What is the sum of all 4 digit numbers which are = to an integral multiple of the product of their digits?

Question

I'm timing this....

i.e., if a number x has digits a,b,c,d, then:
x = 1000*a + 100*b + 10*c + d
what is the sum of all 4 digit numbers x such that x = n*a*b*c*d where n is some integer?

I would really like to know why anyone voted thumbs down to this question. Please email me.

note to: I ♥ AUG
the product of the digits of 1250 is 0.

nice 1 pascal...
what language?

we have a winner...
but to all you programmers: give it a shot anyway.
.

nice compact program Michael M

Answer 1

The numbers themselves are:

1111, 1112, 1113, 1115, 1116, 1131, 1176, 1184, 1197, 1212, 1296, 1311, 1332, 1344, 1416, 1575, 1715, 2112, 2144, 2232, 2916, 3111, 3132, 3168, 3171, 3276, 3312, 3915, 4112, 4224, 4416, 6144, 6624, 6912, 7112, 7119, 8112, 8832, 9315, and 9612

Their sum is 136,479.

Edit: a computer algebra system called maxima. It's open source - the website is http://maxima.sourceforge.net/ . It's fairly nice, but systems like Mathematica and Maple are much better. The main advantage of this one is the price :p

Edit 2: Thanks. Just in case you're interested, here's the actual commands I used:

f(x):=1000*x[1]+100*x[2]+10*x[3]+x[4];
g(x):=if (mod(f(x), x[1]*x[2]*x[3]*x[4])=0) then x else [1,1,1,1];
q:{1,2,3,4,5,6,7,8,9};
s: map (f, map (g, cartesian_product(q,q,q,q)));
sum (listify (s)[i], i, 1, cardinality (s));

The output of step 4 gives you the numbers, and step 5 sums them.

Edit 3: Actually, now that I look at it, it would have been slightly more eficient to define g(x):=if (mod(f(x), x[1]*x[2]*x[3]*x[4])=0) then f(x) else 1111; and then use s: map (g, cartesian_product(q,q,q,q)); in line 4.

Answer 2

10

Answer 3

haha re your additional comments; that cracked me up! :D

What do you mean by "an integral multiple of the product of their digits"? Do you mean if I had a number like 1250, its product is 10, and since 1250 = 10*125?

If so, then YIKES! I think that problem would take too long! Or, at least it would take me too long this late at night!

Good luck, kid, but hey thanks for making me have to think there for a few minutes! :-)

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Oops! I guess the product of 1250 is 0 not 10! I don't think you want me helping you tonight; my brain is clearly not functioning properly! I'll be thinking about this problem though, it's a fun one for sure!

Answer 4

enable P(a, b) = ? x for all a < x < b Lemma a million: by utilising the undemanding regulations of Set concept, for any valuable series a < b < c, P(a, c) = P(a, b) * b * P(b, c). All of my arguments carry on with from this assumption. ----------- (2) ----------- P(0, 2) = P(0, a million/2) * a million/2 * P(a million/2, a million) * a million * P(a million, 2) for each x, a million/2 < x < a million, there exists precisely one counter-area a million/x, such that a million < a million/x < 2. for the reason that this could be a bijection, we are able to precise P(a million/2, a million) * P(a million, 2) as: ? (a million < x < 2) x * (a million/x) = ? (a million < x < 2) (a million) = a million by utilising substitution, P(0, 2) = a million/2 * P(0, a million/2) P(0, a million/2) procedures 0, so P(0, 2) procedures 0 I justify P(0, a million/2) = 0 by utilising the reality that P(0, a million/2) is a limiteless manufactured from components all <= a million/2. So, for any valuable integer n, P(0, a million/2) <= (a million/2)^n. (a million/2)^n -> 0 as n -> infiinity, so P(0, a million/2) -> 0 ---------- (3) ---------- i'm at a loss for words as to why you think of this concern is unsuitable - what's incorrect with it? The "Riemann" approach works: enable f(0) = a million, f(x) = x * f(x - a million/n), n -> infinity f(ok/n) = ok! / ok^n, f(2) = (2n)! / (2n^2n) -> 0 as n -> infinity additionally, how can we exhibit this actual countless product as a sum? not sure the thank you to try this. @Scythian: you have hit the mark - the finished complicated crux of this concern. the ingredient is, you are able to map any era (a, b) onto (c, d) utilising the bijection f(x) = c + (b-x)*((c-d)/(b-a)). in actuality, we are able to map any finite era (a, b) onto R = (-infinity, +infinity) by utilising the bijection: f(x) = 0 whilst x = (a + b) / 2, a million / (2x - a - b) in any different case Does this make the priority unsolvable? i think of not - i think of it in simple terms ability we could desire to consistently be careful how we interpret the "countability" of countless units. it quite is organic for us human beings, dwelling in a Cartesian, Euclidean international, to anticipate that linearity is a valid and inevitably suitable mapping for any set of numbers. although, that may not the case, as is shown interior out by utilising rigorous learn of Set concept. i can not pretend to be an authority in this field, yet i've got heard adequate from experienced mathematicians to nicely known the basics.

Answer 5

Pascal is right. 136479.