Question

I am a 3 digit number. I am 700 rounded to the nearest 100. If You round me to, I round down not up. My tens digit is odd but I am even. Who am I?

Answer 1

712 is the correct answer!

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Answer 2

The numbers satisfying the above conditions are
710, 712, 714, 716, 718, 730, 732, 734, 736, 738.
Since we need to give only one single answer, we shall go for the first lowest number.
∴ The number rounded down to 700 has an odd number at its ten's digit and is an even number = 710.

Given:

• A 3-digit number
• Rounded to 700.
• Rounded down not up.
• Ten's digit odd.
• Even number.

To find: The number

Step by step explanation:
Let the number to be found = x

1. According to the given condition the number is rounded to nearest 100 i.e., 700 and also it is given that the number is rounded down and not up.
   ⇒ From the above we can conclude that the number is greater than 700 and less than 750 as the numbers between this lower and upper bound respectively will round down to nearest 700.
   700 < x < 750.
   
2. Now it is given that the ten's digit of the number is odd.
   In the range 700 - 750 the ten's digit is odd when the numbers are 1 and 3.
   i.e., the numbers in the range 710 - 719 and 730 - 739 will be the required range for ten's digit.
   
3. The last condition says that the number is an even number.
   We know a number is even when its unit digit ends with 0, 2, 4, 6, 8.
   Therefore, the list of numbers in the range 710 - 719 and 730 - 739 whose unit digit ends with 0, 2, 4, 6, 8 will be the required numbers.

So, the numbers satisfying the above conditions are
710, 712, 714, 716, 718, 730, 732, 734, 736, 738.
Since we need to give only one single answer, we shall go for the first lowest number.
∴ The number rounded down to 700 has an odd number at its ten's digit and is an even number = 710.

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