Math Is Fun Forum

If the sides of the rectangle are x and y, the perimeter is

The area is

By AM–GM,

In your example, P = 34; work out the upper bound for A from the inequality above.

Bumping this because after two and a half years nobody appears to have a solution yet.

For the related problem with squares instead of rectangles, it turns out that the number of squares visible on an n×n square grid is

For a 3×3 grid in particular, the answer is 14. This actually came up in the final of University Challenge 2009–10, shown on Easter Monday. One idiot contestant actually buzzed with the ridiculous answer 6. I hate idiots!!

SCORE THREE GOALS

Proof:

If n is even, the sum is equal to

If n is odd, evaluate the sum to n−1 (which is even) and add n[sup]2[/sup].
­

IN GENERAL:

Test for Divisibility by a number of the form n = 10b+a, where b is a positive integer and a ∈ {1,3,7,9}

NB: n need not be prime for this to work. It only needs to be coprime with 10 (which such a number necessarily is). We consider the cases for a.

a = 1:
— Subtract b times the last digit from the remaining truncated number.

a = 3:
— Add 3b+1 times the last digit to the remaining truncated number.

a = 7:
— Subtract 3b+2 times the last digit from the remaining truncated number.

a = 9:
— Add b+1 times the last digit to the remaining truncated number.

Examples:
To test for divisibility by 53, add 3(5)+1 = 16 times the last digit to the truncated number. Thus, for 44944: 4494 + 16(4) = 4558; 455 + 16(8) = 583; 58 + 16(3) = 106; as 53∣106, 44944 is divisible by 53.

To test for divisibility by 91, subtract 9 times the last digit from the the truncated number. Thus for 55601: 5560 − 9(1) = 5551; 555 − 9(1) = 546; 54 − 9(6) = 0; as 91∣0, 55601 is divisible by 91.

Note that 91 = 7×13 is not prime. However, 91 is coprime with 10 and so the method works, enabling us to save time by not having to test for 7 and 13 separately.

Right, now we know the trick, let’s use it to devise more divisibility tests!

Test for Divisibility by 17
— Subtract five times the last digit from the remaining truncated number. If the result is divisible by 17, so will be the number.
Example: 12342
Steps:
1234 − 5(2) = 1224
122 − 5(4) = 102
Since 17∣102, the number is divisible by 17.

Test for Divisibility by 19
— Add twice the last digit to the remaining leading truncated number. If the result is divisible by 19, so will the number.
Example: 10203
Steps:
1020 + 2(3) = 1026
102 + 2(6) = 114
11 + 2(4) = 19
Since 19∣19, the number is divisible by 19.

It’s an octahedron with vertices at the points (±1,0,0), (0,±1,0), (0,0,±1).

In other words, A takes 12 hours to paint the house, B takes 6 hours to paint the house, and C takes 24 hours to remove all the paint from the house.

Of course you can write ƒ(x) = 0. But you also have to specify that the domain is

and not

. Whoever said you can’t probably meant that you can’t just write

and leave it there, but also have to indicate that the function is not defined at 0.

The answer should be IBERIAN. That’s SIBERIAN or LIBERIAN minus the first letter.