[GuruBen]

Originally Posted by maxximus

Today's funda...it's a combination of few of my old posts...



300! is expanded and expressed in base x. find the no. of zeroes at the right end of this expression when x=

98

answers... 21


maxximus

The answer shud be 24

[maxximus]

Originally Posted by GuruBen

The answer shud be 24

yeah...lil calculn mistake while adding 7s...didnt take muliples of 49...so the answer for base 35 n 98 wud change...i've rectified them...thanx for the post..

n am delighted than u've got the concept...seems the thread is working...

keep pouring in ...

cheers!
maxximus

[vallisri]

but 16 is not in options
the options are
48, 38, 54, 60

[maxximus]

Originally Posted by vallisri

but 16 is not in options
the options are
48, 38, 54, 60

okay...the question is 24s in 150!...sorry i thought the no. of 2s u have counted are 48...didnt see the original question...there's a slight error in ur calculatn but the answer is correct...to do it quickly...keep dividing the quotients when 150 is successively divided by 2 till its divisible.

so, the no. of 2s... 75 + 37 + 18 + 9 + 4 + 2 + 1 = 146

highest power of 24 = [146/3] = 48

where, [] is the greatest integer function...

regards
maxximus

[blitzz]

Great Initiative..................
Thanks a lot...

[charly]

Hi Maximuss,
truly great initiative taken.Pls keep up the good job.

i hv a dbt i'd like to get clarified. A question regarding :
256! is expanded and expressed in base 576 . how many zeroes will this expression have on its right end?
u said its the same finding power of 576 in 256! But lets try for 4! exprsd in base 2.
we know 4!= 24=11000(base 2) --> 3 zeroes.
But going by ur method,
4!/2= 2 zeroes only.

Same discrepancy in answer for 5!(again 3 zeroes) base 2.

Kindly explain.

[maxximus]

Originally Posted by charly

Hi Maximuss,
truly great initiative taken.Pls keep up the good job.

i hv a dbt i'd like to get clarified. A question regarding :
256! is expanded and expressed in base 576 . how many zeroes will this expression have on its right end?
u said its the same finding power of 576 in 256! But lets try for 4! exprsd in base 2.
we know 4!= 24=11000(base 2) --> 3 zeroes.
But going by ur method,
4!/2= 2 zeroes only.

Same discrepancy in answer for 5!(again 3 zeroes) base 2.

Kindly explain.

@ charly...

no man..u've not got the concept...read the main post 2-3 times...its an important concept...

think of 32 being expressed in base 2...its 100000...5 zeroes...because 32 is successively divisible by 2...five times. if u express 96...its 1100000 for same reason...if u express 162 in base 3...its 20000 bcoz 162 is four times divisible by 3.

4! has three 2s... 4/2 =2
2/2 =1
2+1 = 3
hence, there'll be 3 zeroes

so does 5! as 5/2 = 2
2/2 = 1

2+1 = 3

similarly here...v need to check how many times will 576 successively divide 256!

576 = 2^6 x 3^2...

2s will be the constraint...total 2s in 240! = 120+60+30+15+7+3+1 = 236
hence answer wud be [236/6]=39

please go thru my post properly...basic concepts are not in place...do try the problems i've mentioned and match ur answers

regards
maxximus

[maxximus]

Today's concept: Finding out smallest no. which leaves specific remainders with specific divisors.

Type # 1.

find smallest no. other than k, that leaves remainder k when divided by w,x,y...

to solve such questions, take lcm of w,x,y...and add k to it.

e.g. find Smallest no. other than 4, that leaves remainder 4 when divided by 6,7,8 or 9...

take lcm of 6,7,8,9 and add 4

i.e. 504 + 4 = 508

Type # 2

find smallest no. that leaves remainder 3,5,7 when divided by 4,6,8 respectively.

unlike last case, this time the remainder is not constant. but if u see carefully, difference b/w divisor n remainder is constant. i.e. 4-3=6-5=8-7=1

in such questions, take lcm of divisors n subtract the common difference from it

here, the answer wud be lcm of 4,6,8 i.e 24 - 1 = 23

Type # 3

Smallest no. that leaves remainder 3,4,5 whn divided by 5,6,7 respectively and leaves remainder 1 with 11,

we have just seen a way to tackle the first 3 conditions...the no. wud be lcm of 5,6,7 - 2 = 208

now we have one more condition...remainder 1 with 11.

concept => to a no. if v add lcm of divisors...the corresponding remainders dont change.

i.e to 208, if v keep adding 210 ... the first 3 conditions will continue being fulfilled.

so, let 208 + 210k be the no. that will satisfy the 4th condition...viz (208 + 210k)% 11 = 1

208%11 = 10

210k%11 = k

therefore, 10 + k shud leave remainder 1 when divided by 11.

