April 2022
Problem of the Month

Digits Problem from 1915
Submitted by Alan Rench

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:

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

Send any comments or questions to: David Pleacher