self Maths Puzzle

find 3 consecutive odd integers, all less than 100, such that the sum of their squares forms a Unidigit (that is, all its digits are the same). Thus, for integer abc, a^2 + b^2 +c^2 equals a Unidigital.

Read Solution (Total 2)

41,43,45

41^2 + 43^2 + 45^2 = 5555
12 years agoHelpfull: Yes(1) No(2)

If numbers are x,x-2,x+2, then
x^2+(x+2)^2 =(x-2)^2 = x^2+2x^2+8 = 3x^2+8

Now 3x^2+8 should be unidigital.
checking different combinations , we get
x=43.
so numbers are 41,43,45
12 years agoHelpfull: Yes(1) No(2)

self Other Question

A square is cut from the centre of a rectangle. The length of the square's side is exactly half the length of the rectangle's longest side. After cutting out the square, it is found that the rectangle's area has been reduced by exactly 38%. The sides of the rectangle and square are all two-digit integers. Find their values. Find a 6-digit integer abcdef with the following properties:

Take the rightmost digit and place it in front of the integer to form a new number (abcdef --> fabcde).

The new one is the product of the old one and the shifted digit (fabcde = f*abcdef).

What was the original integer?