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by Chrystelle » Fri Dec 15, 2006 3:14 am
Does somebody has an idea?

What will be the remainder when 13^7 + 14^7 + 15^7 + 16^7 is divided by 58?

How many times is the digit "3" written while numbering the pages of a book from 1 to 1000?

In a class comprising boys and girls, there were 45 hand shakes amongst the girls and 105 hand shakes amongst the boys. How many hand shakes took place between a boy and a girl, if each member of the class shook hands exactly once with every other student in the class?

10. If a circle, regular hexagon and a regular octagon have the same area and if the perimeter of the circle is represented by "c", that of the hexagon by "h" and that of the octagon by "o", then which of the following is true?
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by aim-wsc » Sun Dec 17, 2006 8:14 pm
What will be the remainder when 13^7 + 14^7 + 15^7 + 16^7 is divided by 58?
I call these problems as speedbraker.
clearly GMAC don t want you to caldute the final figure...so you just have to catch the trend:

you must notice that
13+14+15+16=58

you can write the same as 13^1 + 14^1 + 15^1 + 16^1 = 58

now take some simple/ small numbers say 2,3,4
you will notice that 2^n + 3^n +4^n is alsways divisible by (2+3+4)

gotcha: in the same way,
the desired figure is also divisible by 58

Answer: the remainder is ZERO '0'
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by aim-wsc » Sun Dec 17, 2006 8:54 pm
How many times is the digit "3" written while numbering the pages of a book from 1 to 1000?
no. of times 3 came in 1-100: 10 in unit place + 10in second place =20

No. of times 3 came in 1-1000 : (20 X 10 )+ 100 times from 300-399=200+100
Answer: 300
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by aim-wsc » Sun Dec 17, 2006 9:25 pm
In a class comprising boys and girls, there were 45 hand shakes amongst the girls and 105 hand shakes amongst the boys. How many hand shakes took place between a boy and a girl, if each member of the class shook hands exactly once with every other student in the class?
Lets take a general scenario

[Note I am not solving the problem here but giving you general idea figures: ]
if 4 boys there (say I am among them) I ll shook hand with htree other boys. So the other boys

But no. of handshakes will be. 3+2+1=6

If you go on increasing the no. you will find
The series as: (where former figure is no. of persons)
5-10
6-15
7- 21
8-28
9-36
10-45
.
.
similarly for
15-105

(I didn’t spent so much time since I already knew 10-45 &11-55 I just had to add some more to know whats for 105)

back to the problem:

That means there were 10 girls and 15 boys.
That means 15 boys will have 10 girls to shake there errmm hands ;) or should I say 10 girls will have 15 boys… case is same.
ie 15 X 10
Answer: 150

PS: my apologies for such emirical explanation, I know there must be some easy fromula or path to solve such nuts. I havent checked math book from ages so from Today I ll check permutation & combination…& if I find some easy (this one is also easy route bytheway) I ll post it.
Thankyou.
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by aim-wsc » Sun Dec 17, 2006 9:30 pm
If a circle, regular hexagon and a regular octagon have the same area and if the perimeter of the circle is represented by "c", that of the hexagon by "h" and that of the octagon by "o", then which of the following is true?


:shock: :shock: :shock: :shock: :o :o :o :o :roll: :roll: :roll:

something is missing.

One suggestion:
Avoid placing multiple questions in the same topic.
it gets difficult to discuss in group.
Use separate topic for each problem next time onward.
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by thankont » Tue Dec 19, 2006 12:01 am
if we speak about handshakes then we speak about combinations (order does not matter) so for girls we have xC2=45 or x(x-1)=90 answer is x=10 and for boys yC2=105 or y(y-1)=210 answer is y = 15
so 10*15 = 150
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by aim-wsc » Tue Dec 19, 2006 12:44 am
thanks for providing simpler solution. :)
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pls clarify

by anuroopa » Thu Mar 08, 2007 9:00 am
Hi aiim-wsc,
This is with reference to the 2nd Q - u say 3 appears 300 times btween 1 to 1000 - is it possible that you have double counted some of the numbers between 300 and 399 - Should we not subtract that from the total - pls clarify - i am confused
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Re: pls clarify

by aim-wsc » Thu Mar 08, 2007 10:51 pm
Sorry for replying late, Anuroopa :(
anuroopa wrote:Hi aiim-wsc,
This is with reference to the 2nd Q - u say 3 appears 300 times btween 1 to 1000 - is it possible that you have double counted some of the numbers between 300 and 399 - Should we not subtract that from the total - pls clarify - i am confused
I dont think so.
Numbers of times '3' printed/typed from 300-399 is ( 100+ 20 )
And for rest its just 20. therefore 20 X 9= 180

In total 120+180=300

Answer: 300

Why are you confused? Where you stuck?
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understood

by anuroopa » Fri Mar 09, 2007 2:16 am
Hey,
thnks for responding - lok - i figured it out - while calculating - i went into thinking no of numbers between 1 - 1000 containing 3 rather than no of 3's

thanks a ton
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by Cybermusings » Tue Mar 27, 2007 2:07 am
Thanks for the solutions!
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by 800GMAT » Thu Aug 16, 2007 3:23 pm
3 once------3C1 * 9 * 9= 243
3 twice------3C2 * 9 = 27
3 thrice------1


243+ 27*2 + 3

=243 + 54 + 3 = 300
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by beny » Thu Aug 16, 2007 5:36 pm
aim-wsc wrote: now take some simple/ small numbers say 2,3,4
you will notice that 2^n + 3^n +4^n is alsways divisible by (2+3+4)
2^2 + 3^2 + 4^2
= 4 + 9 + 16
= 29

29 is NOT divisible by 9...
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by givemeanid » Thu Aug 16, 2007 6:58 pm
beny wrote:
aim-wsc wrote: now take some simple/ small numbers say 2,3,4
you will notice that 2^n + 3^n +4^n is alsways divisible by (2+3+4)
2^2 + 3^2 + 4^2
= 4 + 9 + 16
= 29

29 is NOT divisible by 9...
Correct.
That is true when the exponent is odd.
x^n + y^n + z^n is divisible by (x+y+z) when n is odd.
So It Goes
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by aim-wsc » Fri Aug 17, 2007 3:12 am
beny wrote:
aim-wsc wrote: now take some simple/ small numbers say 2,3,4
you will notice that 2^n + 3^n +4^n is alsways divisible by (2+3+4)
2^2 + 3^2 + 4^2
= 4 + 9 + 16
= 29

29 is NOT divisible by 9...
oops! :p
Apologies :(
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