@emanresu ok got it .thanks 😃 i faced this issue in lot of problems.i drew the diagram and got it :)
find the smallest no. that leaves remainders 1 with 5, 4 with 7, 6 with 11 and 7 with 13.
@nole said:find the smallest no. that leaves remainders 1 with 5, 4 with 7, 6 with 11 and 7 with 13.
( 13k+7) mod 11 = 6
(2k+7)mod11 = 6
k=5,16,27..
5+11n-11 = 11n-6
13(11n-6) + 7
=143n-71
(3n-71) mod 7 = 4
n=25,32..
143*39-71
5506 ?
@nole said:find the smallest no. that leaves remainders 1 with 5, 4 with 7, 6 with 11 and 7 with 13.
501....trial/error + the chinese remainder approach
regards
scrabbler
regards
scrabbler
@krum said:( 13k+7) mod 11 = 6(2k+7)mod11 = 6k=5,16,27..5+11n-11 = 11n-613(11n-6) + 7=143n-71(3n-71) mod 7 = 4n=25,32..143*39-715506 ?
Note that there will be 1 such number every (LCM of 5, 7, 11, 13) numbers, i.e. every 5005 numbers. So if 5506 works, then 5506 - 5005 = 501 will also work.
regards
scrabbler
regards
scrabbler
@krum
@scrabbler
answer is 501. but can anyone of u explain by normal ak+b method.without chinese remainder method.i took this from pg here is the link of the page
http://www.pagalguy.com/forums/quantitative-ability-and-di/concepts-total-fundas-t-23536/p-784613?page=1
maximus explained using normal intitutive method,but had some doubt so asked here.
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.
in the highlighted portion it is given (7+13k) mod 11 gives remainder 7 +2k. I didn't get that part.can anyone explain.
@scrabbler
answer is 501. but can anyone of u explain by normal ak+b method.without chinese remainder method.i took this from pg here is the link of the page
http://www.pagalguy.com/forums/quantitative-ability-and-di/concepts-total-fundas-t-23536/p-784613?page=1
maximus explained using normal intitutive method,but had some doubt so asked here.
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.
in the highlighted portion it is given (7+13k) mod 11 gives remainder 7 +2k. I didn't get that part.can anyone explain.
@scrabbler said:Note that there will be 1 such number every (LCM of 5, 7, 11, 13) numbers, i.e. every 5005 numbers. So if 5506 works, then 5506 - 5005 = 501 will also work.regardsscrabbler
thanks sir, should have checked lcm :splat:
@scrabbler to cut short,i mean (13k+7)mod 11 =6 ,here 13k mod 11 gives 2k as remainder.
if i take k=6 then 78/11 rem= 1 but 2(6) gives 12 yes i can again divide to get 1, but m not able to understand this part.
if i take k=6 then 78/11 rem= 1 but 2(6) gives 12 yes i can again divide to get 1, but m not able to understand this part.
@nole said:Chinese remainder is just a formal boring way of writing ak+b :P@krum@scrabbleranswer is 501. but can anyone of u explain by normal ak+b method.
smallest no. that leaves remainders 1 with 5, 4 with 7, 6 with 11 and 7 with 13
I said let me satisfy 1st 2 conditions. Tried 7k+4 numbers , 4 11, 18....and the first which also was 5m+1 was 11. So now all numbers of form 35n+11 satisfy both these (35 is LCM of 7, 11) such as 11, 46, 81, 116, 151....and the first such satisfying 11p+6 is 116. So now all numbers of form 385q + 116 satisfy all 3 conditions (385 being LCM of 5, 7, 11) such as 116, 501, 886 and so on and the first one which also is 13r+7 form is 501.
regards
scrabbler
@nole said:@scrabbler to cut short,i mean (13k+7)mod 11 =6 ,here 13k mod 11 gives 2k as remainder.
