Pradesh, India
8. Mohamed Sheriff (MEDDORA) Freetown, Western area
Sierra Leone, West Africa