hence, k = 2. and the no. is 208 + 210 x 2 = 628


e.g. find the smallest no. that leaves remainder 2 when divided by 3,4 or 5 and is divisible by 7

for first 3 conditions....no. is 120 + 2 = 122

hence, 122 + 120k is the required no. which reduces to 3 + 2k when divided by 7...now 3+2k shud be a multiple of 7...easily, k=2 and the required no. is 122 + 120 x 2 = 362

Type # 4

What if there is no relation between divisors n remainders?

e.g. find the smallest no. that leaves remainders 1 with 5, 4 with 7, 6 with 11 and 7 with 13.

we can c...there's no relation among these divisor-remainder sets...neither is the remainder constant...nor is the difference b/w divisor n remainder a constant.

in such cases...take 1 case n target another case...
e.g. i take the case 7 with 13...and target 6 with 11.

which is the smallest no. that leaves 7 with 13? 7 itself...right? so all nos of the form 7 + 13k will give 7 rem with 13.

now am targeting 6 with 11...so i divide 7 + 13k by 11...i get remainder 7 + 2k...now 7 + 2k = 6,17,28,39,50...so that the remainder with 11 is 6.

a no. that gives integral value of k is 17 i.e. 7 + 2k = 17. hence, k =5 and the no. that satisfies these two conditions is 7 + 13 x 5 = 72

now that 2 conditions are fulfilled, lets target a third condition...say 4 with 7.

to 72, if v add lcm of 11, 13 i.e 143, 2 conditions awready satisfied wud continue being satisfied...

hence the no. is of the form 72 + 143 k.

72 + 143k % 7 = 2 + 3k

now 2 + 3k shud be = 4,11,18,25,32... to satisfy the condition of 4 rem with 7..

a no. that gives integral soln is 11..i.e. 2 + 3k = 11, k = 3.

hence, the no. that satisfies all 3 conditions is 72 + 143 x 3 = 501.

now if v see carefully...4th condition...remainder 1 with 501 has already been satisfied...so the no. v have been looking for is 501.

For ease of calculation, start from biggest divisor n gradually move to smaller ones...u'll always see that last 1-2 conditions will be satisfied automatically.

there are theorems for solving above questions...viz chinese theorem etc...but i solve such questions by the way i've suggested...i find this approach very practical as the flow of nos. is very much visible...n i believe i can tackle any twist in the question devised by cat makers thru this method...there are lotsa other questions based on this concept which i'll soon post but the basic concept remains the same...


few points to be noted

*you can always re-check ur answer

**at times, u can use options to solve such questions.

***dont let concepts go away believing such questions can be easily dealt with thru options...the question may not always be find the smallest no. which...... at times it may be ..find the sum of integers of smallest no. which leaves remainders...blah blah...

****there may be a case when they put an option which satisfies all the conditions but is not the smallest poss value...n put another option...our favorite...none of these!!! lets not undermine genius of cat makers!!

questions for practice...


find the smallest no. that leaves remainder (s)

Q1. 2 when divided by 3,5,6 or 9 (other than 2)
Q2. 2,5,7 when divided by 7,10 and 12 respectively
Q3. 1,2,3,4 with 3,4,5,7 respectively.
Q4. 6 with 7,8,9,10 and 3 with 11.
Q5. 3 with 6, 0 with 11, 3 with 5, 7 with 8
Q6. 2 with 5, 7 with 8, 3 with 4, 5 with 7&11
Q7. 1 with 11, 4 with 5, 9 with 10, 7 with 9.

hope the post helps...puys...come up with ur own...sweet methods...which are confined in ur sharp brains...i'll also appreciate if u can come up with feedbacks/suggestions...like today i thought its better to give answers a day late...


regards
maxximus

[iyervani]

Originally Posted by maxximus

find the smallest no. that leaves remainder (s)

Q1. 2 when divided by 3,5,6 or 9 (other than 2)
Q2. 2,5,7 when divided by 7,10 and 12 respectively
Q3. 1,2,3,4 with 3,4,5,7 respectively.
Q4. 6 with 7,8,9,10 and 3 with 11.
Q5. 3 with 6, 0 with 11, 3 with 5, 7 with 8
Q6. 2 with 5, 7 with 8, 3 with 4, 5 with 7&11
Q7. 1 with 11, 4 with 5, 9 with 10, 7 with 9.

regards
maxximus

1) 92
2) 415
3) 298
4) 495
5) 1683
6) 687
7) 529

the 5th one took me a long time though.....:( ...have not posted the method....same as what maxximus has explained....useful post maxximus thnxxx...hope i got'em all right....

[vineet.nitd]

Maxximus ..you are doing and amazing job .Bahut achche .

Keep the spirits and the discipline up :satisfie: