Suppose chosen digits are a, b, c, d, e are the choosen digits such that a > b > c > d > e, then possibilities are:-abedcacedbadecbbcedabdecacdebaSo, 6*C(10, 5) = 6*252 = 1512 such numbers
The five distinct digits can be selected in C(10, 5) ways. And the five digits are in order say a > b > c > d > e.Clearly e has to be central digit and first two digits can be selected in C(4, 2) ways and arranged in exactly one way. Also remaining two digits can be arranged at last two places in exactly one way again.Thus total possible numbers are = C(10, 5)*C(4, 2) = 1944.Pls check the final calculations..Team BV - Kamal Lohia
A queen on an English chessboard is able to attack in the same row, column and diagonal. Find the probability that 2 randomly placed queens on an 8 by 8 chessboard will be able to attack each other .
Same row or same column
No of ways = 64*14/2 = 448
Same diagonal
No of ways = 2*C(8, 2) + 4{C(7, 2) + C(6, 2) + ... + C(2, 2)} = 280
Probability = (448 + 280)/C(64, 2) = 728*2/(64*63) = 13/36
How many five digit natural numbers with distinct digits are there such that the first three digits are in descending order while the last three are in ascending order.eg: 98234plz post the method
A queen on an English chessboard is able to attack in the same row, column and diagonal. Find the probability that 2 randomly placed queens on an 8 by 8 chessboard will be able to attack each other .
number of attacking positions will be different 2 wrt the position of one queen.
at four corners 7+7+7=21
at adjacent to corners 7+7+8=23 and so on.
At center, attacking positions = 7+7+7+7=28
=> average attacking position = 14 (for same row and column) + 140/16 (across) =91/4
Hence, probability of attacking each other = (91/4)/63 = 13/36
Getting a bad feeling about this answer. what's oa?
Sir what was the logic that you were talking about in ABC + DEF + GHI = n question ??
There are many small things in this..
First -
If we add 2 and 3, the sum is 5 i.e. 2 + 3 = 5, which has same sum for LHS as well as RHS as 5, right.
Now when we add 5 and 7, the sum is 12. i.e. 5 + 7 = 12 Let's compare 'sum of digits'. It is (5 + 7) i.e. 12 for LHS but (1 + 2) i.e. 3 for RHS. Now this is not same.
I hope you can easily gauge the reason for this.
Yes..reason is the carry which we have taken while adding. We have taken carry of 1 because of 10...i.e. it was counted as 10 in LHS but it is being counted as 1 only in RHS, creating a decrease of (10 - 1) i.e. 9 in 'sum of digits' of RHS in coparison with LHS.
Now question is, whether there will always be decrease of 9 in 'sum of digits' whenever there is a carry in the addition process.
Answer is YES. Whenever there is a carry of '1' in the addition process, it'll create a decrease of '9' in the 'sum of digits' of RHS in comparison with LHS.
Or we can state it now as "the difference in 'sum of digits' of LHS and RHS is always going to be multiple of 9 i.e. of the form 9n where 'n' gives the number of carries being taken in the addition process"
And we can also infer that LHS and RHS are always equal modulo-9.
Second -
In this particular question, LHS is comprised of 9 single digit distinct positive integers whose sum is 1 + 2 + 3 + 4 + ...+ 8 + 9 = 45.
So clearly according to above discussion, 'sum of digits' of RHS should be multiple of 9. And also the difference of sum of digits will tell about the exact number of carries being taken.
Coming to questions now -
(i) ABC + DEF + GHI = 800 .............straight forward not possible (I am sure you know the reason much easily now 😃 )
(ii) ABC + DEF + GHI = 900...........possible and difference is sum of digits = 45 - 9 = 36 = 9(4). So exactly 4 carries had been used. Clearly addition of hundred's digit don't give a carry here and by adding 3-single digit numbers we can't have more than 2 carries. So we can conclude, straight forward, that C + F + I = 20, B + E + H = 18 and A + D + G = 7. I am sure, it is clear. :)
(iii) ABC + DEF + GHI = 1000..........no need to tell again that it's impossible. :)
a + 2b + 3c + 4d + 5e = 405a + b + c + d + e = 100 we need to maximize C 3c + 4d + 5e = 405 c + d + e = 100 -------- *5 2c + d = 95 => as C has to be a integer max. of C = 47
My approach is different though and easier as per me.
Sum of all the sets i.e. 405 is basically sum of the choices exercised by 100 students, of which some like exactly one, some liked exactly two and so on and that which Abhishek has mentioned as a, b, c, d, e respectively.
I am working in terms of I, II, III, IV, V which means number of students which play "at least" one, two, three and so on sports. And these symbols will mean I = a + b + c + d + e II = b + c + d + e III = c + d + e IV = d + e V = e in terms of the symbols used in above quoted solution.
Now my method is easy and based on visualisation.
I know that I + II + III + IV + V = 405 and also I ≥ II ≥ III ≥ IV ≥ V. Though there are some more rules and regulations to it but that's not required here.
