Elementary number quiz.
 

:smile:
Elementary number quiz.
1) (a) 7 does not divide into 9 or into 99 or into 999. What no.s (abbr. numbers) made up entirely of 9’s can be divided by 7 with no remainder?
What is the result of each division?
(b) which no.s ,composed of 9’s can be divided exactly by 13? By 17, by 23? By 24?
(c )Which no.s made up of 4’s can be divided exactly by 7?

Quiz 2: (a) Which no.s less than 100 have exactly 3 factors including themselves and 1
(b)What no. less than 100 has the largest no. of factors?

Quiz 3 (a) If you double 6 then double it again and again and again like this 6, 12, 24, 48, 96, 192 you will never get to a no. which is a perfect square and explain why?
(b)Many no.s can be written as the diff. of two perfect squares. For example 5=9-4, 20=36-16, 21=25-4.
Which no.s cannot be written as the diff. of two squares?

Have a pleasant evening and refrain from spoilers.
[ref: David Wells]
Mally :coffee:

For 1a, [spoiler](10^(6*n)-1)/7 = "142857" x n
(that is, that string concatenated n times.)[/spoiler]
I haven't tried 1b, 1c, or 3b yet.
But my guess for 1b is that [spoiler]something similar happens for 13, 17,
and 23, but 24 seems to get stuck on 15.[/spoiler]

[quote]Quiz 2: (a) Which no.s less than 100 have exactly 3 factors including themselves and 1
(b)What no. less than 100 has the largest no. of factors?[/quote]Do you mean all factors, unique factors, prime factors, or unique prime factors?

Obviously, the answers to both parts of 2 would seem to depend on this question.

In the first case, I'm pretty sure the answer to 2a is
[spoiler]zero, since two numbers must multiply to get the number[/spoiler].

In the unique factors case for 2a, I'm guessing the answer is
[spoiler]squares of primes, namely 4, 9, 25, and 49[/spoiler].
But clearly, in the unique prime factors case,
[spoiler]this isn't all of them, since 16 = 2^4 is another answer.[/spoiler].

For 3a, these numbers [spoiler]are of the form 3*2^n. So the number any
would be a perfect square of would be sqrt(3)*2^(n/2). sqrt(3) is irrational,
so none of these numbers will be a perfect square of an integer.[/spoiler]

Elementary number quiz.
 

:rolleyes:
Good going so far Ken g6 and please keep it up.

Regards quiz 1(a) Please note the property of the remainder i.e. 142857.Its the value of 1/7 with decimal repeating i.e.142857 142857......
Discarding the decimal the multiplication of it by any number from 1 to 6 is cyclic eg.142857*6= 857142. This is a trick often used by 'lightning calculators' as the no. itself looks random and they roll off any multiplication from 1 to 5 i.e. 142857 * 5 = 714285.
The rule of thumb is that if you pick 5 then start with the no. just greater than it i.e. in this case 7 and go in order to 7142857 (7>5)
Similarly 142857 * 4 =571428 (5>4).
The procedure can be extended to make it look more of a feat
eg: 142857 142857 *4 =571428 571428 and the audience id none the wiser if the 'random no.' is picked by a helper in the audience!

Regards 1 (b) try old fashioned long division and you will be surprised at the length of the answers .
Yes 24 never divides a number composed of 9's

Ans. to 1 (c) is 444 444

2 ) ( a ) Please read the question carefully. It clearly says 3 factors including itself and 1. These therefore are the squares of prime no.s i.e.
4 , 9 , 25 , 49
3) (a ) You are right as the factor 3 is always present.

