Brunch with DigitsPro Problems > Math > Logic > Puzzles
Brunch with Digits
Three distinct digits got together for brunch, and discussed their various relationships. "Do you realize," one of them said, "that if you multiply us together, you get a multiple of three?"
"That's nothing," said another. "If you turn us into a three digit number, that's a multiple of three also!"
The third digit took a long sip of tea while pondering this news, then announced, "Depending on how you arrange us, we would form a three digit number that's also a multiple of four!"
"But never a multiple of five," the first digit countered through a mouthful of raspberry danish.
The second digit looked at his friends and said, "I know we're sitting evenly spaced around the table now, but did you realize that if we were seated on a number line, we would still be evenly spaced?"
"Lines, circles, big deal!" exclaimed the third number. "I want to talk about triangles! Did you know that two of us as a two digit number can form a triangular number?"
"Don't be such a square," retorted the first digit, to which the third digit replied, "I'm not! I'm a cube!"
"Friends! Don't argue," exclaimed the second number. "If you can just get along for a moment, I'll go pay the bill." He paid with a twenty dollar bill, and his change was the largest value (in dollars and cents) that could be formed with the three digits.
How much money did the friends spend on brunch?
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I am a three digit number, and the sum of my digits is 13. My first and last digits differ by 1. The product of my last two digits is less than 20 but bigger than 10. I have no repeated digits. What numbers could I be?
All the integers from zero to nine are contestants on the new reality TV show, "The Digital Age." Every week a new problem is posed, and any digit which cannot "solve" the problem is booted off the show. Below are the questions. Which digit will win?
I am a side of a triangle, and my other two side lengths are 4 and 5.
I am not the last digit in any perfect square.
I am not a solution of x2 - 14x + 48 = 0.
If all the digits still in the game are multiplied together, I am not a digit of the result.
If all the digits which are no longer in the game are added together, I am a digit of the result.
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