@bodhi_vriksha said:Common guys the following is a challenge problem.Investigate all seven-digit numbers that have only the digits 5 and 7 and that are divisible by 35.
@bodhi_vriksha said:Common guys the following is a challenge problem.Investigate all seven-digit numbers that have only the digits 5 and 7 and that are divisible by 35.
I have a feeling it is 8. Seems too small an answer though. Have to rush off abhi, will investigate further when I hit the comp again. Answer options hai?
regards
scrabbler
regards
scrabbler
@scrabbler said:I have a feeling it is 8. Seems too small an answer though. Have to rush off abhi, will investigate further when I hit the comp again. Answer options hai?regardsscrabbler
I would like to say that your estimate is in the ball park range:)
Let me create a few nice options for you :)
a) 8 b)9 c>10 d>12 e>15
But the question is about finding those numbers and not just the number of such numbers!
@jain4444 said:Out of 200 in an aquarium, 99% are red. How many red fish must be removed in order to reduce the percentage of red fish to 98%?
Initially 200 fishes ==> 99% red fishes ==> 198 red fishes .
Now , (198 - x ) = (200 - x ) * 0.98 ==> 198 -x = 196 -0.98x ==> 2 = 0.02 x ==> x = 100 ;
so 100 fishes must be removed :)
Now , (198 - x ) = (200 - x ) * 0.98 ==> 198 -x = 196 -0.98x ==> 2 = 0.02 x ==> x = 100 ;
so 100 fishes must be removed :)
@saurabhlumarrai said:I am not getting the approach...can you please elaborate that why you have taken all 2,3,4,8 and left 1,5,6,7..???
see, if we consider all modulo values, we have 20 each of 1,2,3,4,5,6,7 and 21 each of 0 and 8. so, we should take 20 each of (2 or 7), (3 or 6), (4 or 5)...since each of these pairs adds up to 9, we can take either of them but not both....thereafter we have the option of taking either 20 1s or 21 8s, so we choose the latter...finally, we take a single 0, to arrive at the total of 82
@fire123 said:65% of students in a class have taken subject A, 80% B, 75% C & 85% D. At least what percentage of students have taken alll four subjects??Guys help me out with gud explanation
See the attached graphical solution. It can also be interpreted as the following:
100 - {(100-85)+(100-80)(100-75)+(100-65)} =100 -95=5%
@jain4444 said:Out of 200 in an aquarium, 99% are red. How many red fish must be removed in order to reduce the percentage of red fish to 98%?
100 ?
(99/100*200 - x)/(200 - x) = 98/100
x = 100
@bodhi_vriksha said:I would like to say that your estimate is in the ball park range Let me create a few nice options for you a) 8 b)9 c>10 d>12 e>15 But the question is about finding those numbers and not just the number of such numbers!
i could figure out only 4 no's :(
7775775
5577775
7757575
7755755
7775775
5577775
7757575
7755755
@bodhi_vriksha said:I would like to say that your estimate is in the ball park range Let me create a few nice options for you a) 8 b)9 c>10 d>12 e>15 But the question is about finding those numbers and not just the number of such numbers!
mujhse nahi ho raha....i quit....wasted 20 mins of my office time... 
@bodhi_vriksha said:I would like to say that your estimate is in the ball park range Let me create a few nice options for you a) 8 b)9 c>10 d>12 e>15 But the question is about finding those numbers and not just the number of such numbers!
Is the answer 9 ??
Let the number be abcdef5
Now a*10^6 + b*10^5 + c*10^4 + d*10^3 + e*10^2 + f*10 +5 mod 7 = 0
We can see that
5*10^6 mod 7 = 5
5*10^5 mod 7 = 4
5*10^4 mod 7 = 6
5*10^3 mod 7 = 2
5*10^2 mod 7 = 3
5*10^1 mod 7 = 1
(Taken 5 because we are required to use 5 in the number and without which it was getting difficult to solve further)
Now digits can be 5 or 7 only.
=> We have to make diff possible arrangements such that abcdef5 mod 7 = 0
Now,
All 7s --> not possible
one 5---> 5+2 = 7 --> 1 possibility
two 5---> 5+4+5=14, 5+6+3 =14 --> 2 ways
three 5---> 5+1+2+6=14, 5+1+3+5=14, 5+3+2+4=14 --> 3 ways
four 5---> 5+5+4+6+1 =21, 5+6+5+3+2 =21 --> 2 ways
five 5 ---> 5 + 1+2+3+4+6 = 21 --> 1 way
Hence a total of 9 possibilities ? ..
Nos are: 7775775, 5577775, 7757575, 7755755, 57777555, 7575575, 5557755, 5755575, 7555555 ..
P.S. Only I know kaha kaha se dhua nikla iske baad..
@bodhi_vriksha said:I would like to say that your estimate is in the ball park range Let me create a few nice options for you a) 8 b)9 c>10 d>12 e>15 But the question is about finding those numbers and not just the number of such numbers!
After racking my brains for almost 10 mins got only 4 numbers....
Let the number be a_ _ _ _ _ _ 5
Used a=7 first...
Sum of triplets at odd - sum of ttiplets at even = 0
So first 3 from rhs must be the same as the next 3...they cancel out..and leave the left most 7...
Ab aage combination lekar to is gonna be time taking...
Tried arithmetic series.as well...nothing clicked there...
Anyway its a good question....request you not to put the oa right now....
