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**ZHero****Real Member**- Registered: 2008-06-08
- Posts: 1,889

Ooops!

Sorry.. Forgot to include a '2' in the answer above!

In general.. If any number

Now.. Write 45 as a product of as many numbers as possible (simply put PRIME)

45=3*3*5

which gives

p+1=3 => p=2

q+1=3 => q=2

r+1=5 => r=4

So we may have only (any) THREE values of a, b & c!

Thus we may have Max of THREE PRIME factors!

*Last edited by ZHero (2008-07-22 02:08:03)*

If two or more thoughts intersect with each other, then there has to be a point.

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**mathsyperson****Moderator**- Registered: 2005-06-22
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*This was to Jane, who has since deleted her post*

I don't see any nonsense in ZHero's post. It may not be as formal as you'd like, but everything in there is correct. That method also gives a lower maximum.

Side note: The smallest number with 45 factors is 3600. Also, any number with that amount of factors must be a square.

Why did the vector cross the road?

It wanted to be normal.

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**Sudeep****Member**- Registered: 2008-07-21
- Posts: 20

Find the remainder when 9*5^43 divided by 1000.

Do we have any source to find all sorts of such questions....help will be really appriciated

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**TheDude****Member**- Registered: 2007-10-23
- Posts: 361

To find the remainder we simply need to find the last 3 digits of 9*5^43. First find the last 3 digits of 5^43, which is fairly simple when you notice the following pattern:

X Last 3 digits of 5^X

1 005

2 025

3 125

4 625

5 125

6 625

... ...

For any even number n >= 4, the last 3 digits of 5^n are 625. For any odd number k >= 3, the last 3 digits of 5^k are 125. Since 43 is odd we know the last 3 digits of 5^43 are 125. Multiply this by 9 to get 1125. This means our remainder when dividing by 1000 will be 125.

Wrap it in bacon

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**JaneFairfax****Member**- Registered: 2007-02-23
- Posts: 6,868

,

You can prove this by induction.

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**Sudeep****Member**- Registered: 2008-07-21
- Posts: 20

The following table gives the details of the ages and heights of the boys and girls attending a hobby camp. The height (in cm) of each boy/girl is an integer.

Age Group Boys Girls

(Age(A),in years)

A<=10 8(120,130,150) 10(110,130,140)

11<=A<=13 10(130,150,160) 8(120,140,150)

14<=A<=16 6(150,160,180) 6(140,150,170)

A>16 5(170,180,200) 8(150,160,180)

In the above table, the number mentioned outside the brackets in each cell gives the number of boys (or girls) in that age group and the values mentioned inside the brackets give the minimum height, the average height and the maximum height of the persons in that group, in that order.

For example, of the eight boys who are of age not more than 10 years, the minimum height of any boy is 120 cm, the average height of all the boys is 130 cm and the maximum height of any boy is 150 cm.

1)A team is to be selected from the boys in the camp who were of age 13 years or below. If only boys who are 140 cm or more in height can be selected, the number of boys who can be selected is at most.

2)If a team of girls of age 13 years or less is to be selected, with the condition that all the girls who are 130 cm or more in height must be selected, what is the minimum number of girls selected?

Please give your approach while giving the solution.

*Last edited by Sudeep (2008-08-05 13:49:59)*

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**Sudeep****Member**- Registered: 2008-07-21
- Posts: 20

What is the sum of the sum of the sum of the digits of 55!?

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**mathsyperson****Moderator**- Registered: 2005-06-22
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55! is the product of 55 numbers: 1, 2, 3, ..., 54 and 55.

All of these numbers are less than 100, and so 55! < 100^55 = 10^110.

So 55! has no more than 110 digits. The possible sum of its digits is maximum when they are all 9, and so it maximum possible sum-of-digits is 990.

It's easy to see that the number with the highest sum-of-digits that is not higher than 990, is 899 with a digit-sum of 26.

Similarly, the number between 1 and 26 with the highest digit-sum is 19, which has a sum of 10.

Therefore, the sum of the sum of the sum of the digits of 55! is bounded above by 10.

The divisibility-by-9 test says that a number is divisible by 9 iff its digits sum to a multiple of 9.

55! has 9 as a factor, and so the sum of its digits (and so, the sum-of-sum and the sum-of-sum-of-sum) will also be a multiple of 9.

This reduces the possibilities for the sum-of-sum-of-sum to either 0 or 9.

Clearly 55! has at least one non-zero digit, and so its digit sum will be more than 0. Hence its sum-of-sum and sum-of-sum-of-sum will also be non-zero, which leaves 9 as the only possibility.

**The sum of the sum of the sum of the digits of 55! is 9.**

Why did the vector cross the road?

It wanted to be normal.