13k / 11 = (11k+2k) /11
Now 11k gives remainder 0 so the remainder is equiv to that of 2k. I guess that is what is meant.
regards
scrabbler
Now 11k gives remainder 0 so the remainder is equiv to that of 2k. I guess that is what is meant.
regards
scrabbler
@scrabbler 13k/11 gives remainder 2k. now k =6 , so 13(6)/11 gives remainder 2k. i.e 12 (when actually it is 1). now we are keeping 2k in mind to further progress in the question. won't that create any problem ? this was my doubt.
@nole said:@scrabbler 13k/11 gives remainder 2k. now k =6 , so 13(6)/11 gives remainder 2k. i.e 12 (when actually it is 1). now we are keeping 2k in mind to further progress in the question. won't that create any problem ? this was my doubt.
Might. You have to find the minimum number. You might get a larger one with this (as happened to krum sir abhi) in which case - reality check karo. 1st values will lie below LCM of all divisors given (here 5005) so if answer > that, - 5005 till we get to the required range.
That's why I don't take variables if I can avoid. I don't trust the buggers.
regards
scrabbler
That's why I don't take variables if I can avoid. I don't trust the buggers.
regards
scrabbler
In a triangle ABC, three points D,E and F are choosen on sides BC,AB and AC respectively such that DE is parallel to AC and DF is parallel to AB .If the area of the triangle BDE is 36 and that of quadrilateral AEDF is 60,find the ratio of perimeter of triangle CDF to that of triangle ABC.
@nole said:In a triangle ABC, three points D,E and F are choosen on sides BC,AB and AC respectively such that DE is parallel to AC and DF is parallel to AB .If the area of the triangle BDE is 36 and that of quadrilateral AEDF is 60,find the ratio of perimeter of triangle CDF to that of triangle ABC.
is oa 1:2
A person was asked to measure the volume of 2 different spheres. He found it out correctly but forgot the answer... But he remembers that both the radii were integers.Can you tell him the exact answer?a) 379272 b) 349272 c) 739272 d) 439272
in the solution it is given
here's the solution...or lets say approach...
4/3x22/7xR^3 + 4/3x22/7x r^3 is the sum of volumes...the terms in denominator shud be cancelled with R^3 as well as r^3. but terms in numerator cannot be cancelled since radii are integers...
in denominator 3 getting cancelled i got that , how 7 will get cancelled? can anyone explain.
in the solution it is given
here's the solution...or lets say approach...
4/3x22/7xR^3 + 4/3x22/7x r^3 is the sum of volumes...the terms in denominator shud be cancelled with R^3 as well as r^3. but terms in numerator cannot be cancelled since radii are integers...
in denominator 3 getting cancelled i got that , how 7 will get cancelled? can anyone explain.
@nole said:A person was asked to measure the volume of 2 different spheres. He found it out correctly but forgot the answer... But he remembers that both the radii were integers.Can you tell him the exact answer?a) 379272 b) 349272 c) 739272 d) 439272in the solution it is givenhere's the solution...or lets say approach...4/3x22/7xR^3 + 4/3x22/7x r^3 is the sum of volumes...the terms in denominator shud be cancelled with R^3 as well as r^3. but terms in numerator cannot be cancelled since radii are integers...in denominator 3 getting cancelled i got that , how 7 will get cancelled? can anyone explain.
use options
4/3*22/7(R3+r3)= one of the options
option should atleast be divisible by 11 as 22=2*11
i think only option b satisfies.
4/3*22/7(R3+r3)= one of the options
option should atleast be divisible by 11 as 22=2*11
i think only option b satisfies.
logic R3+r3= option*3*7/(22*4)
since lhs is integer rhs is integer
for rhs to be integer option should be divisible by 8 & 11 as 22*4=8*11
every option divisible by 8 as last 3 digit is divisible and same for all.
only option b divisible by 11.
@emanresu for 88 2,4,8,11 are the factors but we checking only for 8 and 11 coz if a number is divisible by 8 it should be divisible by 2 and 4 also. that's why we cecking with only 8 ( and not 2,4 ) rite?