Math Is Fun Forum

Let

be a three-digit number.

Then

.

If

, we write

.

Then the sum of digits is

.

If

, we have

.

If

, the sum of digits is

.

If

, then

and the sum of digits is

.

So the sums are

(i) 28 if

(ii) 19 if

and

(iii) 10 if

and

concubinage

pelargonium

preannouncing

Yeah, as I said I made a mistake.

Let

,

,

, where

are positive integers (so

themselves are not necessarily integers). This gives

which is what is required to be solved.

I haven't gone very far, have I?

Let the sides be

. The perimeter is

and the area is

by Heron's formula.

Hence

Setting

:

Since

are sides of a triangle, we can make the substitution

,

,

, which gives

Thus

are integers, as

are integers; they are also positive as the sum of two sides of a triangle must be strictly greater than the third side. Substituting in the above equation:

So the RHS is divisible by 4. This means either (i) two of

equal 2, or (ii) one of

equals 4. In case (i) we have (taking

WLOG)

, an absurdity. So case (ii) must hold. WLOG let

. Then

Thus

must be an integer

or

, which gives

or

.

Substituting back gives

or

. Hence the only triangle with equal area and perimeter is the one with sides 6, 8, 10.

The last two digits of

are the last two digits of

. Hence the ten's digit of

(call it N) is the last digit

plus any carry over from

. If

,

is a single-digit number so there is no carry-over: N is just the last digit of

. If

,

so the carry-over is 2: N is the last digit of

.

It should be

Agnishom is seeing this image:

That was what I initially saw as well. That's because neither of us had visited the site http://a1star.com/ first. Once I went to the site, I didn't see the remote-linking error image any more.

Some sites which restrict remote linking work like this. The problem is that you can see the correct image because you've visited the site, but others who haven't visited the site see a different image. They have to visit the site in order to see the correct image.

You're welcome.

Suppose the number is

NB:
(i) Since the number must be even, I write

for the last digit;

.

(ii)

and

cannot be zero but the other

can be 0.

So we want

i.e.

As the LHS is divisible by 19, so must the RHS, and as

we must have

divisible by 19. The smallest such n is

– in other words, the smallest solution is an 18-digit number.

PS: Yes, there are infinitely many solutions, because there are infinitely many n such that

is divisible by 19.

What derivatives?

Suppose the centre of the rotation is (a,b). This rotation takes the point (2,−1) to (0,3). Then the same rotation about the origin would take the point (2−a,−1−b) to (−a,3−b). (And the other points correspondingly, but we only need one point to work out a and b.)

The matrix representing a clockwise rotation of 90° about the origin is

Hence

Solving the simultaneous equations gives

.

(a) is false. For example

is a non-cyclic Abelian group of order 27. It' s not cyclic because every non-identity element has order 3.

Suppose at time t the dog is running at angle θ to the cat's direction of motion. The component of the cat's velocity parallel to that of the dog is ucosθ so the velocity of the dog relative to that of the cat is

where s the separation between the animals at time t. (Note that the RHS is positive as

.) Suppose the dog catches the cat after time T. Integrating gives

Now, the component of the dog's velocity parallel to that of the cat is vcosθ, so when the dog catches the cat we have

Substituting the integral into the previous equation gives