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**Identity****Member**- Registered: 2007-04-18
- Posts: 934

Very clever mathsyperson

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**mathsyperson****Moderator**- Registered: 2005-06-22
- Posts: 4,900

Thanks!

I cheated a bit though, at first I used MathsIsFun's full precision calculator to display 55!, then worked out the answer by literally adding all the digits, then the digits of the answer and so on.

I didn't like that method much, so I made the more analytical one after knowing what the answer was.

Why did the vector cross the road?

It wanted to be normal.

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**ZHero****Real Member**- Registered: 2008-06-08
- Posts: 1,889

I love honest people!!

If two or more thoughts intersect with each other, then there has to be a point.

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**Sudeep****Member**- Registered: 2008-07-21
- Posts: 20

How many natural numbers having at most six digits have the sum of their digits as at most six?

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**Sudeep****Member**- Registered: 2008-07-21
- Posts: 20

How many numbers less than 10^5 have the sum of digits equal to 10

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**Sudeep****Member**- Registered: 2008-07-21
- Posts: 20

In a certain system product of 44*11 is represented s 2124 find how 1453 can be represented in decimal system ?

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**Muggleton****Member**- Registered: 2009-01-15
- Posts: 64

Sudeep wrote:

How many numbers less than 10^5 have the sum of digits equal to 10

Such a number cannot be one digit long; it must have at least two digits. The number of such two-digits numbers is 9 (namely 19, 28, 37, 46, 55, 64, 73, 82, 91).

For the three-digit numbers, I group them into nine groups, according to the starting number:

**Group 1:** 109, 118, 127, 136, 145, 154, 163, 172, 181, 190**Group 2:** 208, 217, 226, 235, 244, 253, 262, 271, 280**Group 3:** 307, 316, 325, 334, 343, 352, 361, 370**Group 4:** 406, 415, 424, 433, 442, 451, 460**Group 5:** 505, 514, 523, 532, 541, 550**Group 6:** 604, 613, 622, 631, 640**Group 7:** 703, 712, 721, 730**Group 8:** 802, 811, 820**Group 9:** 901, 910

So there are 10 group-1 numbers, 9 group-2 numbers, etc. Total number of three-digit numbers: 54.

Now for the four-digit numbers. The number of four-digit numbers ending __00 is clearly the same as the number of two-digit numbers, namely 9.

For the four-digit numbers not ending __00, look at the first two digits. The number of four-digit numbers whose first two digits add up to *n* is the same as the number of group-*n* numbers. For example, the number of four-digit numbers beginning 72__ is the same as the number of group-9 numbers, the number of four-digit numbers beginning 34__ is the same as the number group-7 numbers, etc.

**Group 1:** 10__**Group 2:** 11__, 20__**Group 3:** 12__, 21__, 30__**Group 4:** 13__, 22__, 31__, 40__**Group 5:** 14__, 23__, 32__, 41__, 50__**Group 6:** 15__, 24__, 33__, 42__, 51__, 60__**Group 7:** 16__, 25__, 34__, 43__, 52__, 61__, 70__**Group 8:** 17__, 26__, 35__, 44__, 53__, 62__, 71__, 80__**Group 9:** 18__, 27__, 36__, 45__, 54__, 63__, 72__, 81__, 90__

Therefore there are 1×10 + 2×9 + 3×8 + 4×7 + 5×6 + 6×5 + 7×4 + 8×3 + 9×2 = 210 four-digit numbers not ending __00.

Hence the total number of natual numbers less than 10000 with digits adding up to 10 is 9 + 54 + 9 + 210 = 282.

*Last edited by Muggleton (2009-01-18 06:11:19)*

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**JaneFairfax****Member**- Registered: 2007-02-23
- Posts: 6,868

Er
I think Sudeep wrote:

10^5

So you need to include 5-digit numbers as well.

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**Muggleton****Member**- Registered: 2009-01-15
- Posts: 64

Oh dear! You're right. I made a mistake.

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**JaneFairfax****Member**- Registered: 2007-02-23
- Posts: 6,868

Well, youve done a good job so far; Ill finish it for you.

The number of 5-digit numbers ending ___00 is equal to the combined total number of 2- and 3-digit numbers, namely 63.