Try 3 (b) Try it as it is quite a theoretical one. :smile:
Best of luck
Mally :coffee:

1a) [spoiler]any with 6n digits as stated before[/spoiler]

1b) [spoiler]13 - 6n digits of 9, 17 - 16n digits of 9, 23 - 22n digits of 9, 24 - never[/spoiler]

2a) [spoiler]4 (1,2,4) - 9 (1,3,9) - 25 (1,5,25) - 49 (1,7,49) [/spoiler]

2b) [spoiler]2^3*3^2=72 and 2^5*3=96 have 12 factors (1,2,3,4,6,8,9,12,18,24,36,72 and 1,2,3,4,6,8,12,16,24,32,48,96)[/spoiler]

3a) [spoiler]doubling in this manner gives numbers of the form 3*2^k, none of which satisfiy the p[sub]1[/sub][sup]2q[sub]1[/sub][/sup]*p[sub]1[/sub][sup]2q[sub]2[/sub][/sup]*... form required for a perfect square[/spoiler]

3b) [spoiler]numbers congruent to 2 mod 4 - for the other cases...
0 mod 4 : (n+1)[sup]2[/sup]-(n-1)[sup]2[/sup] gives 4n+0
1,3 mod 4 : (n+1)[sup]2[/sup]-n[sup]2[/sup] gives 2n+1
[/spoiler]

Elementary number quiz.
 

Tom11786 You are right all the way. Very well done.
:rolleyes:
I give a few pointers to aid in understanding these problems further.
.
2(a)Two different prime numbers (no.s) multiplied together (such as 2*3) always have 4 factors (1,2,3,6)
Hence the no.s are squares of primes no.s
2(b) Numbers have many factors when they have at least two primes which are repeated as often as possible like 72 and 96.

3(a) Write the numbers as prime factors multiplied together.
Because the factors of perfect no.s come in pairs but the no.s in the sequence have a single factor 3
3(b) All no.s can be written as a difference of 2 squares except multiples of 4.
The difference of 2 squares is always the sum of consecutive odd no.s.
eg: 12^2 -7^2 =23+21 +19+17+15
11^2-7^2 =21 +19 +17 +15.

Its worth noting these two theorems.
1)The square of a no. is either divisible by 4 or leaves a remainder 1 when divded by4
2)The square of an odd number is of the form 8q + 1.
Thus 5^2 =3*8 +1 ; 7^2 =6*8 +1.

After seeing the responses I proudly present the solutions presented to me after only hours of my of posting by Akruppa. I could not better the mastery of Akruppa, over this subject. :bow:
Re: Elementary number quiz.
________________________________________
(a)
99...9 = 10^k-1
10^k == 1 (mod 7) <=>
3^k == 1 (mod 7) <=>
6|k (with ord_7(3) = 6)
I.e. 7 | 999999

(b)
ord_13(10) = 6
ord_17(10) = 16
ord_23(10) = 22
24 divides only 10^0-1, as 10^k-1 is odd for k>1.

(c) 44...4 = 4/9*(10^k-1)
4/9*(10^k-1) == 0 (mod 7) <=>
10^k-1 == 0 (mod 7) <=>
6|k as in (a)

2 (a) the squares of primes, {4, 9, 25, 49}
(b) of prime factors, counting multiplicity : 64 and 96 (6 each)
of distinct prime factors : 30, 42, ... (3 each)
of distince positive divisors : 60, 72, 84, 90, 96 (12 each)

3 (a) the sequence is 3*2^k, exponent of 3 is always 1, therefore odd, therefore the number is no square
(b) a^2 - b^2 = c , a > b
(a+b)(a-b) = c
If c is even, a-b and a+b must both be even, thus 4|(a+b)(a-b).
Ergo c == 2 (mod 4) cannot be written as a^2-b^2

Alex
Thank you Alex for the support. :smile:
Mally :coffee:

1 (b) should be "... for k>0".

For some reason I keep confusing zero and one.

Alex

Elementary number quiz.
 

[QUOTE=akruppa]1 (b) should be "... for k>0".

For some reason I keep confusing zero and one.

Alex[/QUOTE] :smile:
Just see if your answer for 2 (b) is correct The only no.s with the largest factors are 72 and 96 (12 factors each)
Mally :coffee:

[QUOTE=mfgoode]:smile:
Just see if your answer for 2 (b) is correct The only no.s with the largest factors are 72 and 96 (12 factors each)
Mally :coffee:[/QUOTE]

If by "no. of factors" you mean the number of positive, not neccessarily prime divisors, 60, 84 and 90 qualify as well:

60: 1 2 3 4 5 6 10 12 15 20 30 60
84: 1 2 3 4 6 7 12 14 21 28 42 84
90: 1 2 3 5 6 9 10 15 18 30 45 90

Alex

Elementary number quiz.
 