Wil try it out at home...abhi in office....full on attempt ghar jaa kar....
@saurav205 said:
Anyway its a good question....request you not to put the oa right now....Wil try it out at home...abhi in office....full on attempt ghar jaa kar....
Araam se. I am waiting to see those numbers on this thread... :)
@viewpt said:TPQ: A writes all the nos. frm 1 to 1000 on paper in order. Find the 2883rd digit written by him?
1*9 + 2*90 + 3*900 = 2889 digits ; 2883rd digit would be 2889 - 3*2 ==> units place of 997 ==> 7 ??
@bodhi_vriksha said:Common guys the following is a challenge problem.Investigate all seven-digit numbers that have only the digits 5 and 7 and that are divisible by 35.
Number has to be divisible by 35, and consists of only 5 and 7 -> Last digit is 5
so, N = abcdef5 = 10*(abcdef) + 5
N mod 5 = 0
N mod 7 = 5 + 3( abcdef mod 7) = 5 +3R
for N mod 7 = 0, => R = 3
=> abcdef mod 7 = 3
=> def - abc mod 7 = 3
Now, def and abc have 8 possibilities each ( only 5 and 7 can come in the digits places)
So, possibilities are 777,775,757,577,755,575,557,555
Now, in total, we have 64 possibilities of abcdef , but we need to find only those where it is of the form 7k + 3
Using this,
1) For def = 777, abc = 557
2) For def = 775, abc = 555
3) For def = 757, abc = 775
4) For def = 557, abc = 757, 575
5) For def = 577, abc = 777
6) For def = 755, abc = 577
7) For def = 575, abc = 775
8) For def = 555, abc = 755
So, there are 9 numbers and they are :
5577775, 5557755, 7757575, 7575575, 5755575, 7775775, 5777555, 7755755, 7555555
@bodhi_vriksha
I got 9 ...approach is a bit lengthy, so will only post the logic i used, not the entire solution.
1. Took the 7 digit number mod 1001, since 1001 is a multiple of 7, and if we divide the number say abcdefg into groups of 3 and subtract the ones at even position from the ones at odd position, the final number would be 1,2 or 3 digit number which is
2. since we know g=5, we can rewrite it as (ef5+a- bcd)mod 7=0, where ef5 and cde are three digit numbers, and a is a singe digit number.
3. Writing it in form of an equation (100(e-b) +10(f-c) +5+(a-d)) mod 7=0
4. since 100 mod 7=2 and 10 mod 7=3. we can rewrite above equation in terms of remainders alone and get (2(e-b) +13(f-c) +(5+a-d)) mod 7=0
5. Now since the number consists of only 5's and 7's, the values inside the brackets can only yield 3 values 0,2 and -2 respectively.
6. now take cases and arrangements. eg.
when 1st and 2nd bracket are 0, 5+(e-a) mod7 should give 0 and and hence the bracket value of +2 fits here. So e=b f=c, a=7, d=5. and the ea pair, and fd pair can each take 2 values of both being 5 or 7 respectively, so it can be
7555555
7. Continue on with the various other cases and you arrive at 9 solutions, i may have missed out a case so please confirm the OA.
I got 9 ...approach is a bit lengthy, so will only post the logic i used, not the entire solution.
1. Took the 7 digit number mod 1001, since 1001 is a multiple of 7, and if we divide the number say abcdefg into groups of 3 and subtract the ones at even position from the ones at odd position, the final number would be 1,2 or 3 digit number which is
2. since we know g=5, we can rewrite it as (ef5+a- bcd)mod 7=0, where ef5 and cde are three digit numbers, and a is a singe digit number.
3. Writing it in form of an equation (100(e-b) +10(f-c) +5+(a-d)) mod 7=0
4. since 100 mod 7=2 and 10 mod 7=3. we can rewrite above equation in terms of remainders alone and get (2(e-b) +13(f-c) +(5+a-d)) mod 7=0
5. Now since the number consists of only 5's and 7's, the values inside the brackets can only yield 3 values 0,2 and -2 respectively.
6. now take cases and arrangements. eg.
when 1st and 2nd bracket are 0, 5+(e-a) mod7 should give 0 and and hence the bracket value of +2 fits here. So e=b f=c, a=7, d=5. and the ea pair, and fd pair can each take 2 values of both being 5 or 7 respectively, so it can be
7555555
7. Continue on with the various other cases and you arrive at 9 solutions, i may have missed out a case so please confirm the OA.
@YouMadFellow said:
So, there are 9 numbers and they are :7775575, 7755555, 7577755, 5577575, 5575755, 5777775, 7555775, 5757755, 5557555
A valiant effort. However, none of your numbers is divisible by 35.
Spot your mistake....
@bodhi_vriksha said:A valiant effort. However, none of your numbers is divisible by 35. Spot your mistake....
Sorry, I exchanged the def and abc 
The new numbers are :
5577775, 5557755, 7757575, 7575575, 5755575, 7775775, 5777555, 7755755, 7555555
@YouMadFellow said:Sorry, I exchanged the def and abcThe new numbers are :5577775, 5557755, 7757575, 7575575, 5755575, 7775775, 5777555, 7755755, 7555555
Liked your approach. Well done :)
@bodhi_vriksha said:A valiant effort. However, none of your numbers is divisible by 35.Spot your mistake....
he has identified the abc and def correctly but put the number in the wrong order...
currently it is in defabc5 format
shud me abcdef5 format
so numbers becom 5577775, 5557755