- yes its 0
say:
N =111
M =111
It’s 99 (which happens to divide 0; also, the question-writer was unduly skittish about digits being equal to zero).
Let the original number be A * 100 + B * 10 + C. Its reverse is then C * 100 + B * 10 + A. The difference will be (A - C) * 99. (And taking its absolute value won’t change what divides it). So 99 always divides the difference, and if |A - C| = 1, nothing greater will.
Oops I missed that the question mentions “positive difference” , so the difference cant be zero. Hence A cant be same as C:smack:.
You are right.
A CAN be the same as C, and the answer would still be 99, in the sense that 99 divides 0. The question is distractingly, unnecessarily paranoid in its wording.
This is the correct answer, but it’s not clear to me that we can get away with only counting factors of 7 (and not the other prime factors of 7!) just because it’s the largest prime from 1 to 7.
For example, 7 is also the largest prime from 1 to 10. But 10![sup]34[/sup] doesn’t divide 213!. The largest power of 10! which divides 213! is only 10![sup]25[/sup].
Right, I know. Since 7 was a smaller number I took smallest way for explaining.
Otherwise one would need to count factors of every prime number. like in 10! :
2’s factor comes 8 times
3’s factor comes 4 times
5’s factor comes 2 times
7’s factor comes once.
Count factors in 213! for each of these
For 2:
213/2+213/4+213/8+213/16+213/32+213/64 + 213/128
Divide it by 8. Call it a
For 3
213/3+213/9+213/27+213/81
Divide it by 4. Call it b
For 5
213/5+213/25+213/125
Divide it by 2. Call it c.
For 7
213/7+213/49
Divide it by 1. Call it d.
The minimum of a,b,c,d is the answer. Turns out b=c=25 is the minimum here.
Could there be another shorter procedure??
Right, that’s what I would’ve done (and I would’ve done the same thing for the multiple prime factors of 7!). But, like you, I am curious if there is a more efficient, shorter procedure.
For example, I know it is always the case that a! * b! * … * z! divides k! whenever k >= a + b + … + z. [Since (a + b + … + z)!/(a! * b! * … * z!) has a combinatoric interpretation as the number of ways to arrange in sequence a set comprised of a indistinguishable items of one color, b items of another color, etc.]. From this, we know that b![sup]n[/sup] always divides (nb)!, without having to tediously check all the prime factors; thus, we immediately know that 7![sup]30[/sup] and 10![sup]21[/sup] divide 213!. And, of course, by checking just the largest prime factor, we obtain upper bounds on the exponent. But tightening to the precise cutoff, I don’t yet know how to do efficiently.
Good job.
Here’s what I could understand till now:
-
For x! when x is prime, there is only need to check factors of x and that is the anwser. For ex. in 13! the anwser is simply 213/13+213/169
-
For x! when x is non-prime, find the largest prime smaller than x.
For ex. in 33! it is 31, one needs to check for 31 i.e. 213/31. Now go to: 32=2[sup]5[/sup] and 33=3*11. In this case one needs to only check for 2,3,11,31 and no other prime factors.
I might now look at this topic tomorrow only as it is 1:30 am here:).
I’m not sure I understand your #2; are you saying that, if p is the largest prime <= x, you only need to check the prime factors of p, p + 1, …, x? Why is that? (Indeed, even for the special case of #1, which I find very plausible, I’d like the assurance of a proof)
Yes, you understood #2 correctly.
#1 can be concluded if one visualizes 213! as blocks of x. For ex., for 13!, as blocks 1 to 13, 14 to 26 and so on.
#2 can be arrived upon using #1 and the method we used in post#46.
I’m sorry, I need more detail than that. Visualizing 213! as 30 blocks of 7 plus a little left over makes it clear that 7! divides 213! at least 30 times, since any block of 7 consecutive numbers has a product divisible by 7! (since their ratio is a binomial coefficient). But how does this visualization help me conclude the stronger result that the highest power of 7 dividing 210! is also the highest power of 7! dividing 210!?
Take 13!, For every prime from 2 to 13, one got to find the ratios of factors in 213! to factors in 13!. The minimum of all ratios is your answer.
If one visualizes 1 to 13 , then 1 to 26 , then 1 to 39 and keeping on incrementing 13 , one can easily see that the desired ratio for any prime can not be less than the ratio for 13. Hence there’s no need to check for primes other than 13. This will work for all x! when x is prime.
Start with a simple problem and seek a formula. Assume only 3 beads, red white and blue. The possibilities are RWB RBW, and turning it upside down, there is only one. Which is 2!/2, or (n-1)!/2, but one can be expressed in a lot of other algebraic permutations using the same digits, too. so let’s test it on a higher number.
So let’s use 4. RWBY. RWYB. RBWY. RBYW. RYWB. RYBW. That’s 6 = 3!. But RWYB is the same as RBYW, so each triplet after R is the same in both directions, so it’s only 3, which is 3!/2. Or, (n-1)!/2. Each sequence is the same, in an infinite repetition, regardless of where you start, so WBYR is already covered.
From that analogy, there should be 7!/2, which is 2520.
As for the definition, it is presumed that “eight different colored beads” means exactly 8 beads, each of which is a different color, and not “an unknown number of beads that are eight different colors, some of them multicolored and some of them identical to each other” or various other possible syntactical possibilities. The problem is obviously insoluble, unless we assume that we’re talking about exactly eight beads, each one a different color, and if anything else is implied, there is insufficient data to even begin thinking about the problem.
OK new one.
If all numbers in the expression below are in base 4 then what is the base 4 value of this expression:
2^11 + 3^10 + 2^23 + 11^10
These are really tough considering it’s for middle schoolers.
Can’t you at least take a crack at it yourself first?
Not having fun with these?
That’s not a problem that requires any cleverness. Just convert everything to base 10 and go.
This is just grinding through arithmetic. Nothing fun here.
I’ll take some minor shortcuts, but they’re not in any way essential.
2^11 = (2^2)^2 * 2 = 10^2 * 2 = 200
3^10 = ((10 - 1)^2)^2 = (101 - 20)^2 = 10201 + 1000 - 10100 = 1101
2^23 = (2^11)^2 * 2 = 200^2 * 2 = 200000
11^10 = (11^2)^2 = (121)^2 = 21301
200 + 1101 + 200000 + 21301 = 22302
Actually, there are fairly nice patterns for all the exponentiations here, which I should have taken advantage of (and to some extent did):
The powers of 2 are 1, 2, 10, 20, 100, 200, etc.
The powers of 11 are 1, 11, 121, 1331, 14641 [which, after converting the digits greater than 10, is 21301], etc. The rows of Pascal’s triangle.
The powers of 3 are the powers of 1(-1): 1, 1(-1), 1(-2)1, 1(-3)3(-1), 1(-4)6(-4)1 [= 1101], etc. The rows of Pascal’s triangle with alternating sign on the digits.
And, of course, one could take advantage of the cancellation in (10 - 1)^n + (10 + 1)^n: 14641 + 1(-4)6(-4)1 = 2 * 10601 = 23002. It’s then easy to add that to the powers of 2 to get 223202. [There was a typo above when I wrote 22302]