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JulsSmile [24]
3 years ago
14

Find a 3 digit number with all these properties: all 3 digits are different, the 1st digit is the square of the second digit in

the 3rd digit on one more than twice the second digit
Mathematics
1 answer:
DanielleElmas [232]3 years ago
8 0

Answer:

425 or 937

Step-by-step explanation:

First, I listed out all the possible numbers for the first digits, namely, the squares under 10.

1*1=1

2*2=4

3*3=9

Since all the digits have to be different, the first digit cannot be 1 because 1 squared is 1. So that leaves us 4 and 9 to work with, which I tried out one at a time.

Starting with 4:

2 squared is 4.

42

2 times 2 plus 1 equals 5.

425

Starting with 9:

3 squared is 9.

93

2 times 3 plus 1 equals 7.

937

So here we have two numbers that both work and meet the requirements (unless I understood the problem wrong at the part where it says "...the second digit in the 3rd digit on one more than twice the second digit")!

I hope this helped! :D

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