[QUOTE=akruppa]If by "no. of factors" you mean the number of positive, not neccessarily prime divisors, 60, 84 and 90 qualify as well:

60: 1 2 3 4 5 6 10 12 15 20 30 60
84: 1 2 3 4 6 7 12 14 21 28 42 84
90: 1 2 3 5 6 9 10 15 18 30 45 90

Alex[/QUOTE]
:surrender
Hats off to you akruppa! Im amazed at your total dedication in getting to the bottom of a probem and not leaving a post half baked. The shear tenacity is awesome!

I have to concede that you are perfectly right in your deduction of several no.s more than was required to satisfy any quiz master.

My original question was 'What number less than 100...(singular). I found two off hand 72 and 96 and was content and so overlooked the others.I must confess it was sheer slugnishness on my part. I think ken g6 and Tom 11784 also gave similar answers. Its hard to check with the spoilers thrown in. I dont agree one can benefit by it, anyone as a matter of fact, including school boys but thats for another post.
I had the nagging doubt that the old sexagesimal base system was picked exactly for this reason of having many factors. Now you have by sheer perseverance taken out the balance from your magicians hat! :surprised
:rolleyes:
By the way akruppa I believe there is a number less than 1000 which has a number with no less than 27 factors! I have as yet not cracked it out.
Could you please give it a shot? (This is not a homework question!) I would imagine its a multiple of the no.s you have already given but no obligations. If you want to leave it alone well and good enough for me :smile:
Mally :coffee:

[QUOTE=mfgoode]I believe there is a number less than 1000 which has a number with no less than 27 factors! I have as yet not cracked it out.
[/QUOTE]

That one is easy == 900

Look at it this way:

Let C be some composite number that factors as p1^n1 * p2^n2 * p3^n3 ...

Now consider the number of combinations that can be formed from those, and only those, prime factors.

In the case of p1, it can appear 0, 1, ..., n1 times.
Independently, p2 can appear 0,1, ..., n2 times.
etc.
Therefore the total number of combinations is (n1+1)*(n2+1)*...*(nn+1)

2^2*3^2*5^2 = 900 has 3*3*3 =27 factors

But 2^4*3^2*5 = 720 has 5*3*2 =30 factors
and 2^3*3*5*7 = 840 has 4*2*2*2 = 32 factors

Let p_1 ^ v_1 * p_2 ^ v_2 ... p_n ^ v_n be the prime factorisation of k with the p_i all distinct. Then the number of positive divisors of k is

(v_1 + 1) * (v_2 + 1) * ... * (v_n + 1)

by choosing how many of the prime p_i (from zero to v_i) we put in the divisor.

For example with 60 = 2^2 * 3^1 * 5^1 we have (2 + 1) * (1 + 1) * (1 + 1) = 12

You want at least 27 distinct divisors and n<1000. Since 2*3*5*7*11 = 2310, we can have at most 4 distinct prime divisors.

Starting with 2*3*5*7 = 210 we find 2*2*2*2 = 16 divisors.
With 2^2*3*5*7 = 420, 3*2*2*2 = 24 divisors. Same with 630.
With 2^3*3*5*7 = 840, 4*2*2*2 = 32 divisors. This one qualifies.
Other multiples of 210 are >1000.

Other n with 4 distinct primes, say multiples of 2*3*5*11, become too large too fast. I.e. 660 has only 24 divisors.

With 3 distinct primes:
Multiples of 2*3*5 = 30 that qualify are 2^6*3*5 = 960 with 7*2*2 = 28 divisors and
2^4*3^2*5 = 720 with 5*3*2 = 30 divisors. Multiples of 2*3*7=42 get too large to quickly, the smallest one I found with enough divisors is 1008 with 30 - but alas! slightly too large.

With 2 distinct primes:
We'd need (5 + 1) * (4 + 1) = 30 divisors. But 2^5*3^4 = 2592, too large.
Or maybe (6 + 1) * (3 + 1) = 28 divisors, but 2^6*3^3 = 1728, still too large.

I think 840 with 32 divisors, 720 with 30 and 960 with 28 are the only ones that match your criteria.

Alex

PS: Wacky beat me to it, and I missed 900 :sad: