Math CentralQuandaries & Queries


Question from Con, a parent:


My name is Con and my son is required to answer the following questions for his maths class.

He has attempted Q1 through trial and error and has found the answer to 72453. Is this correct?

He has attempted to draw the triangles described in Q2 in a number of ways and has found that BE can not equal ED and is dependent of angle BAC. Therefore, he claims that the triangle can not be drawn/practical. Is this correct or is there a slolution?

Digits 2, 3, 4, 5 and 7 are each used once to compose a 5-digit number abcde such that 4 divides a 3-digit number abc, 5 divides a 3-digit number bcd and 3 divides a 3-digit number cde. Find the 5-digit number abcde.

Let ABC be a triangle with AB=AC. D is a point on AC such that BC=BD. E is a point on AB such that BE = ED = AD. Find the size of the angle EAD.

Your assistance is greatly appreciated.

Kind Regards.



For Q2, don't give up so easily. There is indeed a triangle ABC with D between A and C and E between A and B that satisfies all the conditions. The only tools needed to find that triangle are (1) the sum of the angles of a triangle equals 180 degrees, and (2) the base angles of an isosceles triangle are equal (where an isosceles triangle has two sides equal). If you are not good at algebra, guess either the angle at A or at C, then fill in all the other angles; then adjust your guess up or down as needed.

For Q1 there are simple divisibility rules for 3, 4, and 5.
A number is divisible by 5 if and only if its final digit is 0 or 5.
A number is divisible by 3 if and only if the sum of its digits is divisible by 3 (in your example 4+5+3 = 12, which is divisible by 3).
A number is divisible by 4 if and only if the number formed by its final two digits is divisible by 4 (in your example, 24 is divisible by 4 thus so is 724; so is 472, but then 2+5+3 = 10 and the final condition would fail to be satisfied).

It is easy to see that your number is correct, and almost as easy to check whether or not it is the only correct answer


Con wrote back


Is the answer to the 50 deg?



You're close, which is good when playing horseshoes, but not so good when playing math. The only way to see what's wrong is to check the angles: Draw a picture while you follow my computations.

If you want the angle at A to be 50 degrees then

in triangle AED, E would be 50 and D would be 80. (We're using angle A = angle E because DA = DE; moreover, the sum of all three angles must be 180.) Next,

in triangle EBD (with ED = EB) E is 130 while B = D = 25.

Next, the angles at D must sum to 180; we already have 80 and 25, so the angle at D in triangle DBC must be 180 - 105 = 75.

Next, we use AB = AC to get the angle at C must be half of 180 - 50; that is, the angle at C must be 65.

Now we're in trouble: We want BD = BC, whence in triangle BCD the angles at C and D should be equal; unfortunately one is 65 while the other is 75.

We conclude that the guess that A = 50 was TOO BIG. Why not try A = 45 and go through the calculation once more.


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