[Guest]

Show that the square of any integer leaves a remainder of 0, 1, 4 or 7 when divided by 9.

Use this to establish the following condition that a number which is a perfect square must satisfy the following:

For a number that is a perfect square, add up its digits to form a second number. if that number has more than one digit, add up its digits to form a third number. continue until you obtain a single digit number. that final number must be 1, 4, 7 or 9.

good luck i certainly haven't had any with it get.

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[Guest]

Show that the square of any integer leaves a remainder of 0, 1, 4 or 7 when divided by 9.

Use this to establish the following condition that a number which is a perfect square must satisfy the following:

For a number that is a perfect square, add up its digits to form a second number. if that number has more than one digit, add up its digits to form a third number. continue until you obtain a single digit number. that final number must be 1, 4, 7 or 9.

good luck i certainly haven't had any with it get.

use spoilers

show working

do the remainders have to be whole numbered remainders or can they be the same digit repeating?

im no good at maths....but im trying

[Prime]

1) Any number x > 9 may be expressed as x = 9*k + r, where r < 9 (smallest integer remainder from division by 9) and k is some integer.

x² = 81k² + 18kr + r². As evident, the r² determines the remainder from division by 9 for x². (All other terms are divisible by 9). Therefore the remainder from division by 9 of any square is going to be one of the remainders from squares of 0 through 9.

Furthermore, we can establish that the remainder for 5² is the same as for 4², 6² same as 3², and so on. (For that we could use the same trick with x = 9 - r. And then notice that 4 = 9 - 5, 6 = 9 - 3, and so on.) Thus to find all possible remainders we only have to check for remainders of the squares of 0, 1, 2, 3, and 4. Which are: 0, 1, 4, 0, and 7.

2) There is that old divisibility by 9 check, used by schoolchildren all over the world. It states that in decimal representation, the sum of digits of a number is its remainder from division by 9 (or 3). It is easy to see why. An n-digit number written in decimal representation is actually:

d_n-1*10^n-1 + d_n-2*10^n-2 + ... + d₀*10⁰. Where each "d" is a digit. If we substituted each "10" by (9+1) in that exp​ression, raised each to the requested power and multiplied by corresponding d, we'd find that all terms are divisible by 9, with possible exception of each individual "d".

[Prime]

A side note.

The same rule holds for any number N, not just 9. That is the remainder from the division of any square by N is one of the remainders from division of squares of 0 through [N/2] (integer part of N/2) by N.

E.g. remainder of division of any square by 5 would be 0, 1, or 4. But never 2, or 3.