For the 5-digt numbers not ending ___00

**Group 1:** 100__

**Group 2:** 101__, 110__, 200__

**Group 3:** 102__, 111__, 120__, 201__, 210__, 300__

**Group 4:** 103__, 112__, 121__, 130__, 202__, 211__, 220__, 301__, 310__, 400__

**Group 5:** 104__, 113__, 122__, 131__, 140__, 203__, 212__, 221__, 230__, 302__, 311__, 320__, 401__, 410__, 500__

**Group 6:** 105__, 114__, 123__, 132__, 141__, 150__, 204__, 213__, 222__, 231__, 240__, 303__, 312__, 321__, 330__, 402__, 411__, 420__, 501__, 510__, 600__

**Group 7:** 106__, 115__, 124__, 133__, 142__, 151__, 160__, 205__, 214__, 223__, 232__, 241__, 250__, 304__, 313__, 322__, 331__, 340__, 403__, 412__, 421__, 430__, 502__, 511__, 520__, 601__, 610__, 700__

**Group 8:** 107__, 116__, 125__, 134__, 143__, 152__, 161__, 170__, 206__, 215__, 224__, 233__, 242__, 251__, 260__, 305__, 314__, 323__, 332__, 341__, 350__, 404__, 413__, 422__, 431__, 440__, 503__, 512__, 521__, 530__, 602__, 611__, 620__, 701__, 710__, 800__

**Group 9:** 108__, 117__, 126__, 135__, 144__, 153__, 162__, 171__, 180__, 207__, 216__, 225__, 234__, 243__, 252__, 261__, 270__, 306__, 315__, 324__, 333__, 342__, 351__, 360__, 405__, 414__, 423__, 432__, 441__, 450__, 504__, 513__, 522__, 531__, 540__, 603__, 612__, 621__, 630__, 702__, 711__, 720__, 801__, 810__, 900__

So the total number of 5-digit numbers not ending in two zeros is 1×10 + 3×9 + 6×8 + 10×7 + 15×6 + 21×5 + 28×4 + 36×3 + 45×2 = 660.

Hence total number of natual numbers less than 100000 with digits adding up to 10 is 282 + 63 + 660 = 1005.

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**Muggleton****Member**- Registered: 2009-01-15
- Posts: 64

Thanks!

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**mathsyperson****Moderator**- Registered: 2005-06-22
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I'm pretty sure that the amount of _ _ _00 numbers is the same as the amount of just the 3-digit numbers, so your answer should be 9 less.

I did this a different way. I might've made it overcomplicated though.

Look at this length-10 line:

----------

By placing exactly 4 markers along it, you can represent every number less than 10^5 whose digits add up to 10.

eg. 23131 = --'---'-'---'-

In theory, you can use a combinatorial formula to work out the total number of marker placements, and hence the total amount of numbers with the required property.

However, we get into trouble with this method because combinatorics doesn't work when two markers can be placed on the same point.

According to our symbolism, 12070 would be -'--''-------', but this isn't a valid combination.

Time for a workaround!

Now, we consider a length-15 line and place markers along it, and each line length means one more than the corresponding digit.

eg. 23131 = ---'----'--'----'--

12070 = --'---'-'--------'-

Now 0 digits can be represented by the system, and so the representation works.

There are 14 places to put the markers (15 lines, so 14 gaps between them) and 4 markers, so our answer is 14C4 = 1001.

...Nearly.

One final tweak is needed, because this representation will also throw out numbers like (10)0000.

Luckily, it's easy to count those (there are 5) so after ignoring them, the final answer is **996**.

Why did the vector cross the road?

It wanted to be normal.

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**JaneFairfax****Member**- Registered: 2007-02-23
- Posts: 6,868

mathsyperson wrote:

I'm pretty sure that the amount of _ _ _00 numbers is the same as the amount of just the 3-digit numbers, so your answer should be 9 less.

Youre correct. It should be only 995.

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**Subhamoy****Member**- Registered: 2009-04-30
- Posts: 3

M=10^5(m5)+10^4(m4)+10^3(m3)+10^2(m2)+10^1(m1)+10^0(m0) ≤ 10^5

⇒m5=0

⇒∑(10^i)mi ≤ 10^5 (i=1,2,3,4)...(1)

∑mi=10....(2) So the inequalility holds as max. value of M is 91000

∴No of M's satisfying (1)&(2) is = coefficient of x^10 in (x^0+x^1+x^2+.....upto ∞ terms )^5 - (5 c 4) (as mi can't take value 10;0≤mi≤10;0<x<1)

= ,, ,, ,, ,, (1-x)^(-5) - (5 c 4)=(14 c 4) - 5=1001-5=996

*Last edited by Subhamoy (2009-05-02 17:54:55)*

soc

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 15,327

Find the number(s) n such that:

for every integer k.

*Last edited by anonimnystefy (2011-06-05 10:01:37)*

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**gAr****Member**- Registered: 2011-01-09
- Posts: 3,478

Hi anonimnystefy,

"Believe nothing, no matter where you read it, or who said it, no matter if I have said it, unless it agrees with your own reason and your own common sense" - Buddha?

"Data! Data! Data!" he cried impatiently. "I can't make bricks without clay."

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 15,327

correct.yeah actually every number of the form 13m or 13m+1.

Here lies the reader who will never open this book. He is forever dead.

Taking a new step, uttering a new word, is what people fear most. ― Fyodor Dostoyevsky, Crime and Punishment

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