1 and 9 are both perfect squares that each consist of only odd digits. Are there any other perfect squares consisting of only odd digits?
For 1-digit numbers it is obvious. Lets try 2-digit numbers, ab=10a+b, ab^2= 100a^2+20a*b+b^2=20k+b^2, for any k>0, it will have even digit (s), for k=0 we have only the given numbers- 1 & 3. There are no more perfect squares consisting only odd digits.
144 = 12^2 1444 = 38^2 Are there any other numbers 144..44 (starting with 144 and ending in 44) that are perfect squares?
There is a 6 digit-number consisting of the digits: 1, 1, 2, 2, 3, and 3.
The ones are 1 digit apart, the twos are 2 digits apart, and the threes are 3 digits apart.
What is the number?
Find a way to write the digits 1 through 9 in sequence, so that the numbers determined by any two consecutive digits are divisible either by 7 or by 13.
You are provided with a grid (as shown in the picture). Can you fill the squares with numbers 1-8 in a manner that none of the two consecutive numbers are placed next to each other in any direction (vertically, horizontally or diagonally?)