Now to maximise "exactly" three i.e. (III - IV), I must maximise III and minimise IV. (Recall I am drawing five lines representing I, II, III, IV and V one above other and I being at bottom and also following above two restrictions)
So III can be at most equal to II which at most can be equal to I which at most can be equal to 100. So keep I = II = III = 100 each. Now IV + V = 105 and we need to minimise IV. So best way is to keep III and IV as close as possible (recall that V cannot be more than IV). So best arrangement is IV = 53 and V = 52.
Thus maximum value for "exactly" three = III - IV = 100 - 53 = 47. :)
Once you are through with all the jargon, this method should appeal you more than anything else. At least I feel so. :)
@bodhi_vriksha Yes sir, definitely liked your approach....I tried with the line method....but then was missing the last part which you have explained....the part where you minimise iv and keep both iv and v as close as possible...aslo was missing on the concept that iv > v...
can you explain the concept again, or direct me to the page where u had explained it earlier. i have somewhat got the hold of this but need a more descriptive explanation....
Basically sum of consecutive integers is = number of terms*average of the sequence
If number of terms are odd, then average is an integer which is located at the middle of the sequence. For example 100 = 5*20 = (5 terms)*(Average of 20) = 18 + 19 + 20 + 21 + 22
If number of terms are even, then the average is a fractional number of the form xx.5 which lies between two middle terms of the sequence. For example 100 = 8*12.5 = (8 terms)*(Average of 12.5) = 9 + ..+ 12 + 13 + ..+ 16
Every prime number p > 3 is of the form 6k ± 1.So p² = (6k ± 1)² = 1 mod 12 Ext. Qs. find the remainder wen d square of 4 digit prime no. is divided by 24?Team BV - Kamal Lohia
Still 1.
One way is by saying that (6k+1)^2 = 36k^2 + 12k + 1, if k is even then 36k^2 + 12kdiv by 24, if odd then 36k^2 + 12k = 12 (3k^2 +k) which is again 12 * even and divisible by 24. regards scrabbler
Every prime number p > 3 is of the form 6k ± 1.So p² = (6k ± 1)² = 1 mod 12 Ext. Qs. find the remainder wen d square of 4 digit prime no. is divided by 24?Team BV - Kamal Lohia
Every prime number p > 3 is of the form 6k ± 1.So p² = (6k ± 1)² = 1 mod 12 Ext. Qs. find the remainder wen d square of 4 digit prime no. is divided by 24?Team BV - Kamal Lohia
Every prime number p > 3 is of the form 6k ± 1.So p² = (6k ± 1)² = 1 mod 12 Ext. Qs. find the remainder wen d square of 4 digit prime no. is divided by 24?Team BV - Kamal Lohia
A second way which I like, though longer:
If the prime is n, then n^2 - 1 = (n+1)(n-1) Now n is a prime > 2 so is odd. So n+1 and n-1 are even, and that too consecutive even numbers so one of them will also be divisible by 4. Hence the product will be divisible by 2*4. Also one out of any 3 consecutive numbers iwll be divisible by 3. Now in n-1, n, n+1,the prime number n cannot be the one so one of the others must have a factor of 3 as well.
So overall (n+1)(n-1) = n^2 - 1 is of the form 24k. So n^2 = 24k+1 and will hence leave a remainder 1 when divided by 24. regards scrabbler
Every prime number p > 3 is of the form 6k ± 1.So p² = (6k ± 1)² = 1 mod 12 Ext. Qs. find the remainder wen d square of 4 digit prime no. is divided by 24?Team BV - Kamal Lohia
I guess still 1...
expanding the term will give you
36k^2+12k+1
36k^2-12k+1
so,
12k(3k+1) + 1
if k is odd , then
12 * odd( odd + 1) + 1
12*odd (even) + 1
so the 1st term will be divisible by 24..so remainder 1...
if k is even
12k(3k+1) + 1
12*even(odd) + 1
so still the 1st term is divisible by 24....
If 12k(3k-1) + 1
then
k odd
12*odd(odd-1) + 1
12*odd(even) + 1
1st term is divisible by 24
so R will be 1
if k even
then
12*even(3*even-1) + 1
12*even(odd) +1
still the 1st term is divisible by 24..hence R = 1..
A second way which I like, though longer:If the prime is n, then n^2 - 1 = (n+1)(n-1)Now n is a prime > 2 so is odd. So n+1 and n-1 are even, and that too consecutive even numbers so one of them will also be divisible by 4. Hence the product will be divisible by 2*4.Also one out of any 3 consecutive numbers iwll be divisible by 3. Now in n-1, n, n+1,the prime number n cannot be the one so one of the others must have a factor of 3 as well.So overall (n+1)(n-1) = n^2 - 1 is of the form 24k. So n^2 = 24k+1 and will hence leave a remainder 1 when divided by 24.regardsscrabbler
