As I must’ve said by now, I enjoy playing around with numbers, calculators and the like. And as I also must’ve made clear by now, I am fascinated with coïncidences.
Anyway as you will see, the two are related, for this thread at least.
The year 1976 is perfectly divisible by both 13 and 76. Does this mean anything? Probably not. It is ironic that it was the Bicentennial. And there were 13 original colonies. But outside of that, I can’t think of anything special about the year. President Ford was in office. I don’t remember that, although technically I was there. But I think the year otherwise was uneventful.
Anyhow, that has nothing to do with my question. I was however recently wondering what other numbers are perfectly divisible by 13 and 76, and how you could find this out.
I vaguely remember something like this as a story problem in HS. It probably involves algebra. And it may involve Cartesian coördinates too:confused:.
I’ll tell you where I stand now on the problem. I took y=13x and y=76x. Then I thought I should add them together, getting 13x=76x. But that makes little sense. Also I should tell you I am looking for whole numbers FWIW. Although an equation that gives fractions would work too I suppose.
An obvious example is, of course, 1376 = 988, and all multiples of it (of which 1776 is one: 9882).
As for an algorithm to find others: 13 is a prime, and it does not feature in the factorisation of 76, so 13 and 76 are coprimes. I’m not sufficiently strong in number theory to find a formal proof whether there are numbers other than multiples of 988 which are divisible by both 13 and 76, but the coprimality makes me suspect that this is not the case.
Actually some elementary mathematics suffices to prove this.
Suppose two numbers are coprime, e.g., x = 13 and y = 76. Their greatest common divisor is 1, so write 1 = Ax + By (exact values for A and B are not important so no need to calculate them). Then, if n is divisible by x and y, say n = yq, we see that q = Axq + Byq is divisible by x. In other words, n is divisible by the product xy, q.e.d.
So all the integers divisible by both 13 and 76 are just of the form 988k where k is an arbitrary integer.
There must be a typo there somewhere, because if Y and Z are coprime then Z cannot be divisible by XY ( these are supposed to be arbitrary integers). However, if Y and Z are coprime and Z divides XY, then Z divides X.
nb it is not really necessary to consider prime numbers here, for which if P divides XY then P divides X or P divides Y. Maybe you could, but that is more complicated than what is going on here.
Sure, your (neat) proof also works for any other pair of numbers which are coprimes; they don’t need to be primes themselves. It’s just that in the present case, the fact that one of the two numbers is itself a prime, and not a divisor of the other, makes it evident that the two must be coprimes.
Jim B., let’s go back to the equations in your OP. You have a number y. It is 13 times some other number, which we will call x. So y = 13x. You also know that y is 76 times some other number (which is not the same as x), which we will call z. So y = 76z. Then you can combine those two equations. Note that you are not adding those two equations. Rather, you are combining them. When you combine them, you say that y = 13x and y = 76z, so because the right half of each equation equals y, it must be true that the two right halves equal each other. So 13x = 76z. Notice that 76 = 2 * 2 * 19. So 13x = 2 * 2 * 19z. Both sides equal y. So y must be divisible by 2 * 2. Also, y must be divisible by 13. And, finally, y must be divisible by 19. Those are all prime numbers. So y must be divisible by 2 * 2 * 13 * 19. So there must be a number w so that y = 2 * 2 * 13 * 19w. This means y = 2 * 2 * 13 * 19w. But that just says that y = 13 * 19w, which means that y = 988z. z can equal any integer, so there are an infinite number of values for y.
The numerology in the OP seems like it is motivated by 1976 = 2 × 13 × 76. But, instead of merely investigating multiples of 988, you could relax the 13 and consider other multiples of 76, e.g., 101 × 76 = 7676. Or keep the 13 and change the 76… there are a lot of things you might try
I don’t know if it’s still taught in primary school, but we’re essentially discussing the concept of a Least Common Multiple.
Once you have the least common multiple, all multiples of the LCM satisfy requirements. And as noted above, the LCM for two coprime numbers (as 13 and 76 are) is their product.
More interestingly, we’re diving into the Fundamental Theorem of Arithmetic, which appears trivially easy to a lot of people but has deep implications.
Wendell and Antibob have it right.
76 factors to primes 2x2x19
13 has no prime factors other than 13.
So any number divisible by 76 and 13 must have factors 2x2x19x13 - essentially multiples of 76x19
There will only be numbers divisible by both and NOT a multiple of their product if there are common factors.
(i.e. AxB where say, A and B share a common factor:say 10 and 25 - factors 2 or 5, a 5 is shared.
Then numbers divisible by both A and B will be multiples of the LCM - 50
10=2x5 and 25=5x5 so 2x5x5 is divisible by both - i.e. 50 - because it includes factors 2x5 and 5x5)
Thank you for re-explaining it in a way that makes most sense to me. I found DPRK’s proof a little hard to follow - which is no fault of theirs, it simply reflects why I gave up formal study of mathematics after failing the first year of my Bachelor’s degree - that (and this) was roughly the point when the concepts involved stopped clicking with me. But I know just about enough to enjoy these threads, if not contribute meaningfully to them.
